ICSE Coaching Classes for Std. IX & X in Mumbai, Navi ...
Transcript of ICSE Coaching Classes for Std. IX & X in Mumbai, Navi ...
GROUP (A)- CLASS WORK PROBLEMS
Q-1) If a ,b ,c are the position vectors of the
points A, B, C respectively Such that
3 5 8a b c+ =+ =+ =+ = , then find the ratio in which
i) C divides AB, ii) A divides BC
Ans. i) We are given that
3 +5 = 8a b c
∴∴∴∴3 +5 5 +3
= =8 5 + 3
a b b ac
∴∴∴∴ C divides AB internally in the ratio 5 : 3
ii) We are given that
3 +5 = 8a b c
∴∴∴∴ 3 = 8 – 5a c b
∴∴∴∴8 – 5 8 – 5
= =3 8 – 3
c b c ba
∴∴∴∴ A divides BC externally in the ratio 8 : 5.
Q-2) Show that the points whose position
vectors are 5 + 6 + 7a b c , 7 + 6 + 9a b c and
3 +6 +5a b c are collinear.
Ans. Let p = 5 + 6 +7a b c ,
q = 7 + 6 + 9a b c ,
r = 3 + 6 +5a b c ,
PQ = –q p
= ( ) ( )7 + 6 + 9 – 5 + 6 + 7a b c a b c
= 2 +2a c ... (i)
QR = –r q
= ( ) ( )3 + 6 +5 – 7 + 6 + 9a b c a b c
= 4 – 4a c
= ( )–2 2 + 2a c
= –2PQ [using i]
i.e. QR is scalar multiple of PQ
∴ PQ & PR belong to same line
∴ points P, Q and R are collinear
Q-3) ABCD is a quadrillateral M and N are
mid points of the diagonal AC and BD
respectively. Show that
+ + + = 4AB AD CB CD MN .
Ans. Let , , , , ,a b c d m n be the position vectors of
vertices , , , , ,A B C D M N respectively.
Since M and N are midpoints of diagonal
AC and BD respectively.
m =+
2
a c,
n =+
2
b d
2m = +a c & 2n = +b d …(i)
Now + + +AB AD CB CD
= – + – + – + –b a d a b c d c
= 2 +2 – 2 – 2b d a c
= ( ) ( )2 + – 2 +b d a c (using i)
= ( ) ( )2 2 – 2 2n m
= 4 – 4n m
= ( )4 –n m
= 4MN
∴∴∴∴ + + +AB AD CB CD
= 4MN
Q-4) G and G′′′′are centroids of ∆∆∆∆ABC and ∆∆∆∆A′′′′B′′′′C′′′′.
Show + + = 3AA BB CC GG′ ′ ′ ′′ ′ ′ ′′ ′ ′ ′′ ′ ′ ′ .
Ans. Let , , , ', ', ',a b c a b c g and 'g are the position
vectors of A,B,C,A′′′′,B′′′′,C′′′′,G and G′′′′ respectively
with respect to some origin O.
G and G′′′′ are the centroid of ∆∆∆∆ABC and ∆∆∆∆A′B′C′
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∴∴∴∴ g =+ +
3
a b c
∴∴∴∴ 3g = + +a b c ...[i] And
∴∴∴∴ 'g ='+ '+ '
3
a b c
∴∴∴∴ 3 'g = '+ '+ 'a b c ...[ii]
L.H.S. = '+ '+ 'AA BB CC
= ( ) ( ) ( )'– + '– + '–a a b b c c
= ( ) ( )'+ '+ ' – + +a b c a b c
= 3 '– 3g g
= 3 'GG
Q-5) D, E, F, are mid - point of sides BC, CA, AB,
respectively of ∆∆∆∆ABC, show
i) + + =OE OF DO OA
ii) 2 1 1
+ + =3 3 2
AD BE CF AC
Ans. Let , , , ,a b c d e and f be position vectors of
, , , ,A B C D E and F respectively w. r. t.point
O in the plane Here D,E,F are the mid-points
of BC, CA and AB.
∴∴∴∴ Using mid-point formula,
d =+
2
b c...(i);
e =+
2
c a...(ii)
f =+
2
a b...(iii)
(i) L.H.S. = + +OE OF DO
= + –OE OF OD
= + –e f d
AA
B C
EF
D
=+ + +
+ –2 2 2
c a a b b c
...[from (i), (ii) and (iii)]
=1
+ + + – –2
c a a b b c
= ( )12
2a = a
= OA = R.H.S.
Hence proved.
(ii) L.H.S.
=2 1
+ +3 3
AD BE CF
= ( ) ( ) ( )2 1– + – + –
3 3d a e b f c
= + 2 + 1 +
– + – + –2 3 2 3 2
b c c a a ba b c
=+ – 2 2 + – 2
+2 3 2
1 + – 2 +
3 2
b c a c a b
a b c
=( )2 + – 2+ – 2 + – 2
+ +2 6 6
c a bb c a a b c
=3 +3 – 6 + 2 – 4 + – 2
6
b c a a b b c
=1
3 – 36
c a
=1
–2
c a
=1
2AC
= R.H.S. Hence proved
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Q-6) AB and CD are two chords of a circle
intersecting at right angles in P. Show that
+ + + = 2PA PB PC PD PO where O is centre
of the circle.
Ans. Draw OQ ⊥⊥⊥⊥ AB and OR ⊥⊥⊥⊥ CD
Let , , , , ,a b c d p q and r are the position
vectors of A,B,C,D,P,Q and R respectively with
respect to centre O. Q and R are the mid-
points of AB and CD respectively.
∴∴∴∴ Using mid-point formula,
q =+
2
a b
∴∴∴∴ 2q = +a b ...(i)
r =+
2
c d
∴∴∴∴ 2r = +c d ...(ii)
L.H.S.
= + + +PA PB PC PD
= ( ) ( ) ( ) ( )– + – + – + –a p b p c p d p
= ( ) ( )+ + + – 4a b c d p
= 2 +2 – 4q r p [from (i) and (ii)
L.H.S. = ( )2 + – 2q r p ...(i)
Now, ���� OQPR is a rectangle, hence it is also
parallelogram
∴∴∴∴ by parallelogram law
OP = +OQ OR
∴∴∴∴ p = +q r ...(ii)
Substituting (ii) in (i) we get,
A BQ
O
D
R
C
P
= ( )2 – 2P P
= – 2 = 2P PO
Hence proved.
Q-7) If , ,a b c and d are the position vectors of
the points A, B, C and D respectively such
that 2 + 7 = 5 + 4a b c d . Prove that AB and
CD intersect.
Ans. Given 2 + 7a b = 5 + 4c d
∴∴∴∴2 +7
2+7
a b
=5 + 4
5 + 4
c d
= e (say)
Let e is the position vector of E.E .
By section formula,
E divides AB internally in the ratio 7 : 2 and
CD internally in the ratio 4 : 5
∴∴∴∴ E is the point on both AB and CD.
∴∴∴∴ AB and CD intersect in E.
Q-8) In ∆∆∆∆ABC, O is circumcenter, H is orthocenter
then prove that
i) + + =OA OB OC OH
ii) (((( ))))+ + = 2 + +HA HB HC AO BO CO
Ans. Let , , ,a b c o and h are p.v. of A,B,C,O and H
respectively.
We know that, if O,G and H are collinear then
G divides OH in the ratio 1:2 where G is
centroid and g is its position vector.
By section formula
g =+2
1+2
h o
∴∴∴∴ 3g = + 2h o … (i)
i) L.H.S.
= + +OA OB OC
= ( ) ( ) ( )– + – + –a b b o c o
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= + + – 3a b c o
=+ +
3 – 3 =3
a b cg o g
∴
= + 2 – 3h o o from equation (i)
= –h o
= OH
= R.H.S. Hence proved
ii) L.H.S.
= + +HA HB HC
= – + – + –a h b h c h
= + + – 3a b c h
= 3 – 3g h
= + 2 – 3h o h from eq. (i)
= 2 – 2o h
= ( )2 –o h
= 2HO
R.H.S. = ( )2 + +AO BO CO … (ii)
= ( ) ( ) ( )2 – + – + –o a o b o c
= ( )2 3 – + +o a b c
= 2 3 – 3o g
= ( )2 3 – + 2o h o
= 2 3 – – 2o h o
= 2 –o h
= 2HO … (iii)
From equations (ii) and (iii)
L.H.S. = R.H.S. Hence proved.
Q-9) ABCD is a quadrilateral M and N are mid -
points of the diagonal AC and BD
respectively. Show that + = 2AD BC MN .
Ans. Let , , , , ,a b c d m n be the position vectors of
points A,B,C,D,M,N w.r.t a fixed point
∴∴∴∴ M and N are mid points of AB and CD
respectivrly.
m =+
2
a band
n =+
2
c d
2m = +a b and 2n = +c d …(i)
L.H.S. = +AD AC
= – + –d a c b
= ( ) ( )+ – +c d a b
= 2 – 2n m
= ( )2 –n m
= 2MN
Q-10)If +3 +ai j bk is unit vector, prove that
2 2+ + 8 = 0a b
Ans. Let
r = + 3 +ai j bk
⇒⇒⇒⇒ r = 2 2+ + 9a b
But r = 1
∴∴∴∴ 2 2+ + 9a b = 1
∴∴∴∴ 2 2+ + 9a b = 1
∴∴∴∴2 2+ + 8a b = 0
Q-11)If , ,a b c are position vectors of AB and C
where A ≡ ≡ ≡ ≡ (0, 2, –1), B ≡ ≡ ≡ ≡ (0, – 1,3),
C≡ ≡ ≡ ≡ (0, 1,2) Find x and y such that
= +c xa yb
Ans. a = 2 –j k ;
b = – + 3j k ;
c = + 2j k
∴∴∴∴ c = +xa yb
+ 2j k = ( ) ( )2 – + – + 3x j k y j k
+ 2j k = ( ) ( )2 – + – + 3x y j x y k
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∴∴∴∴ 2x – y = 1 ... (i) and
–x + 3y = 2 ... (ii)
Solving equations (i) and (ii) x = 1 and y = 1
Q-12)If A(3, 2, -1), B(2, 3, 4), C(0, 5,14) then show
that the points A, B, C are collinear and
find the ratio in which B divides segment
AC.
Ans. AB = –b a
= – + +5i j k
AC = –c a
= –3 +3 +15i j k
AC = 3AB
–c a = ( )3 –b a ⇒⇒⇒⇒ –c a
= 3 – 3b a ⇒⇒⇒⇒ +2c a
= 3b
∴∴∴∴ b =+2
3
c a
=1. + 2
1+2
c a⇒⇒⇒⇒ B
divides seg AC internally in the ratio 1:2.
∴∴∴∴ A, B, C are collinear.
Q-13) If A(3,1,–1), B(1,5,2) are the vertices and
G(3,3,1) is the centroid of the triangle ABC
then by Vector method, find the mid-point
of the BC.
Ans. Let C be ( )1 1 1, , .x y z
Then the postion vectors , , ,a b c g of the points
A, B, C, G are
a = 3 + –i j k ,
b = +5 +2i j k ,
c = 1 1 1+ +x i y j z k ,
g = 3 +3 +i j k ,
Since G is the centroid of the ∆∆∆∆ ABC, by the
centroid formula,
g =+ +
3
a b c
∴∴∴∴ 3g = + +a b c
∴∴∴∴ ( )3 3 + +i j k
= ( ) ( )
( )1 1 1
3 3 + – + +5 +2
+ + +
i j k i j k
x i y j z k
∴∴∴∴ 9 +9 +3i j k
= ( ) ( ) ( )1 1 13+1 – + 1+5 + + –1+2+x i y j z k
∴∴∴∴ 9 +9 +3i j k
= ( ) ( ) ( )1 1 1+ 4 + +6 + +1x i y j z k
By equality of vectors, we have,
9 = x1 + 4, 9 = y
1 + 6 and 3 = z
1 + 1
∴∴∴∴ x1 = 5,y
1 = 3 and z
1 = 2
∴∴∴∴ C(5,3,2)
∴∴∴∴ c = 5 +3 +j j k
Let P be the midpoint of BC.
Then by midpoint formula, we have,
∴∴∴∴ p =+
2
b c, where p is the position vector
of P.
∴∴∴∴ p =( ) ( )+5 +2 5 + 3 + 2
2
i j k i j k
= ( )1
6 + 8 + 4 3 + 4 + 22
i j k i j k
∴∴∴∴ midpoint of the side BC is (3,4,2).
GROUP (A)- HOME WORK PROBLEMS
Q-1) A and B are two points with position
vectors a and b . If point C and D have
position vectors 5 4−−−−a b and 3 5
,8
++++a b
prove A, B, C, D are collinear.
Ans. Let c and d respectively be position vectors
of points C and D
c = 5 – 4a b
d =3 +5
8
a b
c =5 – 4
5 – 4
a b
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d =3 +5
3+5
a b
⇒⇒⇒⇒ point C divides seg. AB externally in the ratio
4 : 5
⇒⇒⇒⇒ points A, B, C are collinear points ...(i)
⇒⇒⇒⇒ point D divides seg. AB internally in the ratio
5 : 3. ...(ii)
∴∴∴∴ points A, B, C, D are collinear points from
(i, ii)
Q-2) Show that the point P, Q, R with position
vectors – 3 + ,a b c –2 + 3 + 4a b c and
– + 2b c are collinear.
Ans. Let , ,p q r respectively be position vectors of
points P, Q, R w.r.t a fixed point.
p = – 3 +a b c
q = –2 + 3 + 4a b c
r = – +2b c
PQ = –q p
= ( ) ( )–2 + 3 + 4 – – 3 +a b c a b c
PQ = –3 + 6 +3a b c
PQ = ( )–3 – 2 –a b c
–3
PQ= – 2 –a b c ...(i)
QR = –r q
= ( ) ( )– + 2 – –2 + 3 + 4b c a b c
= – + 2 +2 – 3 – 4b c a b c
= 2 – 4 – 2a b c
= ( )2 – 2 –a b c
= 23
−
PQ
QR =–2
3PQ
i.e QR is scalar multiple of PQ
∴∴∴∴ QR and PQ parallel or collinear
and point Q is common point
∴ PQ and QR are collinear
∴ points P, Q, R are collinear.
Q-3) If points P, Q, R, S have position vectors
, , ,p q r s such that (((( ))))– = 2 – .p q s r Show
that the lines QS and PR trisect each other.
Ans. –p q = 2 – 2s r
+2p r = +2q s
+2
3
p r=
+ 2
3
q s
+2
1+2
p r=
+ 2
1+2
q s= m
above is position vector of a point which
divies PR and QS internally in the ratio 2 : 1
i.e. point M (say) divides PR and QS internally
in the ratio 2 : 1
∴∴∴∴ PR and QS trisect each other.
Q-4) A, B, C are three points with position
vectors , ,a b c w.r.t origin O. If
7 = 4 + 3 ,c a b prove that A, B, C are
collinear.
Ans. 7c = 4 + 3a b
c =4 +3
7
a b
c =4 +3
4+3
a b
⇒⇒⇒⇒ point c divides seg AB internally in the ratio
3 : 4
∴∴∴∴ points A, B, C are collinear points.
Q-5) P, Q, R are three points with position
vectors , ,p q r w.r.t some origin O. If
2 = 5 – 3r p q then prove that P, Q, R are
collinear.
Ans. 2r = 5 – 3p q
r =5 – 3
2
p q
r =5 – 3
5 – 3
p q
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⇒⇒⇒⇒ point R divides seq. PQ externally in the
ratio 3 : 5.
⇒⇒⇒⇒ points P, Q, R are collinear.
Q-6) D, E, F are mid-points of the sides BC, CA
and AB respectively of ∆∆∆∆ABC. O is any point
in the plane of ∆∆∆∆ABC Show that,
i) + + = 0AD BE CF
ii) + + = + +OA OB OC OD OE OF
iii) = 2BC FE
Ans. Let , , , , ,a b c d e f respectively be position
vectors of points A, B, C, D, E, F write a fixed
point .
point D. E. F are mid points of sides BC, AC
and AB respectvely
d =+
2
b c,
e =+
2
a c,
f =+
2
a b
i) + +AD BE CF
= – + – + –d a e b f c
= + + +
– + – + –2 2 2
b c a c a ba b c
= 2 – 2 +2 – 2 +2 – 2
2
a a b b c c
= 0
2
= R.H.S
ii) R.H.S= + +OD OE OF
= + +d e f
= + + +
+ +2 2 2
b c a c a b
= 2 + 2 +2
2
a b c
= ( )2 + +
2
a b c
= + +a b c
= + +OA OB OC
= L.H.S
iii) R.H.S = 2FE
= 2 ( )–e f
= 2+ +
–2 2
a c a b
= + – –
22
a c a b
= –c b
= BC
= L.H.S
A
B CD
EF
Q-7) The position vectors of four points A, B, C
and D are , , 2 3 , 2+ −+ −+ −+ −a b a b a b
respectively. Express the vectors ++++AC DB
and BC in terms of a and b
Ans. Let position vectors of pts. A, B, C, D be
, , ,a b c d
then
c = 2 + 3a b ,
d = – 2a b
AC = –c a
= 2 +3 –a b a
= + 3a b
DB = –b d
= ( )– – 2b a b
= 3 –b a
BC = –c b
= 2 +3 –a b b
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= 2 + 2a b
Q-8) A divides seg PQ internally in the ratio
1 : 3 and B divides seg QR externally in
the ratio 5 : 2. Express AB in terms of ,p q
and r .
Ans. Let , , ,a b p q and r be position vectors of points
A, B, P, Q and R w.r.t a fixed point.
∵ point A divides seg PQ internally in the ratio
1 : 3
∴∴∴∴ a =( )1 +3
1+3
q p
a =+3
4
q p...(i)
∵ point B divided seg QR externally in the
ratio 5 : 2
b = 5 – 2
5 – 2
r q
b =5 – 2
3
r q ...(ii)
∴∴∴∴ AB = –b a
=5 – 2 +3
–3 4
r q q p
=20 – 8 – 3 – 9
12
r q q p
= ( )1
20 –11 – 912
r q p
Q-9) The points C and D divides AB in ratio x : 1
inernally and externally respectively. Show
that A and B divides CD in ratio 1 –
1+
x
x
±±±±
Ans. Let , , ,a b c d be position vectors of point A,
B, C, D respectively C divides AB internally
in ratio x : 1.
∴∴∴∴ By internall division formula,
c =+
1+
a xb
x.... (i)
D ddivides AB externally in ratio x : 1
∴∴∴∴ By esternal division formula,
d =+
1 –
a xb
x.... (ii)
From (i) and (ii)
( )1+c x = +a xb .... (iii)
( )1 –d x = –a xb .... (iv)
Adding equation (iii) and (iv)
( ) ( )1+ + 1 –c x d x = 2a
∴∴∴∴ a =( ) ( )1+ + 1 –
2
x c x d
a =( ) ( )
( ) ( )
1+ + 1 –
1+ + 1 –
x c x
x x
∴∴∴∴ A divides OC in ratio
(1 + x) : (1 – x)
or A divides CD internally in ratio
(1 – x) : (1 + x)
i.e.1 –
1+
x
x
Now, equation (iii) - (iv)
∴∴∴∴ ( ) ( )1+ – 1 –c x d x = 2xb
∴∴∴∴ b =( ) ( )1+ – 1 –
2
x c x d
x
∴∴∴∴ b =( ) ( )
( ) ( )
1+ – 1 –
1+ – 1 –
x c x d
x x
∴∴∴∴ B divides CD externally in ratio
(1 – x) : (1 + x) i.e.1 –
1+
x
x
∴∴∴∴ A and B divides CD in ratio ±±±±1 –
1+
x
x
Q-10)If the vectors ���� ����2 – 3i q j k++++���� and
���� ����4 – 5 6i j k++++���� are collinear, then find q.
Ans. Let a = � �2 – + 3i q j k� and
b = � �4 – 5 + 6i j k�
Since the vectors a and b are collinear, the
components of i� , �j and �k in a and b are
proportional.
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∴∴∴∴2
4=
–
–5
q=
3
6
1
2=
5
q=
1
2
∴∴∴∴ q =5
2
Q-11) If the points A(3, 0, p), B(–1, q, 3) and
C(–3, 3,0) are collinear, then find
i) the ratio in which the point C divides
the line segment AB
ii) the values of and q.
Ans. Let ,a b and c be the position vectors of A, B
and c respectively.
Then a = � �3 +0. +i j pk� ,
b = � �– + + 3i q j k� and
c = � �3 + 3 + 0.i j k�
i) As the points A, B, C, are collinear,
suppose the point C divides line segment AB
in the ratio λλλλ : 1.
∴∴∴∴ by the section formula,
c = � �λλλλ
λλλλ
. +1
.+1
b a
∴∴∴∴ � �–3 + 3 + 0.i j k�
=
� �( ) � �( )λλλλ
λλλλ
– + + 3 + 3 +0. +
+1
i q j k i j pk� �
∴∴∴∴ ( ) � �( )λλλλ +1 –3 + 3 +0.i j k�
=� �( ) � �( )λ λ λλ λ λλ λ λλ λ λ– + +3 + 3 +0. +i q j k i j pk� �
∴∴∴∴ ( ) ( ) �λ λλ λλ λλ λ–3 +1 + 3 +1 + 0.i i k� �
= ( ) ( ) �λ λλ λλ λλ λ– + 3 + 3 + 3 +i qi p k� �
By equality of vectors, we have,
–3(λ λ λ λ + 1) = – λ λ λ λ + 3 .... (i)
3(λ λ λ λ + 1) = λλλλ .... (ii)
0 = 3λλλλ + p .... (iii)
From equation (i), 3λ λ λ λ – 3 = –λλλλ + 3
∴∴∴∴ – 2λ λ λ λ = 6
∴∴∴∴ λ λ λ λ = – 3
∴∴∴∴ C divides segment AB externally in the ratio
3 : 1.
ii) Putting λ λ λ λ = –3 in equation (2), we get
3(– 3 + 1) = –3q
∴∴∴∴ –6 = –3q
∴∴∴∴ q = 2
Also, putting λ λ λ λ = –3 in equation (3), we get,
0 = –9 + p
∴∴∴∴ p = 9
Hence p = 9 and q = 2
Q-12) The position vectors of the points A and B
are ���� ����2 – +5i j k���� and ����–3 +2i j���� respectively.
Find the position vector of the point which
divides the line segment AB internally in
the ratio 1 : 4
Ans. The position vectors a and b of the points A
and B are
a = � �2 – +5i j k� and
b = �–3 +2i j�
Let C be the point, with position vector ,
divides the line segment AB internally in the
ratio 1 : 4
by the section formula for internal division,
∴∴∴∴ c =
� �1. + 4
1+ 4
b a
=
�( ) � �( )–3 +2 + 4 2 – +5
5
i j i j k� �
=� �( )1
5 – 2 + 205
i j k�
∴∴∴∴ c = � �2– + 45
i j k�
Q-13) If P is orthocentre, Q is the circmentre and
G is the centroid of a triangle ABC, then
prove that = 3QP QG
Ans. Let p and q be the position vectors of P and
G w.r.t the circumcentre Q.
i.e., QP = p and QG = g
we knoe that Q, G, P are collinear and G divides
segment QP internally in the ratio 1 : 2
Vectors
10 Mahesh Tutorials Science
by section formula for internal division,
∴∴∴∴ g =1. + 2
1+2
p q=
3
p
∴∴∴∴ p = 3q
∴∴∴∴ QP = 3QG
Q-14) If A(a, 2, 2), B(a, b, 1) If C(1, 2, –2) are
vertices of ∆∆∆∆ABC and G(2, 1, –c) is its
centroid find a, b and c.
Ans. Let , , ,a b c g be position vectors pf points
A, B, C, G respectively
a = ˆ ˆ ˆ+ 2 +2 ,ai j k
b = ˆ ˆ ˆ+ + ,ai bj k
c = ˆ ˆ ˆ+2 – 2 ,i j k
g = ˆ ˆ ˆ2 + –i j ck
∵∵∵∵ G is centroid of ∆∆∆∆ABC
∴∴∴∴ g =+ +
3
a b c
∴∴∴∴ 3g = + +a b c
( )3 2 + –i j ck
= ( ) ( )6 +3 – 3c = 2 +1 + 4 + +i j k a i b j k
Comparing the coefficeients of , ,i j k on both
sides
we get 2a + 1 = 6
4 + b = 3
–3c = 1
a = 5
2
b = –1
c = –1/3
Q-15) If A ≡ ≡ ≡ ≡ (1, 2, 3), B ≡≡≡≡ (, 4, 5) and C ≡≡≡≡ (p, q, 7) are
collinear points. Find p & q.
Ans. Let , , ,a b c g be position vectors pf points
A, B, C, G respectively
a = +2 +3i j k ;
b = 3 + 4 +5i j k ;
c = + +7pi q j k ;
∴∴∴∴ AB = –b a
= ( ) ( )3 + 4 +5 – + 2 + 3i j k i j k
= 2 +2 +2i j k ... (i)
AC = –c a
= ( ) ( )+ + 7 – + 2 + 3pi p j k i j k
= ( ) ( )–1 + – 2 + 4p i q j k ... (ii)
Given thaty the points are collinear
∴∴∴∴ AB and AC are collinear
AB = mAC [m : nonzero scalar]
2 +2 +2i j k = ( ) ( )–1 + – 2 +m p i q j uk
Comparing the of , ,i j k on both sides we get
4m = 2 m(p – 1) = 2 p – 1 = 4 m (q – 2) = 2
m = 2/4 = 1/2 4/2 (p – 1) = 2 p = 5
( )1
– 22q = 2
q = 6
Q-16) If the three points A(4,5,p). B(q,2,4) and
C(5,8,0) are collinear then find
i) The ratio in which the point C divides
the line AB
ii) The values of p and q.
Ans. Let , ,&a b c be the position vectors of A, B &
C respectively.
∴∴∴∴ a = ˆ ˆ ˆ4 +5 + . ,i j pk
b = ˆ ˆ ˆ+2 +4 ,qi j k
c = ˆ ˆ ˆ5 +8 +0 ,i j k
i) As the point A,B & C are collinear,
Suppose the point C divides line segment
AB in the ratio λλλλ:1
∴∴∴∴ By the section fromula.
c =λ
λ
. +1.
+1
b a
( )λλλλ ˆ ˆ ˆ+1 5 + 8 + 0i j k
Mahesh Tutorials Science 11
Vectors
= ( ) ( )λλλλ ˆ ˆ ˆ ˆ ˆ ˆ+ 2 + 4 + 4 +5 +qi j k i j Pk
( ) ( )λ λλ λλ λλ λˆ ˆ ˆ5 +1 +8 +1 +0.i j k
= ( ) ( ) ( )ˆ ˆ ˆ+ 4 + 2+5 + 4+q i j P kλ λ λ
By equality of vectors, we have
5(λλλλ+1) = λλλλq + 4 ... (1)
8(λλλλ+1) = λλλλ2 + 5 ... (2)
0 = λλλλ4 + P ... (3)
From (1)
5λλλλ+ 5 = λλλλq + 4
5λ λ λ λ – λλλλq = – 1 ... (5)
From (2)
8λλλλ+ 8 = λλλλ2 + 5
6λ λ λ λ = – 3
x = –1
2... (6)
∴∴∴∴ C divides segment AB externally in the
ratio 1 : 2
ii) –1 –1
5 – = –12 2
q
Substitue λ–1
=2
in (5)
–5+
2 2
q= –1=
– 5 + q = – 2
q = 3
0 = λλλλ4 + p
4λλλλ + p = 0
–14
2
= p
p = – 2
∴∴∴∴ p = – 2 and q = 3
Q-17) The position vectors P and Q are ˆˆ ˆ– 2 +i j k
and ˆˆ ˆ+ 4 – 2i j k respectively. Find
coordinates & position vector of (((( ))))ˆR r
which divides the line segment PQ
internally in the ratio 2 : 1.
Ans. Let p and q be the position vectors of the
points P and Q respectively.
Then = – 2 +p i j k and = + 4 – 2q i j k
Since ( )ˆR r divides line segment PQ
internally in the ratio 2 :1, by section formula
for internal division,
r =2 +1.
2 +1
q p
=( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ2 + 4 – 2 + – 2 +
3
i j k i j k
=ˆ ˆ ˆ ˆ ˆ ˆ2 +8 – 4 + – 2 +
3
i j k i j k
= ( )1 ˆ ˆ ˆ3 + 6 – 33
i j k
∴∴∴∴ r = ˆ ˆ ˆ+ 2 –i j k
∴∴∴∴ coordinates of R are (1,2 – 1).
Q-18)Find unit vectors along a, b and AB if
A ≡≡≡≡ (1, 2, 4) and B ≡≡≡≡ (2, –1, 5) .
Ans. A ≡≡≡≡ (1, 2, 4)
B ≡≡≡≡ (2, –1, 5)
Let ,a b be the position vectors of A and B.
w.r.t fixed point.
a = +2 + 4i j k ,
b = 2 – +5i j k
AB = –b a
= 2 – +5 – – 2 – 4i j k i j k
= – 3 +i j k
a = ( ) ( ) ( )2 2 2
1 + 2 + 4
= 1+ 4+16
= 21
b = ( ) ( ) ( )2 2 2
2 + –1 + 5
= 4 +1+ 25
= 30
unit vectors along a =a
a
Vectors
12 Mahesh Tutorials Science
=+ 2 + 4
21
i j k
unit vectors along b = 2 + +5
30
i j k
b
b
AB = ( ) ( ) ( )2 2 2
1 + –3 + 1
= 1+ 9 +1
= 11
unit vectors along AB =AB
AB
=– 3 +
11
i j k
Q-19)If ‘C’ divides AB internally in the raio 5 : 6
and if C (1, 5, 0), A (3, 8, 6) find the co-
ordinates of B.
Ans. Let , ,a b c be the position vectors of points
A, B, C respectively w.r.t a fixed point.
A ≡≡≡≡ (3, 8, 6)
C ≡≡≡≡ (1, 5, 0)
a = 3 +8 +6i j k
c = +5i j
Now, C divides AB internally in the ratio
5 : 6
c =( ) ( )5 + 6
5 +6
b a
+5i j = ( ) ( )5 +6 3 +8 +6
11
b i j k
11 +55i j = 5 +18 + 48 +36b i j k
5b = –7 + 7 – 36i j k
b =7 7 36
– + –5 5 5i j k
co-ordinates of B ≡≡≡≡ –7 7 –36
, ,5 5 5
Q-20) If , ,a b c are position vectors of A, B and C
where A (1, 3, 0), B (2, 5, 0), C (4, 2, 0) such
that = +c xa yb then find x and y.
Ans. A ≡≡≡≡ (1, 3, 0)
B ≡≡≡≡ (2, 5, 0)
C ≡≡≡≡ (4, 2, 0)
a = +3i j
b = 2 +5i j
c = 4 +2i j
Now, c = +xa yb
4 +2i j = ( ) ( )+ 3 + 2 + 5x i j y i j�
4 +2i j = +3 +2 +5xi xj yi y j
4 +2i j = ( ) ( )+2 + 3 +5x y i x y j
comparing co-efficients of i and j on both
sides
x + 2y = 4 ...(i)
3x + 5y = 2 ...(ii)
multiplying equation, (i) by 3 and then
subtracting (ii) form (i).
3x + 6y = 12
3x + 5y = 2
y = 10
substitute y = 10 in equation (i)
x + 20 = 4
x = –16
{x = –16, y = 10}
Q-21)If A (0, 1, 3), B (3, 2, 1) and
ˆˆ ˆ= + +AB xi yj zk then prove x + y + z = 0
Ans. Let a and b be position vectors of points A
and B w.r.t a fixed point.
A = (0, 1, 3)
B = (3, 2, −−−−1)
a = +3j k
b = 3 +2 –i j k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧
AB = –b a
Mahesh Tutorials Science 13
Vectors
+ +x i y j z k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧
= 3 +2 – – – 3i j k j k∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧
+ +x i y j z k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧
= 3 + – 4i j k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧
comparing co-efficients of , ,i j k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧
on both side
x = 3, y = 1, z = – 4
L.H.S = x + y + 2
= 3 + 1 – 4
= 0
= R.H.S.
Q-22) If A (1, 4, 1), B (–2, 3, –5) and C (5, –7, 2) are
vertices of triangle then find the centroid
of the triangle ABC.
Ans. Let , , ,a b c g be position vectors of A, B, C and
G respectively.
Let G be the centroid.
a = ˆ ˆ ˆ+ 4 +i j k
b = ˆ ˆ ˆ–2 +3 – 5i j k
c = ˆ ˆ ˆ5 – 7 +2i j k
∵ G is the centroid of ∆ ABC
g =+ +
3
a b c
3g = + +a b c
= ( ) ( )
( )
+ 4 + + –2 +3 +5
+ 5 – 7 + 2
i j k i j k
i j k
3g = 4 +0 – 2i j k
g =4
3i + 0 j +
2
3k
G 4 –2,0,
3 3
Centroid of ∆∆∆∆ABC
Q-23) If ���� ���� ���� ����= + – 2 , = 2 – + 2a i j k b i j k� �� �� �� � and
����= 3 – ,c i k���� then find the scalars m and n
such that = +c mb nb
Ans. c = +mb nb
∴∴∴∴ �3 –i k� =� �( ) � �( )+ – 2 + 2 – +2m i j k n i j k� �
∴∴∴∴ �3 –i k� = ( ) ( ) �
( ) �
2 + –
+ –2 +
m n i m n j
m n k
+ �
By equality of vectors,
m + 2n = 3 .... (i)
m – n = 0 .....(ii)
and –2m + n = – 1 .....(iii)
To solve equations (i) and (ii),
Subtracting (ii) from (i), we get,
3n = 3
∴∴∴∴ n = 1
Substituting n = 1 in equation (i), we get,
m + 2(1) = 3
∴∴∴∴ m = 1
Substituting m = 1, n = 1 in equation (iii), we
get,
LHS = –2(1) + (1)
= –2 + 1
= –1
= RHS.
Hence, m = 1 and n = 1.
Q-24) If G1 and G
2 are the centroids of the
triangles ABC and PQR respectively, then
prove that 1 2+ + = 3AP BQ CR G G .
Ans. Let , , , , , , ,a b c p q r g g1 2be the position vectors
of the points A, B, C, P, Q, R, G1 and G
2
respectively.
Since G1 and G
2 are the centroids of ∆∆∆∆ABC
and ∆∆∆∆PQR respectively,
∴∴∴∴ g1 =
+ +
3
a b c and g2
= + +
3
p q r
∴∴∴∴ a b c+ + = 3g1 and p q r+ + = 3g2
... (i)
Now, AP BQ CR+ +
= ( ) ( ) ( )– + – + –p a q b r c
= ( ) ( )+ + – + +p q r a b c
= 3 – 3g g2 1... [From (i)]
= ( )3 –g g2 1
= 3 –G G1 2
Vectors
14 Mahesh Tutorials Science
Q-25) If , ,a b c are the position vectors of the
points A (1,3,0), B(2,5,0), C(4,2,0)
respectively and c ma nb= += += += + then find the
values of m and n.
Ans. The position vectors , ,a b c of the points A,
B,C are
ˆ ˆ ˆ ˆ ˆ ˆ3 0. , 2 5 0. ,
ˆ ˆ ˆ4 2 0. ,
= + + = + +
= + +
a i j k b i j k
c i j k
Now c ma nb= +
ˆ ˆ ˆ4 2 0.∴ + +i j k
( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= + 3 + 0. + 2 + 5 + 0.m i j k n i j k
ˆ ˆ ˆ4 +2 + 0.∴ i j k
( ) ( )ˆ ˆ ˆ= + 2 + 3 +5 + 0.m n i m m j k
By equality of vectors,
m + 2n = 4 .....(i)
and 3m + 5n = 2 ......(ii)
Multiplying equation (i) by 3, we get,
3m + 6n = 12
Subtracting equation (ii) from this equation,
we get, n = 10
Subtracting n = 10 in equation (i), we get,
m + 2(10) = 4 ∴∴∴∴ m = – 16
Hence m = – 16 and n = 10.
GROUP (B)- CLASS WORK PROBLEMS
Q-1) Show that the following points are collinear :
i) A(3, 2, –4), B(9, 8, –10), C(–2, –3, 1)
ii) P(4, 5, 2), Q(3, 2, 4), R(5, 8,0).
Ans. i) The position vectors ,a b and c of the
point A, B and C are
a = � �3 + 2 – 4i j k� ,
b = � �9 +8 –10i j k� ,
c = � �2 – 3 +i j k� .
g = 3 +3 +i j k ,
∴∴∴∴ AB = –b a
=� �( ) � �( )9 +8 –10 – 3 + 2 – 4i j k i j k� �
= � �6 + 6 – 6i j k� ....(i)
AC = –c a
=� �( ) � �( )–2 – 3 + – 3 +2 – 4i j k i j k� �
= � �5 – 5 +5i j k�
=� �( )5
– 6 + 6 – 66
i j k�....(ii)
From (i) and (ii),
AC =5
–6
AB
i.e, AC is the scalar multiple of AB .
∴∴∴∴ they are parallel to each other. But they have
the point A in common.
∴∴∴∴ the vectors AB and AC lie on the same line.
the points a, B and C are collinear.
ii) Refer to the solution of Q.1(i)
PR = (–1)PQ .
Q-2) If the vectors ���� ����2 – + 3i q j k���� and ���� ����4 – 5 + 6i j k����
are collinear, then find the value of q.
Ans. Let a = � �2 – +3i q j k� and
b = � �4 – 5 +6i j k� .
Since the vectors a and b collinear, the
components of i� ,�j and �k a and b are
propertional.
∴∴∴∴2
4=
–
–5
q=
3
6
∴∴∴∴1
2=
5
q=
1
2
∴∴∴∴ q =5
2.
Alternative Method.
Let a = � �2 – + 3i q j k� and
b = � �4 – 5 + 6i j k� .
Since the vectors a and b are collinear,
× = 0a b
Mahesh Tutorials Science 15
Vectors
∴∴∴∴
� �
2 – 3
4 –5 6
i j k
q
�
∴∴∴∴ ( ) � ( )� ( )
–6 +15 – 12 –12
+ –10 + 4 = 0
i q j
k q
�
∴∴∴∴ ( ) � � ( ) �–6 +15 – 0 –10 + 4q j j q k
= � �0 +0 +0i j k�
By equalify of vectors,
–6q + 15 = 0 and –10 + 4q = 0
∴∴∴∴ q =15
6=
5
2and
q =10
4=
5
2
Hence, q =5
2
Q-3) Are the four points A(1, –1,1), B(–1,1,1) ,
C(1,1,1) and D(2,–3,4) coplanar? justify
your answer.
Ans. The position vectors , , , ,a b c d of the points A,
B, C, D are
a = � �– +i j k� ,
b = � �– + +i j k� ,
c = � �+ +i j k� ,
d = � �2 – 3 + 4i j k�
∴∴∴∴ AB = –b a
=� �( ) � �( )– + + – – +i j k i j k� �
= �–2 +2i j�
AC = –c a
=� �( ) � �( )+ + – – +i j k i j k� �
= �2 j and
AD = –d a
=� �( ) � �( )2 – 3 + 4 – – +i j k i j k� �
= � �– 2 +3i j k�
If A,B,C, D are coplanar, then there exist
scalars x,y such that
AB = . + .x AC y AD
∴∴∴∴ �–2 + 2i j� =�( ) � �( )2 + – 2 +3x j y i j k�
∴∴∴∴ �–2 + 2i j� = ( ) � �+ 2 – 2 – 3yi x y j yk�
By equality of vecotrs,
y = –2 ....(i)
2x – 2y = 2 ....(ii)
3y = 0 ....(iii)
From (i), y = –2
From (iii), y = 0
This is not possible.
Hence, the points A, B, C, D are not coplanar.
Q-4) If a ,b ,c are non-coplanar vectors then
prove that the vectors
2 – 4 + 4 , – 2 + 4a b c a b c and – + 2 + 4a b c
are collinear
Ans. Let P,Q,R be the points whose position vectors
, ,p q r are given by
p = 2 – 4 + 4a b c ,
q = – 2 +0.a b c ,
r = – +2 + 4a b c .
∴∴∴∴ PQ = –q p
= ( ) ( )– 2 + 4 – 2 – 4 + 4a b c a b c
= – +2 + 0.a b c .... (i)
PR = –r q
= ( ) ( )+ 2 + 4 – 2 – 4 + 4a b c a b c
= ( )3 – + 2 + 0.a b c
∴∴∴∴ PQ = 3PQ .... [By (i)]
∴∴∴∴ ( )+ –3PQ PQ
= 0
Vectors
16 Mahesh Tutorials Science
∴∴∴∴ PQ and PQ are collinear vectors,
Hence the points ( ) ( ),P p Q q and ( )R r are
collinear
Q-5) Express = 3 +2 + 4p i j k as the linear
combination of the vectors
= + , = +a i j b j k and = +c k i
Ans. To express p as a linear combination of
, ,a b c
Let = + +p xa yb zc where , ,x y z
are scalars
∴∴∴∴ 3 +2 + 4i j k
= ( ) ( ) ( )+ + + + +x i j y j k z k i
= ( ) ( ) ( )+ + + + +x z i x y j y z k
+x z = 3;
+x y = 2;
+y z = 4
Solving we get
z =5
2;
y =3
2;
x =1
2
∴∴∴∴ p =1 3 5
+ +2 2 2a b c
Q-6) If ,a b and c are non-coplanar vectors, show
that the vectors 2 – + 4 , – + 4 – 3a b c a b c
and 5 – 6 +11a b c are coplanar.
Ans. p = 2 – + 4a b c ,
q = – + 4 – 3 ,a b c r
= 5 – 6 +11a b c
In order to show that vectors ,p q and r are
coplanar.
To find the scalars x and y such that
= +r x p yq
Now,
r = ( ) ( )2 – + 4 + – + 4 – 3x a b c y a b c
5 – 6 +11a b c
= ( ) ( ) ( )2 – + – + 4 + 4 – 3x y a x y b x y c
2x – y = 5 ...(i)
–x + 4y = –6 ...(ii)
4x – 3y = 11 ...(iii)
From (i) and (ii) x = 2 & y = – 1
For these values of x and y
L.H.S of (iii)
= 4(2) – 3(–1)
= 11
= R.H.S.
Thus, these values satisfy the (iii) equation
also.
∴∴∴∴ The given vectors are coplanar.
GROUP (B)- HOME WORK PROBLEMS
Q -1) a and b are non-collinear vectors. If
(((( ))))= – 2 +c x a b and (((( ))))= 2 +1 –d x a b are
collinear, then find the value of x.
Ans. Since c and d are collinear vectors, there
exists scalar t such that
c = td
∴∴∴∴ ( )– 2 +x a b = ( )2 +1 –t x a b
∴∴∴∴ ( )– 2 +x a b = ( )2 +1 –t x a tb
∴∴∴∴ x – 2
= t(2x + 1) and
t = –1
∴∴∴∴ t = –1
∴∴∴∴ x – 2
= –(2x + 1)
= –2x – 1
∴∴∴∴ 3x = 1
∴∴∴∴ x =1
3
Mahesh Tutorials Science 17
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Q-3) Express ���� ����– – 3 + 4i j k���� as the linear
combination of the vectors
���� ���� ���� ����2 + – 4 ,2 – +3i j k i j k� �� �� �� � and ���� ����3 + – 2i j k���� .
Ans. Let a = � �2 + – 4i j k� ,
b = � �2 – + 3i j k� ,
c = � �3 + – 2i j k� and
p = � �– – 3 + 4i j k� .
Suppose p = + + .xa yb ze
Then, � �– – 3 + 4i j k�
=� �( ) � �( )
� �( )
2 + – 4 + 2 – + 3
+ 3 + – 2
x i j k y i j k
z i j k
� �
�
∴∴∴∴ � �– – 3 + 4i j k�
= ( ) ( ) �
( ) �
2 + + 3 + – +
+ –4 + – 2z
x 2y z i x y z j
x 3y k
�
Q-2) If , ,a b c are non-coplanar vectors then
prove that the vectors 2 – 4 +4 ,a b c
– 2 + 4a b c and – +2 +4a b c are collinear.
Ans. Let P, Q and R be the points whose position
vectors are 2 – 4 + 4 ,p a b c= – 2 + 4 ,q a b c=
– + 2 + 4r a b c=
Now,
( ) ( )– = – 2 + 4 – 2 – 4 + 4PQ q p a b c a b c=
∴∴∴∴ – + 2 +0.PQ a b c= ... (i)
and
( ) ( )= – = – + 2 + 4 – 2 – 4 + 4PR r p a b c a b c
( )= –3 + 6 + 0 = 3 – + 2 + 0.a b c a b c
∴∴∴∴ 3PR PQ= ...[From (i)]
∴∴∴∴ ( )+ –3 0PR PQ =
∴∴∴∴ PR and PQ are collinear vectors.
∴∴∴∴ P, Q and R are collinear.
∴∴∴∴ = 2 – 4 + 4 ,p a b c = – 2 + 4q a b c
= – + 2 + 4r a b c are collinear.
By equality of vectors,
2x + 2y + 3z = –1
x – y + z = –3
–4x + 3y – 2z = 4
Solve these equations by using Cramer’s Rule
D =
2 2 3
1 –1 1
–4 3 –2
= 2(2 – 3) – 2(–2 + 4) + 3(3 – 4)
= – 2 – 4 – 3 = –9 ≠≠≠≠ 0
xD =
–1 2 3
–3 –1 1
4 3 –2
= –1(2– 3) – 2(6 – 4) + 3(–9 + 4)
= 1 – 4 – 15 = –18
yD =
2 –1 3
1 –3 1
–4 4 –2
= 2(6 – 4) + 1(–2 + 4) + 3(4 – 12)
= 4 + 2 – 24 = –18
zD =
2 2 –1
1 –1 –3
–4 3 4
= 2(–4 + 9) –2(4 – 12) –1(3 – 4)
= 10 + 16 + 1 = 27
∴∴∴∴ x = –18
= = 2,– 9
zD
Dy =
–18= = 2,
– 9
yD
D
z = 27
= = – 3– 9
zD
D
∴∴∴∴ = 2 + 2 – 3p a b c
Q-4) Express ���� ����= 3 +2 + 4P i j k���� as the linear
combination of the vectors
���� ���� ���� ����= + , = + , = +a i j b j k c k i� �� �� �� �
Ans. To express p as a linear combination of , ,a b c
Let p = + +xa yb zc
where x, y, z are scalars
∴∴∴∴ � �3 + 2 + 4i j k�
Vectors
18 Mahesh Tutorials Science
Q-5) If a ,b ,c are non-coplanar vectors, then
show that the vectors
– + 3 – 5 , – + +a b c a b c and 2 – 3 +a b c are
collinear
Ans. Let p = – + 3 – 5a b c ,
q = – + +a b c and
r = 2 – 3 +a b c
Then in order to prove that these vectors are
coplanar, we should be also to find scalars x
and y such that
r = +x p yq
Now, +x p yq
= ( ) ( )– + 3 – 5 + – + + 5x a b c y a b c
= ( ) ( ) ( )– – + 3 + + –5 +x y a x y b x y c
and r = 2 – 3 +a b c
∴∴∴∴ 2 – 3 +a b c
= ( ) ( ) ( )– – + 3 + + –5 +x y a x y b x y c
By equality of vectors,
–x – y = 2 .... (i)
3x + y = –3 .... (ii)
and –5x + y = 1 .... (iii)
Adding (i) and (ii), we get,
2x = –1
∴∴∴∴ x =1
–2
∴∴∴∴ –x – y = 2 gives 1
–2y = 2
∴∴∴∴ y =1
– 22
=3
–2
For these values of x and y,
–5x + y =1 3
–5 – + –2 2
=5 3
–2 2
= 1
Thus, these values satisfy the third equation
also.
∴∴∴∴ r =1 3
– + –2 2p q
∴∴∴∴ the given vectors are coplanar.
Q-6) Express ���� ����– – 3 + 4i j k���� as the linear
combination of the vectors
���� ���� ���� ����2 + – 4 ,2 – +3i j k i j k� �� �� �� � and ���� ����3 + – 2i j k���� .
Ans. Let a = � �2 + – 4i j k� ,
b = � �2 – + 3i j k� ,
c = � �3 + – 2i j k� and
p = � �– – 3 + 4i j k� .
Suppose p = + + .xa yb zc
Then, � �– – 3 + 4i j k�
= � �( ) � �( )� �( )
2 + – 4 + 2 – + 3
+ 3 + – 2
x i j k y i j k
z i j k
� �
�
∴∴∴∴ � �– – 3 + 4i j k�
= ( ) ( ) �
( ) �
2 + + 3 + – +
+ –4 + – 2z
x 2y z i x y z j
x 3y k
�
=�( ) � �( ) �( )+ + + + +x i j y j k z k i� �
= ( ) ( ) � ( ) �+ + + + +x z i x y j y z k�
∴∴∴∴ x + z = 3;
x = y = 2;
y + z = 4
solving, we get
z =5
2,
y =3
2,
x =1
2
∴∴∴∴ p =1 3 5
+ +2 2 2a b c
Mahesh Tutorials Science 19
Vectors
By equality of vectors,
2x + 2y + 3z = –1
x – y + z = –3
–4x + 3y – 2z = 4
Solve these equations by using Cramer’s Rule
D =
2 2 3
1 –1 1
–4 3 –2
= 2(2 – 3) – 2(–2 + 4) + 3(3 – 4)
= – 2 – 4 – 3 = –9 ≠≠≠≠ 0
xD =
–1 2 3
–3 –1 1
4 3 –2
= –1(2– 3) – 2(6 – 4) + 3(–9 + 4)
= 1 – 4 – 15 = –18
yD =
2 –1 3
1 –3 1
–4 4 –2
= 2(6 – 4) + 1(–2 + 4) + 3(4 – 12)
= 4 + 2 – 24 = –18
zD =
2 2 –1
1 –1 –3
–4 3 4
= 2(–4 + 9) –2(4 – 12) –1(3 – 4)
= 10 + 16 + 1 = 27
∴∴∴∴ x = –18
= = 2,– 9
zD
Dy =
–18= = 2,
– 9
yD
D
z = 27
= = – 3– 9
zD
D
∴∴∴∴ = 2 + 2 – 3p a b c
Q-7) Express the vector ���� ����+ 4 – 4i j k���� as the linear
combination of the vectors
���� ���� ���� ����2 – +3 , – 2 + 4i j k i j k� �� �� �� � and ���� ����– +3 – 5i j k���� .
Ans. Let a = � �2 – + 3i j k� ,
b = � �– 2 + 4i j k� ,
c = � �– – 3 – 5i j k� and
p = � �+ 4 – 4i j k� .
suppose p = + +xa yb zc .
Then,
� �+ 4 – 4i j k�
= � �( ) � �( )� �( )
2 – + 3 + – 2 + 4
+ – – 3 – 5
x i j k y i j k
z i j k
� �
�
� �+ 4 – 4i j k�
= ( ) ( ) �
( ) �
2 + – + – – 2 + 3
+ 3 + 4 – 5
x x y z i y x y z j
x y z k
�
By equality of vectors,
2x + y – z = 1
–x – 2y + 32 = 4
3x + 4y – 5z = –1
We have to solve these equation by ussing
cramer’s Rule.
D =
2 1 –1
–1 –2 3
3 4 –5
= 2(10 – 12) – 1(5 – 9) – (–4 + 6)
= –4 + 4 – 2 = –2 ≠ ≠ ≠ ≠ 0
Dx
=
1 1 –1
4 –2 3
– 4 4 –5
= 1(10 – 12) –1 (20 + 12) – 1(16 – 8)
= –2 + 8 – 8
Dx
= –2
Dy
=
2 1 –1
–1 4 3
3 –4 –5
= 2(–20 + 12) – 1(5 – 9) – 1(4 – 12)
= –16 + 4 + 8
Dy
= –4
Dz
=
2 1 1
–1 – 2 4
3 4 –4
= 2(8 – 16) – 1(4 – 12) + 1(–4 + 6)
= –16 + 8 + 2
Vectors
20 Mahesh Tutorials Science
Dz
= –6
x =xD
D=
–2
–2= 1
y =yD
D=
–4
–2= 2
z =zD
D=
–6
–2= 3
∴∴∴∴ p = + 2 + 3a b c
ii) ∴∴∴∴ abc =
3 2 1
1 1 2
3 1 2
−
= 1(1) –1(–1)
= 2
GROUP (C)- CLASS WORK PROBLEMS
Q-1) Find [ ]a bc if
= 2 + + 3 , = +2 + 4 ,
= + + 2
a i j k b i j k
c i j k
Ans. abc =
2 1 3
1 2 4
1 1 2
= 2(4 – 4) – 1(2 – 4) + 3(1 – 2)
∴∴∴∴ [ ]abc = –1
Q-2) = – + , = + – 4 , = 2 + +a i j pk b i j k c i j k
find p, if [ ] = 0abc
Ans. abc = 0
∴∴∴∴
1 1
1 1 4
2 1 1
p−
−= 0
∴∴∴∴ 1(1 + 4) + 1(1 +8) + p(1 – 2) = 0
14 – p = 0 ⇒⇒⇒⇒
∴∴∴∴ p = 14
Q-3) Find the volume of the parallelopiped if
= 2 , = 3a i b j and = 4c k are the
coterminus edges of the parallelepiped.
Ans. Volume of parallelepiped = [ ]abc
=
2 0 0
0 3 0
0 0 4
= 2 (12) – 0 + 0
= 24 cubic units.
Q-4) Show that vectors
= + + , = – +a i j k b i j k &
= 2 +3 +2c i j k are coplanar.
Ans. If [ ] = 0abc
then vectors ,a b and c are coplanar.
L.H.S. = a bc =
1 1 1
1 1 1
2 3 2
−
1( – 2 – 3) – 1(2 – 2) + 1(3 + 2)
= – 5 + 5 = 0
= R.H.S.
∴∴∴∴ , ,a b c
are coplanar vectors.
Q-5) Show that A, B, C, D are coplanar, if
A ≡≡≡≡ (2, 1, – 3), B ≡≡≡≡ (3, 3, 0), C ≡≡≡≡ (7, –1, 4) and
D ≡≡≡≡ (2, – 5, – 7).
Ans. We know that, four points A, B, C, and D are
coplanar if
[ ]AB AC AD = 0
AB = –b a
= ( ) ( ) ( )3 – 2 + 3 –1 + 0 + 3i j k
∴∴∴∴ AB = + 2 + 3i j k
AC = –c a
= ( ) ( ) ( )7 – 2 + –1 –1 + 4 + 3i j k
∴∴∴∴ AC = 5 – 2 + 7i j k
AD = –d a
= ( ) ( ) ( )2 – 2 + –5 –1 + –7 + 3i j k ⇒
= –6 – 4j k
∴∴∴∴ L.H.S. = AB AC AD
Mahesh Tutorials Science 21
Vectors
=
1 2 3
5 –2 7
0 –6 –4
= 1(8 + 42) – 2(– 20) + 3 (– 30)
= 50 + 40 – 90 = 0 = R. H. S.
∴∴∴∴ , ,AB AC AD are coplanar vectors.
∴∴∴∴ The points A, B, C and D are coplanar.
Q-6) Find value of m if points (2, –1, 1),
(4, 0, 3), (m, 1, 1), (2, 4, 3) are coplanar.
Ans. Let A ≡ (2, –1,1), B ≡ (4,0,3), C ≡ (m,1,1) and
D ≡ (2,4,3)
Given that the points A, B, C, D are
coplanar
AB = –b a
= ( ) ( ) ( )4 – 2 + 1+1 + 3 –1i j k
AB = 2 + + 2i j k
AC = –c a
= ( ) ( ) ( )– 2 + 1+1 + 1 –1m i j k
AC = ( )– 2 + 2m i j
AD = –d a
= ( ) ( ) ( )2 – 2 + 4 +1 + 3 –1i j k
= 5 + 2j k
∴∴∴∴ AB AC AD = 0
2 1 2
2 2 0
0 5 2
m −= 0
2(4) – 1(2m – 4) + 2(5m –10) = 0 ⇒⇒⇒⇒
∴∴∴∴ 8 – 2m + 4 + 10m – 20 = 0
∴∴∴∴ 8m = 8 ⇒⇒⇒⇒
∴∴∴∴ m = 1
Q-7) If O is the origin, A(1,2,3) B(2,3,4) and
P(x,y,z) are coplanar points prove that
x – x + 2y – z = 0. using vector method.
Ans. The points O, A, B and P are coplanar. (given)
∴∴∴∴ [ ]OA OB OP = 0
∴∴∴∴ OA = a = + 2 + 3i j k
OB = b = 2 + 3 + 4i j k
OP = p = + +xi y j zk
∴∴∴∴ [ ]OA OB OC = 0
∴∴∴∴
1 2 3
2 3 4
x y z= 0
∴∴∴∴ 1(3z –4y) – 2(2z – 4x) + 3(2y – 3x) = 0
∴∴∴∴ 3z –4y – 4z + 8x + 6y – 9x = 0
∴∴∴∴ –x + 2y – z = 0
∴∴∴∴ x – 2y – z = 0
Q-8) If A ≡≡≡≡ (1, 1, 1), B ≡≡≡≡ (2, -1, 3), C ≡≡≡≡ (3, –2, –2)
and D ≡≡≡≡ (3, 3, 4), find the volume of
parallelopiped with segments AB, AC and
AD as concurrent edges.
Ans. We know that, the volume of parallelopiped
with concurrent edges AB, AC and
AD = , ,AB AC AD
AB = –b a
= ( ) ( ) ( )+ ++ ++ ++ +2 –1 –1 –1 3 –1i j k
∴∴∴∴ AB = – 2 + 2i j k
AC = –c a
= ( ) ( ) ( )3 –1 + –2 –1 + –2 –1i j k
∴∴∴∴ AC = 2 – 3 – 3i j k
AD = ( ) ( ) ( )– 3 –1 + 3 –1 + 4 –1d a i j k=
∴∴∴∴ AD = 2 + 2 + 3i j k
∴∴∴∴ volume of parallelopiped
=
1 2 2
2 3 3
2 2 3
−
− −
= 1 (– 9 + 6) + 2 (6 + 6) + 2 (4 + 6)
= – 3 + 24 + 20 = 41 cubic units.
Q-9) Prove that [ + + + ] = 2[ ]a b b c c a abc
Ans. Let = [ + + + ]a b b c c a
= ( ) ( ) ( )××××+ .[ + + ]a b b c c a
= ( ) × × × ×× × × ×× × × ×× × × ×+ .[ + + + ]a b b c b a c c c a
= . [ + + ]
+ . [ + + ]
a b c b a c a
b b c b a c a
× × ×× × ×× × ×× × ×
× × ×× × ×× × ×× × ×
Vectors
22 Mahesh Tutorials Science
××××[ = 0]c c∵
= ( ) ( ) ( )
( ) ( ) ( )
× × ×× × ×× × ×× × ×
× × ×× × ×× × ×× × ×
. + . + .
+ . + . + .
a b c a b a a c a
b b c b b a b c a
= ( ) ( )× ×× ×× ×× ×. + .a b c b c a + 0 + 0 + 0 + 0
(in box product if two vectors are equal , its
value is zero)
= [ ] + [ ]a b c b c a
= [ ] + [ ]a b c a b c
( )∵∵∵∵ =abc b c a
= 2[ ] = . . .a b c R H S
Q-10) If = + 3c a b , show that , = 0abc Also if
2 2 2= + 3 3 +c a ab ab , Prove that the
angle between a and b is 6ππππ
Ans. c = + 3a b
i.e c is linear combination of and a b
∴∴∴∴ , ,a b c are coplanar
∴∴∴∴ abc = 0
c = + 3a b
∴∴∴∴ c = + 3a b
Squaring both sides.
2c =
2+ 3a b
2c = ( ) ( )+ 3 . + 3a b a b
2c = . + 3 . + 3 . + .a a a b b a ab b
2c =
2 2+ 6 . + 9a a b b
2c = ( )
22 π+ 6 6 cos + 9
6a a b
2c =
2 23+ 6 . + 9
2a a b b
c2 =2 23
6 . + 92
a ab b+
c2 = 2 2+ 3 3 + 9a ab b
Q-11) Show that the statement
× ×× ×× ×× ×( – ).[( – ) ( – )] = 2 .a b b c c a a b c
is true only when ,a b and c are coplanar.
Ans. If , ,a b c are coplanar then
[ ]a b c = 0
i.e., ( )××××.a b c = 0
L.H.S.
= ( ) ( ) ( )××××– . – –a b b c c a
= ( ) × × × ×× × × ×× × × ×× × × ×– .[ – – + ]a b b c b a c c c a
= ( ) × × ×× × ×× × ×× × ×– .[ + + ]a b b c a b c a
= × × ×× × ×× × ×× × ×
× × ×× × ×× × ×× × ×
.[ + + ]
– .[ + + ]
a b c a b c a
b b c a b c a
= ( ) ( ) ( )
( ) ( ) ( )
× × ×× × ×× × ×× × ×
× × ×× × ×× × ×× × ×
. + . + .
. + . .
a b c a a b a c a
b b c b a b b c a− +
= ( ) ( )× ×× ×× ×× ×. – .a b c b c a
= [ ] – [ ]a b c b c a
= [ ] – [ ]a b c a b c = 0
R.H.S
= 2 . ×a b c
= 2[ ]a b c
if , ,a b c are planar than [ ]a b c = 0
∴∴∴∴ R.H.S. = (0)
= 0
∴∴∴∴ L.H.S = R.H.S
When , ,a b c are co planar
Q-12) , ,a b c are three non-coplanar vectors. If
= , = , =b c c a a b
p q ra bc a bc a b c
× × ×× × ×× × ×× × ×
then prove that
i) . . . = 3a p b q c r+ ++ ++ ++ +
ii) (((( )))) (((( )))) (((( ))))+ . + + . + + . = 3a b p b c q c a r
Mahesh Tutorials Science 23
Vectors
Ans. i) L.H.S.
= . + . + .a p b q c r
=( ) ( ) ( )× × ×× × ×× × ×× × ×. . .
+ +[ ] [ ] [ ]
a b c b c a c a b
abc abc abc
=[ ] + [ ]+ [ ]
[ ]
abc bc a c ab
abc
=[ ] + [ ]+ [ ]
[ ]
abc abc abc
abc
=3[ ]
[ ]
abc
abc
= 3
ii) L.H.S. ( ) ( ) ( )+ . + + . + + .a b p b c q c a r
= ( )( ) ( ) ( )
( )( )
× ×× ×× ×× ×
××××
+ .. +[ ] [ ]
+ + .[ ]
b c b c c aa b
abc abc
a bc a
abc
+
Since a bc is a scalar.
Let a bc = k
( ).a b p+ = ( )( )+
+ .b c
a bk
=1
+a bc abck
= { }1
+ 0a b ck
=1a b c
k
= ××××1
0 kk
= 1
Similarly ( )+ .b c q = 1
and ( )+ .c a r = 1
( ) ( ) ( )+ . + + . + + .a b p b c q c a r
= 1 + 1 + 1 = 3
iii) . .a p =( )
( )
××××⇒⇒⇒⇒.
b caabc
.c q =( )
( )
××××.c q
cabc
=( )
( )⇒⇒⇒⇒
abc
abc=
( )
( )
c c a
abc
= 1
.b p =( )
( )
××××⇒⇒⇒⇒
.b b c
abc= ( )
0
abc
=( )
( )⇒⇒⇒⇒
bbc
abc
.c v =( )
( )
××××.c a b
abc
=( )
⇒⇒⇒⇒0
abc=
( )
( )
c a b
abc
= 0
⇒⇒⇒⇒ =( )
( )
abc
abc
⇒⇒⇒⇒ .b p =( )
( )
××××.b c a
abc
=( )
( )
bc a
abc ⇒⇒⇒⇒ =
( )
( )
abc
abc
.a r =( )
( )
××××.a c a
abc
=( )
( )
abc
abc
= ( )0
abc = 0
Vectors
24 Mahesh Tutorials Science
Q-13) If the vectors
+ + ; + + ; + 1ai j k i b j k i cka b c≠ ≠ ≠≠ ≠ ≠≠ ≠ ≠≠ ≠ ≠
are coplanar then show that
1 1 1+ + =1.
1 – 1 – 1 –a b c
Ans. Let
p = + +ai j k ,
q = + +i bj k ,
r = + +i j ck
and , ,p q r are coplanar.
∴∴∴∴ [ ]pqr = 0
∴∴∴∴ [ ]pqr =
1 1
1 1
1 1
a
b
c
= 0
Applying C1 – C
2 and C
2 – C
3
1 0 1
0 1 1
1 1
a
b
c c c
−
−
− −
= 0
∴∴∴∴ by taking (a – 1), (b – 1), (1 – c) common
∴∴∴∴ (a – 1)(b – 1)(1 – c ) =
11 0
–1
10 1
–1
1 11 –
a
b
c
c
= 0 ⇒⇒⇒⇒
∴∴∴∴ a b ≠≠≠≠ c ≠1
∴∴∴∴ (a –1) (b – 1) (c – 1) ≠ 0 ⇒⇒⇒⇒
11 0
–1
10 1
–1
1 11 –
a
b
c
c
= 0
1 1– –
1 – – 1 – 1
c
c b a= 0 ⇒⇒⇒⇒
1 1+ +
1 – –1 1 –
c
c b a= 0
1 11+ +
1 – 1 – 1 –
c
c b a
+
= 1 ⇒⇒⇒⇒
1 – + 1 1+ +
1 – 1 – 1 –
c c
c b a= 1
1 1 1+ +
1 – 1 – 1 –c b a= 1 ⇒⇒⇒⇒
1 1 1+ +
1 – 1 – 1 –a b c= 1
Q-14) Show that if four points A, B, C D with
position vectors , , ,a b c d are coplanar then
+ + =bcd cad abd abc
Ans. ∵ four points A, B, C, D are coplanar
∴∴∴∴ AB AC AD are also coplanar
∴∴∴∴ AB AC AD = 0
∴∴∴∴ ( )××××.AB AC AD = 0
( ) ( ) ( )××××– . – –b a c a d a = 0
( ) × × × ×× × × ×× × × ×× × × ×– . – – +b a c d c a a d a a = 0
( ) ( ) ( )
( ) ( ) ( )
× × ×× × ×× × ×× × ×
× × ×× × ×× × ×× × ×
. – . – .
– . + . – .
b c d b c a b a d
a c d a c a a a d= 0
×××× = 0a a
– – –
+ +
bc d bc a bad a c d
a c a aad
= 0
– – – + 0 + 0bc d abc abd c a d
= 0
+ + =bc d abd c a d abc
GROUP (C)- HOME WORK PROBLEMS
Q-1) Find a b c
i) ���� ���� ����
���� ����
= 2 – 2 +3 ; =10 +3 ;
= – +2
a i j k b i k
c i j k
� �� �� �� �
����
ii) = + , = –a i j b j k and = +c k i
Mahesh Tutorials Science 25
Vectors
Ans. i) a = � ��2 – 2 + 3i j k
b = ��10 + 3i k
c = � �� + + 2i j k
abc =
2 2 3
10 0 3
1 1 2
−
−
= 2 (0 + 3) + 2 (20 – 3) + 3 (–10 – 0)
= 6 + 34 – 30
= 10
ii) ∴∴∴∴ abc =
3 2 1
1 1 2
3 1 2
−
= 1(1) –1(–1)
= 2
Q-2) Find ‘p’ if a b c = 0 where,
���� ���� ���� ����= + + , = – +a i j k b i j k� �� �� �� � and
���� ����= 2 +3 +c i j pk����
Ans. i) a = � �� + +i j k
b = � �� – +i j k
c = � ��2 + 3 +i j pk
abc = 0
1 1 1
1 –1 1
2 3 p= 0
1 (–P – 3) – 1 (P – 2) + 1 (3 + 2)
– P – 3 – P + 2 + 5 = 0
– 2 P = –4
P = 2
Q-3) Find the volume of parallelopiped whose
co-terminus edges are
i) ���� ���� ���� ����
���� ����
= – 2 – , = 3 +2 + ,
= + +5
a i j k b i j k
c i j k
� �� �� �� �
����
ii) ���� ���� ����
���� ����
= +2 , = 2 + + ,
= 2 +2 +3
a i k b i j k
c i j k
� �� �� �� �
����
Ans. i) a = � � �– 2 +j j k
b = � � �3 – 2 +j j k
c = � � �+ + 5j j k
abc =
1 –2 –1
3 2 1
1 1 5
= 1 (10 – 1) + 2 (15 – 1) + 1 (3 – 2)
= 9 + 28 + 1
= 38
ii) abc =
1 0 2
2 1 1
2 3 3
= 1 (3 – 2) – 0 (6 – 2) + 2 (4 – 2)
= 1 – 0 + 4
= 5
Q-4) If A ≡≡≡≡ (1, 1, 1), B ≡≡≡≡ (2, 1, 3), C ≡≡≡≡ (3, 2, 2) and
D ≡≡≡≡ (3, 2, 4), find the volume of the
parallelopiped with AB, AC and AD as
concurrent edges.
Ans. A ≡≡≡≡ (1, 1, 1), B = (2, 1, 3), C ≡≡≡≡ (3, 2, 2),
D ≡≡≡≡ (3, 2, 4)
let , , ,a b c d be position vectors of points A,
B, C, D respectively with respect to a fixed
point.
∴∴∴∴ a = � �� + +i j k
b = � ��2 + + 3i j k
c = � ��3 + 2 + 2i k k
d = � ��3 + 2 – 4i j k
AB = –b a
= � �( ) � �( )� �2 + + 3 – + +i j k i j k
= �� + 2i k
AC = –c a
=� �( ) � �( )� �3 + 2 – 2 – + +i j k i j k
= � ��2 + +i j k
vol. of parallelopiped = AB AC AD
Vectors
26 Mahesh Tutorials Science
=
1 0 2
2 1 1
2 1 3
= 1 (3 – 1) – 0 (6 – 2) + 2 (2 - 2)
= 2 – 0 + 0
= 2
Q-5) If the volume of the parallelopiped whose
coterminus edges are
���� ���� ���� ���� ���� ����–3 +2 + , 2 + – , – + 3 +2i j nk i j k i j k� � �� � �� � �� � �
is 7 find n.
Ans. Let a = � ��–3 + 2 +i j nk
b = � ��2 + –i j k
c = � ��– + 3 – 2i k k
( )abc = 7
–3 2
2 1 –1
–1 3 2
n
= 7
–3 (2 + 3) – 2 (4 – 1) + n (6 + 1) = 7
–5 – 6 + 6n + n = 7
7n = 7 + 21
7n = 28
n = 4
Q -6) Show that the following vectors are
coplanar.
i) ���� ���� ���� ���� ���� ����3 – 5 +2 , – + 4 – 3 , +3 – 4i j k i j k i j k� � �� � �� � �� � �
ii) ���� ���� ���� ���� ���� ����3 + +3 , +3 +2 , 2 +6 + 4i j k i j k i j k� � �� � �� � �� � �
Ans. i) To show that, vectors are co-planar we
have to prove that
( )abc = 0
( )abc =
–3 5 2
–1 4 –3
1 3 –4
= 3 (–16 + 9) + 5 (4 + 3) + 2 (– 3 – 4)
= –3 (7) + (7) + 2 (–7)
= –21 + 35 – 14
= 0
( )abc = 0
so,
vectors are co-planar
ii) To show that, vectors are co-planar we
have to prove that
( )abc = 0
( )abc =
3 1 3
1 3 2
2 6 4
= 3 (12 – 12) – 1 (4 – 4) + (6 – 6)
= 0
( )abc = 0
∴∴∴∴ vectors are co-planar
Q-7) If ���� ���� ���� ����= + + , = – +a i j k b i j k� �� �� �� �
and ���� ����= 2 +3 +2c i j k���� find (((( )))).a b c××××
Ans. a = � �� + +i j k
b = � �� – +i j k
c = � ��2 + 3 + 2i j k
××××b c =
� ��
1 –1 1
2 3 2
i j k
= ( ) � ( ) � ( )� –2 – 3 – 2 – 2 + 3i j k
= ��–5 + 5i k
( )××××.a b c =� �( ) �( )� �+ + . –5 + 5i j k i k
= (1) (–5) + (1) (0) + (1) (5)
= –5 + 0 + 5
= 0
∴∴∴∴ , ,a b c are caplaner.
Q-8) Find ‘p’ if the following vectors are coplanar
���� ���� ���� ���� ���� ����= – 3 + 4 , = 2 – 5 + , = – – 6a i j k b i j pk c i j k� � �� � �� � �� � �
Ans. abc = 0
Mahesh Tutorials Science 27
Vectors
1 –3 4
2 –5
1 –1 6
P= 0
1 (30 + P) + 3 (–12 – P) + 4 (–2 + 5) = 0
30 + P – 36 – 3P + 12 = 0
–2P + 6 = 0
P = 3
Q-9) Find x is A (3, 2, -1), B (5, 4, 2), C (6, 3, 5),
D (1, 0, x) are coplanar.
Ans. Let , , ,a b c d be position vectors of pts.
A, B, C, D respectively w.r.t a fixed point.
a = � ��3 + 2 –i j k
b = � ��5 + 4 + 2i j k
c = � ��6 +3 +5i j k
d = �� +i kx
AB = –b a
=� �( ) � �( )� �5 + 4 +2 – 3 + 2 –i j k i j k
= � ��2 +2 + 3i j k
AC = –c a
=� �( ) � � �( )�6 +3 +5 – 3 + 2 –i j k j j k
= � ��3 + + 6i j k
AD = –d a
=�( ) � �( )� �+ – 3 +2 –i kx i j k
=� � �( )+ +–2 – 2 1j j k x
∴∴∴∴ points A, B, C, D are coplanar
, ,AB AC AD will also be coplanar
AB AC AD = 0
�( )
2 2 3
3 1 6
–2 –2 +1k
= 0
∴∴∴∴�( ) � ( ) + + + + + + + =
2 1 12 2 3 3 12 3 –6 2 0k k
∴∴∴∴ � � ( )+ + =2 26 – 6 – 30 –12 0k k
∴∴∴∴ � =–4 –16 0k
∴∴∴∴ �k = – 4
Q-10) If ���� ���� ����= 3 + 2 + , = + + 2a i j k b i j� �� �� �� � and
���� ����= 3 – + 2c i j k���� are the co-terminus edges of
a tetrahedron. Find its volume.
Ans. volume of a Tetrahedron =1
6abc
a bc =
3 2 1
1 1 2
3 1 2−
= 3(4) – 2(– 4) + 1 (– 4)
= 12 + 8 – 4
= 16
volume of a Tetrahedron =1
6abc
=16
6
volume of a Tetrahedron =8
3
Q-11) If = 3 + 2 ,c a b show that = 0abc .
If angle between a and b is 4ππππ prove
that
2 2= 9 + 6 2 + 4 =c a ab b a a and =b b
Ans. We know that,
××××b b = 0
If in a scalar triple product, two vectors are
equal, then the secalar triple product is zero.
a b c = ( )××××a b c
= ( )×××× 3 +2a b a b
= ( )× ×× ×× ×× ×3 + 2a b a b b
= ( )××××3 + 0a b a
= ( )××××3a b a
Vectors
28 Mahesh Tutorials Science
The measure of angle between a & b
is ππππ
4
c2 = C.C
= ( ) ( )3 + 2 . 3 + 2a b a b
= 9 . + 6 . + 6 . + 4 .a a a b b a b b
= ( )∵∵∵∵9 . +12 . + 4 . + . = .a a a b b b a b b a
=
( )
2 2ππππ
∵∵∵∵
9 +12 cos + 44
= & =
a ab b
a a b b
=2 21
9 +12 + 42
a ab b
2 2 29 + 6 2 + 4a ab b
a = &a b = b
Q-12) If = 3 – 2 ,c a b show that = 0abc .
Also find c2 when the measure of the angle
between a and b is 4ππππ .
Ans. We know that,
××××b b = 0
If in a scalar triple product, two vectors are
equal, then the scalar triple product is zero.
abc = ( )××××a b c
= ( )×××× 3 – 2a b a b
= ( )× ×× ×× ×× ×3 – 2a b a b b
= ( )××××3 – 0a b a
= ( )××××3a b a
= 3 × × × × 0 = 0
The measure of angle between a & b
is ππππ
4
c2 = c.c
= ( ) ( )3 – 2 . 3 – 2a b a b
= 9 . – 6 . – 6 . + 4 .a a a b b a b b
= ( )∵∵∵∵9 . – 12 . + 4 . + . = .a a a b b b a b b a
=
( )
2 2ππππ
∵∵∵∵
9 –12 cos + 44
= & =
a ab b
a a b b
=2 21
9 –12 + 42
a ab b
c2 = 2 29 – 6 2 + 4a ab b and
a = &a b = b
Q -13) If ���� ����= – 2 +u i j k���� , ����= 3 +v i k���� and ���� ����= –w j k
Find ( + ).[( ) ( )]× × ×× × ×× × ×× × ×u w u v v w
Ans. u = � �� – 2 +i j k
v = ��3 +i k
w = � �–j k
let a = ( )+u w
= �� –i j
b = ( )××××u v
= � �� + +i j k
=
� ��
1 –2 1
0 1 –1
i j k
= ( ) � ( )+� 1 1i k
c = ( )××××v w
=
� ��
3 0 1
0 1 –1
i j k
= ( ) � ( ) � ( )� –1 – –3 + 3i j k
= � ��– + 3 + 3i j k
To find ( ) ( ) ( )× × ×× × ×× × ×× × ×+ .u w u v v w
c = and
Mahesh Tutorials Science 29
Vectors
≡≡≡≡ ××××.a b c
= abc
=
1 1 0
1 1 1
–1 3 3
= 1(3 – 3) –1(3 + 1)
= – 4
∴∴∴∴ ( ) ( ) ( )× × ×× × ×× × ×× × ×+ .u w u v v w = – 4
Q -14) If four points with position vectors , ,a b c
and d are coplanar, prove that
(((( )))) (((( )))) (((( )))) (((( )))). = . + . .× × × ×× × × ×× × × ×× × × ×a b c b c d c a d a b d
Ans. i.e. To prove that
+ + =bc d c a d abd abc
Let, , ,a b cand d be the position of the points
A, B, C and D respectively. Then,
AB = –b a ,
AC = –c a ,
AD = –d a
The points A, B, C, D are coplanar.
∴∴∴∴ the vectors , ,AB AC AD are coplanar.
∴∴∴∴ the = 0AB AC AD
∴∴∴∴ – – – = 0b a c a d a
∴∴∴∴ ( ) ( ) ( )××××– . – – = 0b a c a d a
∴∴∴∴ ( ) ( )× × × ×× × × ×× × × ×× × × ×– . – – – = 0b a c d c a a d a a
where ×××× = 0a a
∴∴∴∴ ( ) ( )× × ×× × ×× × ×× × ×– . – – = 0b a c d c a a d
∴∴∴∴ ( ) ( ) ( )× × ×× × ×× × ×× × ×. – . – . –b c d b c d b a d
( ) ( ) ( )× × ×× × ×× × ×× × ×. + . + . = 0a c d a c a a a d ...(I)
Now, ( ) ( )× ×× ×× ×× ×. = 0, . = 0,a c d a a d
( )××××– .b a d = ( )××××.b d a
= bda
= abd
( )××××– .a c d = ( )××××.a d c
= a dc
= c a d
Also, ( )××××.b c d = bc d
( )××××– .b c a = ( )××××– .a b c abc
∴∴∴∴ from (I)
∴∴∴∴ + + =bc d c a d abd abc
Q-15) If (((( )))). = 1××××a b c , where ���� ����= ,+ ++ ++ ++ +����a i j k
���� ����= 2 + ++ ++ ++ +����b i q j k and ���� ����= – 4++++����c i j k ,
find q
Ans. Given that
( ).a b c×××× = 1
∴∴∴∴
1 1 1
2 1
1 1 4
q
−
= 1
∴∴∴∴ 1(4q + 1) – 1 (8 – 1) + 1 ( – 2 – 1) = 1
∴∴∴∴ 4q + 1 – 7 – 2 – q = 1
∴∴∴∴ 3q – 8 = 1
∴∴∴∴ 3q = 9
∴∴∴∴ q = 3.
Q-16) Show that, if , ,a b c are non-coplanar
vectors, then the vectors 3 + + 5a b c and
2 – 4 +3a b c are non-coplanar.
Ans. Let = – 2 ,p a c = 3 + + 5 ,q a b c
= 2 – 4 + 3r a b c
∴∴∴∴
1 0 –2
= 3 1 5
2 –4 3
pqr
= 1(3 + 20) – 0(9 – 10) – 2(–12 – 2)
= 23 + 28 ≠≠≠≠ 0
∴∴∴∴ , ,p q r are non-coplanar.
Vectors
30 Mahesh Tutorials Science
Q-17) prove that
i) (((( )))) (((( )))) (((( ))))+ + + = 2 . × ×× ×× ×× ×
b c c a a b a b c
ii) + + + = 0 a b ca b c
iii) (((( ))))+2 – – – –
= 3
××××
a b c a b a b c
a b c
Ans. i) LHS = ( ) ( )××××+ . + +b c c a a b
= + + +b c c a a a
= ( ) ( )+b c a c a b
= ( ) ( )+a b c a b c
= ( )2 a b c
= ( )××××2 .a b c
= R.H.S.
ii) LHS = + + +a b c a b c
= ( ) ( )××××+ + +a b c a b c
=× × × ×× × × ×× × × ×× × × ×
× ×× ×× ×× ×
+ + +.
+ +
b a b b b c c aa
c b c c
= ( ) ( ) ( )
( ) ( ) ( )
× × ×× × ×× × ×× × ×
× × ×× × ×× × ×× × ×
. + . + .
+ . + . + .
a b a a b b a b c
a c a a c b a c c
= 0 + 0 + ( )××××.a b c + 0
– ( )××××.a b c + 0
= 0
= RHS
iii) LHS
= ( ) ( )××××. + 2 – . – – –a a b c a b a b c
= ( )× × ×× × ×× × ×× × ×
× × ×× × ×× × ×× × ×
– –+ 2 – .
– + +
a a a b a ca b c
b a b b b c
= ( )× × ×× × ×× × ×× × ×
××××
0 – + ++ 2 – .
– +
a b c a a ba b c
o b c
= ( ) × ×× ×× ×× ×+ 2 – . +a b c c a b c
=
( ) ( ) ( )
( ) ( ) ( )
× × ×× × ×× × ×× × ×
× × ×× × ×× × ×× × ×
. + . + 2 .
+ 2 . – . – .
a c a a b c b c a
b b c c c a c b c
= ( ) ( )× × ×× × ×× × ×× × ×0 + . + 2 . + 2 0 – 0 – 0a b c b c a
= +a b c b c a
= + 2a b c a b c
= 3 a b c
= RHS
Mahesh Tutorials Science 31
Vectors
GROUP (D)- CLASS WORK PROBLEMS
Q-1) Prove that medians of a triangle are
concurrent
Ans. Let , ,a b c be the position vectors of the
vertices A, B, C and , ,d e f be the position
vectors of the midpoints D, E, F of the sides
BC, CA and AB respectively.
Then by the midpoint formula,
d =+
2
b c,
e =+
2
c a,
f =+
2
a b
∴∴∴∴ 2d = +b c ;
2e = +c a ;
2f = +a b
∴∴∴∴ 2 +d a = + +a b c
∴∴∴∴ 2 +e b = + +a b c
∴∴∴∴ 2 +f c = + +a b c
∴∴∴∴2 +
2+1
d a=
2 +
2+1
e b=
2 +
2+1
f c=
+ +
3
a b c
= g ... (Say)
This show that the point G whose position
vector is g , lies on the three medians AD,
BE and CF dividing each of them internally
in the ratio 2 : 1.
Hence, the medians are concurrent in the
point G and its position vector is
( )+ + /3.a b c This point of concurrence of
A
F E
CB
G
D
the medians of a triangle is called the
centroid of the triangle.
Q-2) Prove that bisectors of angles of a triangle
are concurrent
Ans. Let , ,a b c be the position vectors of the
vertices A, B, C of ∆∆∆∆ABC and let the lengths
of the sides BC,CA and AB be x, y, z,
respectively. If segments AD, BE, CF are the
bisectors of the angles A, B, C, respectively,
then
D divides the side BC in the ratio AB : BC
i.e., z : y,
E divides the side AC in the ratio BA : BC
i.e., z : x,
and F divides the side AB in the ratio AC :
BC i.e., y : x,
Hence by section formula, the postion vectors
of the points D, E and F are
d =+
+
zc yb
z y;
e =+
+
zc xa
z x;
f =+
+
yb xa
y x;
∴∴∴∴ ( )+y z d = +yb zc ;
( )+z x e = +zc xa ;
( )+x y f = + .xa yb
∴∴∴∴ ( )+ +y z d xa = ( )+ +z x e yb
= ( )+ +x y f zc
= + +xa yb zc
∴∴∴∴( )
( )
+ +
+ +
y z d xa
y z x=
( )( )
+ +
+ +
z x e yb
z x y
A
F E
CB D
P
00
xx
Vectors
32 Mahesh Tutorials Science
=( )
( )
+ +
+ +
x y f zc
x y z
=+ +
+ +
xa yb zc
x y z= p ...(Say)
This show that the point P whose position
vector is p , lies on the three bisectors AD,
BE and CF dividing them in the ratios
(y + z) : x; (z + x) : y and (x + y) : z respectively.
Hence, the three bisector segments are
concurrent in the point whose position vector
is + +
.+ +
xa yb zc
x y z
This point of concurrence of the bisectors is
called the incentre of the ∆∆∆∆ABC.
Q-3) Using vector method, prove that the
diagonals of a parallelogram bisect each
other and conversely. OR prove that, A
quadrilateral is a parallelogram if and only
if its diagonals bisect each other.
Ans. i) Let a,b,c and d be respectively the
position vectors of the vertices A, B, C
and D of the parallelogram ABCD then
AB = DC and side AB |||||||| side DC.
∴∴∴∴ AB = DC
∴∴∴∴ –b a = –c d
∴∴∴∴ +a c = +b d
∴∴∴∴+
2
a c=
+
2
b d ... (i)
the position vectors of the mid points of
the diagonal AC and BD are ( )+
2
a c and
( )+
2
b d by (i) they are equal
∴∴∴∴ the mid points of the diagonal AC and
BD are the same. this shows that the
diagonals AC and BD bisect each other.
ii) conversely, suppose that the diagonals
AC and BD of ���� ABCD bisect each other.
i.e, they have the same midpoint.
∴∴∴∴ the position vectors of these midpoints
are equal.
∴∴∴∴+
2
a c=
+
2
b d
∴∴∴∴ +a c = +b d
∴∴∴∴ –b a = –c d
∴∴∴∴ AB = DC
∴ AB= DC and side AB |||||||| side DC
∴∴∴∴ ���� ABCD is parallelogram.
CB
AD
M
Q-4) Prove that the median of a trapezium is
parallel to the parallel sides of the
trapezium and its length is half of the sum
of the lengths of the parallel sides.
Ans. Let a,b,c and d be respectively the position
vectors of the vertices A, B, C and D of the
trapezium ABCD with side AD |||||||| side BC.
Then the vectors AD and BC are parallel.
∴∴∴∴ there exists a scalar k
such that AD = k . BC
∴∴∴∴ AD+BC= .BC +BCk
= (k + 1) BC ... (i)
Let m and n be the position vectors of
the midpoints M and N of the non-
parallel sides AB and DC respectively.
then seg MN is the median of the trapezium.
by the mid point formula.
+=
2
a bm and
+=
2
d cn
∴∴∴∴ MN = –n m
=+ +
–2 2
d c a b
CB
A D
M N
Mahesh Tutorials Science 33
Vectors
∴∴∴∴ = ( )1+ – –
2d c a b
= ( ) ( )1– + +
2d a c d
=AD+BC
2... (ii)
=( )+1 BC
2
k...[(by (i)]
MN is a scalar multiple of BC
∴∴∴∴ MN and BC are parallel vectors
∴∴∴∴ ||MN BC where ||BC AD
∴∴∴∴ The median MN is parallel to the parallel
to sides AD and BC of the trapezium.
Now, AD and BC are collinear.
∴∴∴∴ + = + = +AD BC AD BC AD BC
[∵ AD and BC have same direction]
∴∴∴∴ from (2) we have
∴∴∴∴ MN = +
2
AD BC
∴∴∴∴ MN = ( )1+
2AD BC
Q-5) Using vectors, prove angle subtended in a
semi circle is right angle.
Ans. Let seg AB be a diameter of a circle with
centre C and P be any point on the circle
other than A and B.Then APB∠ is an angle
subtended on a semi-circle.
Let = =AC CB a and =CP r
Then =a r ...(1)
AP = AC+CP
= + ra
= +ar
BP = BC+CP
= –CB+CP
= – +a r
∴∴∴∴ AP.BP = ( ) ( )+ . –r a r a
= . – – + . – .r r r a a r a a
= 2 2– = 0r a
∴∴∴∴ ⊥⊥⊥⊥AP BP
∴∴∴∴ APB is a right angle.
Hence, the angle subtended on a semicircle
is the right angle.
a –a
r
CA B
P
Q-6) Using vectors, prove that altitudes of a
triangle are concurrent.
Ans. Let segment AD and CF be the altitudes of
∆∆∆∆ABC, meeting each other in the point H.
Then it is enough to prove that HB is
perpendicular to AC .
Choose H as the origin and let , ,a b c be the
position vectors of the vertices A, B and C
respectively w.r.t. the origin H.
Then ,HA a HB b= = and HC c= ,
– , –AB b a BC c b= = and –AC c a=
Now HA is perpendicular to BC
∴∴∴∴ . 0HA BC =
∴∴∴∴ ( ). – 0a c b =
∴∴∴∴ . – . 0a c a b = ...(1)
Also HA is perpendicular to AB .
∴∴∴∴ . 0HC AB =
∴∴∴∴ ( ). – 0c b a =
∴∴∴∴ . – . 0c b c a =
∴∴∴∴ . – . 0c b a c = ( ). .c a a c=∵ ...(2)
A
B C
EF
D
H
Vectors
34 Mahesh Tutorials Science
Adding (1) and (2), we get
. – . 0c b a c =
∴∴∴∴ ( )– . = 0c a b
∴∴∴∴ ( ). – = 0b c a
∴∴∴∴ HB.AC = 0
∴∴∴∴ HB is perpendicular to AC .
∴∴∴∴ the altitudes of ∆∆∆∆ABC are concurrent.
Q-7) Using vector method. find the in centre of
a triangle is right angle. are A(0,4,0),
B(0,0,3) and C(0,4,3).
Ans. The position vectors , ,a b c of the vertices A,B
& C are
a = 0 + 4 + 0i j k
b = 0 + 0 + 3i j k
c = 0 + 4 + 3i j k
AB = –b a
= ( ) ( )0 + 0 + 3 – 0 + 4 + 0i j k i j k
= 0 – 4 =3i j k
BC = –c b
= ( ) ( )0 + 4 + 3 – 0 + 0 + 3i j k i j k
= 0 + 4 + 0i j k
AC = –c a
= ( ) ( )0 + 4 + 3 – 0 + 4 + 0i j k i j k
= 0 – 0 +3i j k
Let x = BC ,
y = AC , and
z = AB
x = 2 2 20 + 4 +0
= 16 = 4
y = ( )22 20 + –0 +3
= 9 = 3
z = ( )22 20 + –4 +3
= 16 + 9
= 25 = 5
If H ( )h is the incentre of ∆∆∆∆PQR, then
h =+ +
+ +
xa yb zc
x y z
h =( ) ( ) ( )4 4 +3 3 +5 4 3
4+3 +5
j k j k+
h =16 + 9 + 20 +15
12
j k j k
h = 36 + 24
12
j k
h =0 +36 +24
12
j j k
h = 0 +3 +2i j k
H = (0,3,2)
Q-8) Prove by vector method, sin(α α α α + ββββ) =
sin αααα .cos ββββ + cos αααα . sin ββββ .
Ans.
Let ∠∠∠∠XOP and ∠∠∠∠XOQ be in standard position
and mXOP = –αααα, mXOQ = ββββ.
Takes a pouint A on ray OP and a point B on
ray OQ such that OA = OB = 1.
Since cos (–αααα) = cos αααα
and sin (–αααα) = –sin αααα,
A is (cos (–αααα), sin (–αααα)),
i.e, (cos αααα, –sin αααα)
B is (cos ββββ, sin ββββ)
∴∴∴∴ OA = ( ) ( )α αα αα αα αcos – sin . + 0.i j k
Y
B
Q
X
PA
O –2–2
β
Mahesh Tutorials Science 35
Vectors
OB = ( ) ( )β ββ ββ ββ βcos + sin . + 0.OB i j k
∴∴∴∴ ××××OA OB =
� �
α αα αα αα α
β ββ ββ ββ β
cos – sin 0
cos sin 0
i j k�
= (cos αααα sin β β β β + sin αααα cos ββββ)k
... (i)
The angle between OA and OB is α α α α + ββββ.
Also OA ,OB lie in the XY-plane.
∴∴∴∴ the unit vector perpendicular to OA and OB
is k .
∴∴∴∴ ××××OA OB = [OA.OB sin (α α α α + ββββ)]k
= sin (α α α α + ββββ).k ...(ii)
from (i) and (ii),
sin (α α α α + ββββ) = sin αααα cos ββββ + cos αααα sin ββββ.
GROUP (D) – HOME WORK PROBLEMS
Q-1) By vectors method. prove that the
diagonals of rhombus are perpendicular to
each other.
Ans. Let �OACB be a rhombus and =OA a ,
=OB b . Then OA = OB
∴∴∴∴ =a b
∴∴∴∴ a = b.
Now = +OC a b .
∴∴∴∴ the position vectors of the vertices O, A, B
and C w.r.t. the origin O are 0, ,a b and +a b
respectively.
By the midpoint formula, the position vectors
of the midpoints of the diagonals. BA and OC
( )+ /2a b are and ( )+ + 0a b i.e., ( )+ /2a b .
These are equal.Hence the diagonals OC and
BA have the same midpoint.
∴∴∴∴ these diagonals bisect each other ...(1)
Now OC = +a b
and BA = +BC CA
= –a b
∴∴∴∴ .OC BA = ( ) ( )– . –a b a b
= . – . + . – .a a a b b a b b
= a2 – b2 ( )∵∵∵∵ . = .a b b a
= 0 ( )∵∵∵∵ =a b
∴∴∴∴ ⊥⊥⊥⊥OC BA i.e., the diagonals OC and BA are
at right angles. ...(2)
∴∴∴∴ from (1) and (2), the diagonals of a rhombus
bisect each other at right angles.
Conversely : Let the diagonals OC and BA of
the quadrilateral.OACB bisect each other at
right angles.
Since the diagonals bisect each other, the
quadrilateral OACB is a parallelogram.
Now,
∴∴∴∴ ⊥⊥⊥⊥OC BA
∴∴∴∴ .OC BA = 0
∴∴∴∴ ( ) ( )+ . –a b a b = 0
∴∴∴∴ . – . + . – .a a a b b a b b = 0
∴∴∴∴2 2– . + . –a a b b a b = 0
∴∴∴∴2 2–a b = 0
∴∴∴∴2 2=a b = 0
∴∴∴∴ =a b
∴∴∴∴ ( ) ( )=l OA l OB
i.e., adjacent sides OA and OB of the
parallelogram OABC are equal.
∴∴∴∴ OABC is a rhombus.
Hence, a quadrilateral is rhombus if and only
if diagonals bisect each other at right angle.
O A
CB
a
b
a
b
ab+
a
b–
Vectors
36 Mahesh Tutorials Science
Q-2) Using vectors prove that, if the diagonals
of a parallelogram are at right angles then
it is rhombus.
Ans. Let ABCD be a parallelogram
Let AC = +AB BC
= +a b
and BD = +BA AD
= – +a b
∴∴∴∴ the diagonals AC and BD are perpendicular
∴∴∴∴ .AC BD = 0
⇒⇒⇒⇒ ( ) ( )+ – +a b a b = 0
– . + . – . + .a a a b b a b b = 0
b2 = a2 ⇒⇒⇒⇒ b = a
⇒⇒⇒⇒ l(AB) = l(BC) = l(CD) = l(AD)
∴∴∴∴ the parallelogram ABCD is a rhombus .
Q-3) Show by vector method. If the diagonal of
a parallelogram are congruent then it is a
rectangle.
Ans. Let ABCD be parallelogram
Let = , =AB a AD b
From ∆ ∆ ∆ ∆ ABC, by triangle law,
AC = +AB BC
= +a b
From ∆ ∆ ∆ ∆ ABC, by triangle law,
BD = +BC CD
= –b a
We have to prove that ABCD is a rectangle
i.e. to prove that ⊥⊥⊥⊥a b
Since the diagonals AC and BD are congruent
=AC BD ⇒⇒⇒⇒ 2 2=AC BD
( ) ( )+ . +a b a b = ( ) ( )– . –b a b a
. + . + . + .a a a b b a b b
= . – . – . + .b b b a a b a a
2 2+ + 2 .a b a b
= 22 – 2 . +b a b a
2 . + 2 .a b a b
= 0 ⇒⇒⇒⇒ ( )4 .a b = 0 ⇒⇒⇒⇒ .a b = 0
⊥⊥⊥⊥ ⇒⇒⇒⇒ ⊥⊥⊥⊥ ⇒⇒⇒⇒ ⊥⊥⊥⊥a b AB AD AB AD
∴∴∴∴ The parallelogram ABCD is a rectangle.
b
BA
DC
a
b
a
Q-4) Show by vector method that the sum of the
square of the diagonals of a parallelogram
is equal to the sum of the sequare of its
sides.
Ans. Let � ABCD be a parallelogram.
Let =AB p and =AD q
Since ABCD is a parallelogram,
= =DC AB p and = =BC AD q
= + = +AC AB BC p q
∴∴∴∴ ( ) ( )2 = . = + . +AC AC AC p q p q
= . + . + . + .p p p q q p q q
= p2 + q2 + 2p.q ( ). = .p q q p∵
BD = +BC CD = –BC DC = –q p
∴∴∴∴ BD2 = .BD BD = ( ) ( )– . –q p q p
= . – . – . + .q q q p p q p p
= p2 + q2 + 2p.q ( ). = .p q q p∵
∴∴∴∴ AC 2 + BD 2 = ( )( )
2 2
2 2
+ +2 .
+ + – 2 .
p q p q
p q p q
= p2 + q2 + p2 + q2
A B
C
q
p
q
p
Mahesh Tutorials Science 37
Vectors
= AB2 + BC 2 + CD 2+ DA 2
This show that the sum of the squares of the
diagoanals of a parallelogram is equal to the
sum of the squares of its sides.
Q-5) If A (1, 2, 3); B (3, -1, 5); C (4, 0, -3) are
three non-collinear points by using vector
method then show A is point on circle
having BC as diameter.
Ans. A ≡≡≡≡ (1, 2, 3), B ≡≡≡≡ (3, −−−−1, 5), C ≡≡≡≡ (4, 0, −−−−3)
let , ,a b c be position vectors of points A, B
& C respectively w.r.t a fixed point.
a = +2 +3i j k
b = 3 – +5i j k
c = 4 – 3i k
AB = –b a
= ( ) ( )3 – +5 – + 2 + 3i j k i j k
= 2 – 3 + 2i j k
AC = –c a
= ( ) ( )4 – 3 – + 2 + 3i k i j k
= 3 – 2 – 6i j k
.AB AC = ( ) ( )2 – 3 + 2 . 3 – 2 – 6i j k i j k
= ( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 . + –3 –2 – + 2 –6 .i i j j k k
= 6 (1) + 6 (1) − − − − 12 (1)
= 0
seg AB ⊥⊥⊥⊥ seg AC
i.e. seg BC subtends right angle at A
i.e. seg BC is diameter of a semicircle.
Q-6) Using vector method. find incentre of a
triangle whose vertices are P(1, – 2,1) ;
Q(1, 1, – 3) and R(1, – 2, – 3).
Ans. The position vectors , ,p q r of the vertices
P,Q,R are
p = – 2 +i j k
q = – – 3i j k
r = – 2 – 3i j k
PQ = –q p
= ( ) ( )+ – 3 – – 2 +i j k i j k
= 0 +3 – 4i j k
QR = –r q
= ( ) ( )– 2 – 3 – + – 3i j k i j k
= 0 – 3 +i j Ok
PR = –r p
= ( ) ( )– 2 – 3 – – 2 +i j k i j k
= 0 +0 – 4i j k
Let x = QR ,
y = PR
z = PQ
x = ( )0 + –3 + 022 2
= 9 = 3
y = ( )0 +0 + –422 2
= 16 = 4
z = ( )0 +0 + –422 2
= 9 +16
= 25 = 5
If H ( )h is the incentre of the triangle PQR,
then
h =+ +
+ +
x p yq zr
x y z
h
=3 – 6 +3 + 4 + 4 –12 +5 –10 –15
3 + 4 +5
i j k i j k i j k
=12 –12 – 24
12
i j k
h = – – 2i j k
h = (1, –1, –2)
Vectors
38 Mahesh Tutorials Science
Q -7) Using vectors, prove that the perpendicular
bisectors of the sides of a triangle are
concurrent.
Ans.
Let D, E, F be the midpoints of the sides BC,
CA and and AB of ∆ ∆ ∆ ∆ ABC.
Let the perpendicular bisector of the sides
BC and AC meet each other in the point O.
Choose O as the origin and let , , , ,a b c d e and
f be the position vectors of the points A, B,
C, D, E, F respectively.
Here we have to prove that =OF f is
perpendicular to = –AB b a .
By the midpoint formula,
d =+
2
b c,
e =+
2
c a,
f =+
2
a b,
Now =OD d is perpendicular to = –BC c b .
∴∴∴∴ ( ). – = 0d c b
∴∴∴∴ ( )+. – = 0
2
b cc b
∴∴∴∴ ( ) ( )+ . – = 0c b c b
∴∴∴∴ . + . – . – . = 0c c b c c b b b
∴∴∴∴ c2 – b2 = 0
( )2 2∵∵∵∵ . = , . = . = .c c c b b b andc b b c
∴∴∴∴ c2 – b2 ...(1)
Also, =OE e is pcpendicular to = –AC c a
∴∴∴∴ ( ). – = 0e c a
∴∴∴∴ ( )+. – = 0
2
c ac a
∴∴∴∴ as above c2 – b2 = 0 ...(2)
∴∴∴∴ from (1) and (2), we get
b2 = a2
∴∴∴∴ b2 = a2 = 0
∴∴∴∴ ( ) ( )+ . – = 0b a b a
∴∴∴∴ ( ) ( )+ . – = 0b a b a
∴∴∴∴ ( ). – = 0f b a
∴∴∴∴ =f OF
is perpendicular bisector of the sides of ∆∆∆∆ ABC
are concurrent.
This point of concurrrent of the perpendicular
bisector of the sides of a triangle is called the
circumcentre of the triangle.
B CD
EF
A
Q-8) Prove by vector method, that, a
quadrilateral is a square if and only if
diagonals are congruent and bisect each
other at right angles.
Ans. Let ABCD be a square.
Let , , ,a b c d be the position vectors A, B, C,
D respectively.
Since ABCD is a sequare, we have,
=AB DC
∴∴∴∴ – = –b a c d
∴∴∴∴ + = +b d a c
∴∴∴∴+ +
= =2 2
b d a cp ... (Say)
This shows that the point P whose position
vector is p is the midpoint of BD as well as
of AC.
dD cC
bBaA
Mahesh Tutorials Science 39
Vectors
∴∴∴∴ the diagonals BD and AC bisect each other
at P.
∴∴∴∴ the diagonals of a square bisect each other
...(i)
Now, = = +AC BC BC AB
and ∵∵∵∵= + = + ... =BD BC CD BC BA CD BA
= –BC AB
∴∴∴∴ .AC BD = ( ) ( )+ . –BC AB BC AB
= . – . + . – .BC BC BC AB AB BC AB AB
=2 2– . + . –BC AB BC AB BC AB
= ( ) ( )2 2–l BC l AB
= 0 ... ( ) ( )=l AB l BC ∵
∴∴∴∴ AC is perpendicular to BD
∴∴∴∴ diagonals AC and BD are at right angles
... (ii)
Also, .AC AC = ( ) ( )+ . +BC AB BC AB
= . + . + . + .BC BC BC AB AB BC AB AB
= ( )22
⊥⊥⊥⊥∵∵∵∵+ 0 +0 + ... rBC AB AB BC
= ( ) ( )2 2+BC AB
and .BD BD = ( ) ( )– . –BC AB BC AB
= . – . – . + .BC BC BC AB AB BC AB AB
= ( )22
⊥⊥⊥⊥∵∵∵∵– 0 – 0 + ... rBC AB AB BC
= ( ) ( )2 2+BC AB
∴∴∴∴ .AC AC = .BD BD
∴∴∴∴2 2=AC BD
∴∴∴∴ l(AC) = l(BD) ...(iii)
∴∴∴∴ from (i), (ii) and (iii), the diagonals of a square
are congruent and bisect each other at right
angles.
Conversely : Let the diagonals AC and BD of
the quadrilaeral ABCD are congruent and
bisect each other at right angles.
Since the diagonals bisect each other, the
quadrilateral ABCD is a parallelogram.
Now, r⊥⊥⊥⊥ = 0AC BD
∴∴∴∴ . = 0AC BD
∴∴∴∴ ( ) ( )+ . – = 0BC AB BC AB
∴∴∴∴ . – . + . – . = 0BC BC BC AB AB BC AB AB
∴∴∴∴2 2– . . + . – = 0BC AB BC AB BC AB
∴∴∴∴2 2– = 0BC AB
∴∴∴∴2 2=BC AB
∴∴∴∴ =BC AB
∴∴∴∴ l(BC) = l(AB)
i.e., the adjacent sides AB and BC of the
parallelogram ABCD are equal.
∴ ABCD is a rhombus.
Also, l(AC) = l(BD)
∴ =AC BD
∴2 2=AC BD
∴ . = .AC AC BD BD
∴ ( ) ( ). .BC AB BC AB+
= ( ) ( )– . –BC AB BC AB
∴ . + . + . + .BC BC BC AB AB BC AB AB
= . – . – . + .BC BC BC AB AB BC AB AB
∴ . + .AB BC AB BC
= – . – .AB BC AB BC
∴ 4 . = 0AB BC
∴ . = 0AB BC
∴∴∴∴ r⊥⊥⊥⊥AB BC
i.e., the adjacent sides of a rhombus ABCD
are perpendicular to each other.
Hence, ABCD is a square.
Vectors
40 Mahesh Tutorials Science
Q-9) Using vector method, find the incentre of
a triangle whose vertices are P(0,2,1),
Q(– 2,0,0) and R(–2,0,2).
Ans. The position vectors , ,p q r of the vertices
P,Q,R are
p = � � �2 + , = 2j k q k and
r = �–2 + 2i k�
∴∴∴∴ PQ = –q p
=� �( )–2 – 2 +2i j k�
= � �–2 – 2 –i j k�
QR = –r p
=�( ) ( )–2 + 2 – –2i k i� �
= �2k and
PR = –r p
=�( ) �( )–2 +2 – 2 +i k i k� �
= � �–2 – 2 +i j k�
Let x = QR ,
y = PR and
z = PQ
∴∴∴∴ x = 20 + 0 + 2 = 2
y = ( ) ( )2 2 2–2 + –2 +1 = 3
and y = ( ) ( )2 2 2–2 + –2 +1 = 3
If H ( )h is the incentre of the triangle PQR, then
h =+ +
+ +
x p yq zr
x y z
=
� �( ) ( ) �( )2 2 + + 3 –2 + 3 –2 +2
2+3+3
j k i i k� �
=� � �( )1
4 + 2 – 6 – 6 + 68
j k i i k� �
=� �( )1
–12 + 4 + 88
i j k�
∴∴∴∴ h =� �3 1
– + +2 2i j k�
∴∴∴∴ H =3 1
– , ,12 2
Q-10)Prove that the segment joining the mid-
points of the diagonals of a trapezium is
parallel to the parallel sides and equals half
its difference.
Ans. Let M and N be the midpoints of the diagonals
AC and BD respectively of the trapezium
ABCD in which side AD|| side BC.
Then the vectors AD and BC are collinear.
∴∴∴∴ there exists a non-zero scalar k, such that
AD = .k BC
∴∴∴∴ –AD BC = . –k BC BC = ( )–1k BC ... (i)
Let , , , ,a b c d m and n be the position vectors
of the points A, B, C, D, M and N respectively.
Since M and N are the midpoints of the
digonals AC and BD, by the midpoint formula,
m = +
2
a c and n =
+
2
b d
∴∴∴∴ MN = –n m = + +
–2 2
b d a c
= ( ) ( )1– – –
2d a c b
= ( )1–
2AD BC =
–1
2
kBC ... [By (1)]... (2)
Thus MN is a scalar multiple of BC .
∴∴∴∴ MN and BC are collinear vectors, i.e.,
parallel vectors.
∴∴∴∴ ,MN BC� where .BC AD�
∴∴∴∴ seg MN is parallel to the parallel sides AD
and BC of the trapezium.
Now, AD and BC are collinear vectors.
∴∴∴∴ –AD BC = –AD BC = –AD BC
∴∴∴∴ from (ii) we have,
A D
B C
MN
Mahesh Tutorials Science 41
Vectors
MN = ( )1–
2AD BC
∴∴∴∴ MN = 1
–2AD BC =
1– .
2AD BC
Q-11) If the lengths of two non-parallel sides of
a trapezium are equal then, prove by vector
method that lengths of their diagonals are
equal.
Ans.
Let ABCD be a trapenzium such that AB is
parallel to CD and l (AD) = l (BC)
i.e., =AD BC ...(i)
Let AM and BN be the perpendiculars from
A and B to DC respectively.
∴∴∴∴ ∆∆∆∆AMD ≅≅≅≅ ∆∆∆∆BNC
∴∴∴∴ ∠∠∠∠D =∠∠∠∠C ...(ii)
Now, = +AC AD DC and = +BD BC CD
∴∴∴∴ ( ) ( ). = + . +AC AC AD DC AD DC
= . + . + . + .AD AD AD DC DC AD DC DC
= 2 2
+2 . +AD AD DC DC
( )∵∵∵∵ . = .AD DC DC AD
= ( )2 2
ππππ+2 . cos – +AD AD DC D DC
= 2 2
– 2 . cos +AD AD CD D DC ...(iii)
and ( ) ( ). = + . +BD BD BC CD BC CD
= 22
+2 . +BC BC CD CD
( )∵∵∵∵ . = .BC CD CD BC
= ( )22
ππππ+2 cos – +BC BC CD C CD
= 22
– 2 cos +BC BC CD C CD
=22
– 2 . cos +AD AD CD D CD ...(iv)
[By (i) and (ii)]
DM N
C
BA
π –D π –C
From (iii) and (iv), we get,
. = .AC AC BD BD
∴∴∴∴2 2
=AC BD
∴∴∴∴ =AC BD
∴∴∴∴ l (AC) = l (BD)
Hence, the diagonal of the tranpezium are
euqal.