Post on 13-Mar-2020
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 1
TUTORIAL 3
Question 1
a) 1,3, 4u v
2 2 2
1 3 4
26
u v
b) 2 3 12, 19,12u v
2 2 2
2 3 12 19 12
649
u v
c) 40 34 83 3 3
24 , ,
3v u
2 2 2
40 34 83 3 3
24
3
2705 /17.7012
3
v u
d) 16v u
16v u
Question 2
a) u v b) u v w
c) v w d) w u
Question 3
2
u av bw
i j a i j a i j
Compare LHS and RHS
: 2
: 1
i a b
j a b
Therefore,
3 1,
2 2a b
u +v v
u
v-w
-w
v
w w - u
-u
w
u -v+w
-v
u
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 2
Question 4
Finding angles: .
cosa b
a b
a) 2 2 , 3 4u i j k v i k
2 2 2 2 2 2
1
2, 2,1 3,0,4cos
2 2 1 3 0 4
10
15
10cos
15
48.19
b) 3 7 , 3 2u i j v i j k
2 22 2 2 2
1
3, 7,0 3,1, 2cos
3 7 0 3 1 2
4
4 26
4cos
4 26
101.31
c) 2 2 ,u i j k v i j k
2 22 2 2 2
1
1, 2, 2 1,1,1cos
1 2 2 1 1 1
1
15
1cos
15
104.96
Question 5
a) / /u w since
15 3 3
3 5
3
w i j k
i j k
u
b) None is perpendicular to each other.
0u w and 0v w
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 3
Question 6
a) Line equation: AC tv
Since we know that v is parallel to AB , find v.
3, 2,5
v AB
(Using point A) Substitute into the formula: Let C(x,y,z) be a point on line L
1, , 2 3, 2,5
1 3 , 2 , 2 5
AC tv
x y z t
x t y t z t
(Using point B) Substitute into the formula: Let C(x,y,z) be a point on line L
4, 2, 3 3, 2,5
4 3 , 2 2 , 3 5
BC tv
x y z t
x t y t z t
b) Parallel lines:
Since L1 is parallel to L2, then both lines are parallel to the same vector v.
2 : 3 , 6 , 2 7
3 1
0 6
2 7
1,6,7
L x t y t z t
x
y t
z
v
B(4,-2,3)
A(1,0,-2)
L
L2: x=3-t, y=6t, z=2+7t P(0,1,2)
L1
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 4
Equation of line L1 that passes through point P(0,1,2):
Let Q(x,y,z) be a point on line L1.
, 1, 2 1,6,7
, 1 6 , 2 7
PQ tv
x y z t
x t y t z t
c) Line orthogonal to the plane
Since L is orthogonal to the plane, then L is parallel to the normal vector of the plane, n.
Given the equation of the plane is
3 2 10x y z ,
Therefore, the normal vector is 3, 1, 2n .
Equation of line L that passes through point P(1,4,-2):
Let Q(x,y,z) be a point on line L.
1, 4, 2 3, 1,2
1 3 , 4 , 2 2
PQ tv
x y z t
x t y t z t
d) Line of intersection
Since plane M and N are intersect, vector v is parallel to the line of intersection.
P(1,4,-2)
L n
M
N nM
nN
v = nM x nN
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 5
: 5 4 5,1,1
: 10 6 10,1, 1
M
N
M x y z n
N x y z n
Find v:
5 1 1 2 15 5
10 1 1
M N
i j k
v n n i j k
Find P(x,y,z) which lies on the line of intersection.
: 5 4 1
: 10 6 2
M x y z
N x y z
Assume 0 :z
5 4 3
10 6 4
2, 2
5
x y
x y
x y
Therefore, the point of intersection is 2
, 2,05
P
.
Equation of intersection line that passes through point 2
, 2,05
P
:
Let Q(x,y,z) be a point on the line.
2, 2, 2,15, 5
5
22 , 2 15 , 5
5
PQ tv
x y z t
x t y t z t
Question 7
a) Equation of plane
Find two intersecting vectors:
1, 1, 4 1,3,0AB AC
Find the normal vector, n:
1 1 4 12 4 2
1 3 0
i j k
n AB AC i j k
A C
B
n
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 6
Let Q(x,y,z) be a point on the plane:
0
1, 2, 12,4,2 0
12 12 4 8 0
12 4 20
AQ n
x y z
x y z
x y z
b) Plane orthogonal to the line
Equation of line:
4 1
1 2
0 8
x
y t
z
Since line L is orthogonal to the plane, then we can say
that L is parallel to the normal vector, n of the plane.
1, 2,8n
Let Q(x,y,z) be a point on the plane:
0
3, 2, 1 1, 2,8 0
3 2 4 8 8 0
2 8 7
PQ n
x y z
x y z
x y z
c) Parallel planes
Since the planes are parallel, then 5, 3,2M Nn n .
Plane M passes through point P.
Let Q(x,y,z) be a point on plane M.
0
4, 1, 2 5, 3,2 0
5 20 3 3 2 4 0
5 3 2 27
MPQ n
x y z
x y z
x y z
d) Equation plane that contains 2 lines
Find the normal vector, n:
1 2 1 2 3 13 7 9
1 3 2
i j k
n v v i j k
L: x = 4+t, y = 1-2t, z = 8t
n
n
L1 L2
N: 3x - y + 2z = 10
nN
M
nM
P
P
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 7
P(4,0,1) is the intersection point and let Q(x,y,z) be a point on the plane:
0
4, , 1 13,7,9 0
13 52 7 9 9 0
13 7 9 61
PQ n
x y z
x y z
x y z
e) Two planes orthogonal.
Find the normal vector of plane N
2 1 2 3 8 7
3 2 1
N M
i j k
n PQ n i j k
Let R(x,y,z) be a point on plane N
0
1, 2, 4 3, 8, 7 0
3 3 8 16 7 28 0
13 7 9 41
PR n
x y z
x y z
x y z
Question 8
a) Intersection point between line and plane
2 3 4
2 3 3 1 2 2 4
13 3
3
13
x y z
t t t
t
t
Therefore, the intersection point is 9 19 23
, ,13 13 13
P
n N
P(1,2,4)
Q(-1,3,2)
M : 3x + 2y - z = 4
N
n M = <3,2,1>
L
P
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 8
b) Intersection between lines
1 1 1 1
2 2 2 2
: 1 2 , 3 , 5
: 6 , 2 4 , 3 7
L x t y t z t
L x t y t z t
Equate x and y to find t1 and t2
1 2 1 2
1 2 1 2
1 2
: 1 2 6 2 5 1
: 3 2 4 3 4 2 2
2, 1
x t t t t
y t t t t
t t
Therefore the point of intersection is P(5,6,10).
Question 9
a) Line through two points
2, 2,2
v PQ
Let R(x,y,z) be a point on the line
(i) P(1,2,-1)
1, 2, 1 2, 2,2
1 2 , 2 2 , 1 2
PR tv
x y z t
x t y t z t
Therefore the parametric for the line by using P(1,2,-1) is
1 2 , 2 2 , 1 2x t y t z t .
(ii) Q(-1,0,1)
1, , 1 2, 2,2
1 2 , 2 , 1 2
QR tv
x y z t
x t y t z t
Therefore the parametric for the line by using Q(-1,0,1) is
1 2 , 2 , 1 2x t y t z t .
b) Parallel lines
Since L1 is parallel to L2, then both lines are parallel to the same vector, v.
Let Q(x,y,z) be a point on line L1.
3, 2, 1 2, 1,3
3 2 , 2 , 1 3
PQ tv
x y z t
x t y t z t
Therefore, the parametric for is 3 2 , 2 , 1 3x t y t z t .
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 9
c) Line perpendicular to the plane
Since line L is perpendicular to the plane, then
we can say that L is parallel to the normal vector, n of the plane.
3,7, 5n
Let Q(x,y,z) be a point on the line:
2, 4, 5 3,7, 5
3 3 , 4 7 , 5 5
PQ tn
x y z t
x t y t z t
Therefore, the parametric for is 2 3 , 4 7 , 5 5x t y t z t .
Question 10
a) Parallel planes
Since these two planes are parallel, then 1 2n n .
Let Q(x,y,z) be point on the plane.
0
1, 1, 3 3,1,1 0
3 3 1 3 0
3 1
PQ n
x y z
x y z
x y z
b) Line perpendicular to the plane
Since the line is perpendicular to the plane, then v n
Let P(x,y,z) be a point on the plane.
0
2, 4, 5 1,3,4 0
2 3 12 4 20 0
3 4 34
QP n
x y z
x y z
x y z
Question 11
a) Intersection point between lines
1
2
: 2 , 3 2, 4
: 2, 2 4, 4 1
L x t y t z t
L x s y s z s
L n
P(2,4,5)
n
2
n
1
P(1,-1,-3)
3x + y + z = 7
L: x = 5+t, y = 1+3t, z = 4t n
Q(2,4,5)
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 10
Equate x and y to find t and s
: 2 1 2 2 1 1
: 3 2 2 4 3 2 2 2
0, 1
x t s t s
y t s t s
t s
Therefore the point of intersection is P(1,2,3).
Find the normal vector of the plane:
1 2 2 3 4 20 12
1 2 4
i j k
n v v i j k
Let Q(x,y,z) be a point on the plane:
0
1, 2, 3 20,12,1 0
20 20 12 24 3 0
20 12 7
PQ n
x y z
x y z
x y z
Question 12
a) Distance from a point to the line
From the line equation we can find Q and v.
: 2 , 1 2 , 2
0,1,0 2,2,2
L x t y t z t
Q v
Distance from P to L is
QP vD
v
2 2 2
2 0 1 2 6 4
2 2 2
2 6 4 2 14
i j k
QP v i j k
QP v
2 2 2
2 14 42
32 2 2
D
Q
P(2,1,-1)
v
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 11
b) Distance from a point to the plane
Given the equation of plane is 2 2 4 0x y z , then the
normal vector is 2,1, 2n and P(2,2,3).
2 2 2
2 2 2
2 2 4
2 1 2
2 2 2 2 3 4
2 1 2
8
3
PQ nD
n
x y z
c) Distance between planes
Find a point on plane M. Let x = 0 and y = 0, then .3
5z
2 2 2
53
2 2 2
2 6 1
1 4 6
0 2 0 6 1
1 4 6
9/1.4056
41
x y zD
d) Angles between planes
1 2
1 2
2 2 2 2 2 2
1
cos
1,2,0 2,1, 2
1 2 0 2 1 2
4
3 5
4cos
3 5
53.4
n n
n n
n
P
Q
n
2
nN = <1,2,6>
P(1,-1,-3)
N: x + 2y + 6z = 1
M: x + 2y + 6z = 10 D
n2 = <2,1,-2>
n1 = <1,2,0>
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 12
Question 13
Intersection plane parallel to the line
: 2 2 5 : 5 2 0
1,2, 2 5, 2, 1M N
M x y z N x y z
n n
Find vector, v (intersection line)
1 2 2 6 9 12
5 2 1
M N
i j k
v n n i j k
Given the line equation is:
: 3 2 , 3 , 1 4
2,3,4
L x t y t z t
u
Therefore,
6, 9, 12
3 2,3,4
3
v
u
Since 3v u , therefore the intersection line is parallel to line L.
Question 14
Line intersects to a plane.
: 3 2 , 2 , : 3 4L x t y t z t M x y z
Find t by substitute x,y and z to M
3 2 3 2 4
1
t t t
t
Therefore, the point of intersection is P(1,-2,-1).
Find v:
2 2 1 5 3 4
1 3 1
i j k
u v n i j k
Therefore, the line equation on the plane that passes through P(1,-2,-1) and let Q(x,y,z) be a
point on the line:
L
P
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 13
1, 2, 1 5,3,4
1 5 , 2 3 , 1 4
PQ tu
x y z t
x t y t z t
Therefore, the parametric for is 1 5 , 2 3 , 1 4x t y t z t
Question 15
26 263, 2,4 ,0,
5 5AB CD
Find the normal vector of the plane
26 265 5
52 182 523 2 4
5 5 50
i j k
n AB CD i j k
Let Q(x,y,z) be a point on the plane:
0
52 182 522, , 3 , , 0
5 5 5
52 104 182 52 1560
5 5 5 5 5
2 7 2 10
PQ n
x y z
x y z
x y z
Question 16
Intersection line
: 2 1 : 2 8
1,2,1 1, 1,2M N
M x y z N x y z
n n
Find vector v which is parallel to the line of intersection.
1 2 1 5 3
1 1 2
M N
i j k
v n n i j k
nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 14
Find the point of intersection. Let z =0, then 17 7
,3 3
x y . Therefore, the point of
intersection is 17 7
, ,03 3
P
.
Let Q(x,y,z) be a point the line.
17 7, , 5, 1, 3
3 3
17 75 , , 3
3 3
PQ tv
x y z t
x t y t z t
Therefore, the parametric for is 17 7
5 , , 33 3
x t y t z t .
Question 17
2 3 1 7 3 5
1 1 2
i j k
n u v i j k
Let Q(x,y,z) be a point on the plane:
0
1, 2, 3 7, 3, 5 0
7 7 3 6 5 15 0
7 3 5 14
AQ n
x y z
x y z
x y z