UNIVERSITY OF THE EAST2219 CLARO M. RECTO AVENUE, SAMPALOC, MANILA

COLLEGE OF ENGINEERINGELECTRONICS ENGINEERING DEPARTMENT

ECN – 313 – 1A (VECTOR ANALYSIS)PRACTICE EXERCISES

VIZCONDE, MA. KARYL NIÑA C.20110155812

FERNANDO VICTOR DE VERAJULY 26, 2013

VECTOR ALGEBRA

1.) Given vectors A= - ax + 6az and B = 2ax – 5ay + 4az. Determine the following:

a) |A + 2B| b) 2B – 3A c) The component of A in the direction y. d) A unit vector parallel to 2A – B

a.) A + 2B = (-ax + 6az) + 2(2ax - 5ay + 4az) = (-1 + 4)ax + (0 - 10)ay + (6 + 8)az = 3ax - 10ay + 14az

A + 2B = √32+(−102)+142 A + 2B = √305 ≈ 17.464 units

b.) 2B - 3A = 2(2ax - 5ay + 4az) - 3(-ax + 6az) = (4 + 3)ax + (-10 - 0)ay + (8 - 18)az

2B - 3A= 7ax - 10ay - 10az

c.) A = 0ay

d.) 2A - B = 2(-ax + 6az) - (2ax - 5ay + 4az) = (-2-2)ax + (0 + 5)ay + (12-4)az 2A-B = -4ax + 5ay + 8az 2A - B = (-42) + 52 + 82 = 105

a2A-B=(2 A−B )│2 A−B│

=−4 ax+5 ay+8az

√105 a2A-B = -0.390ax+0.488ay+0.781az

2.) Three points S(2, -6, 1), T(-3, 0 -4) and Q(5, -1, 8) forms a triangle. Find the length of the sides and all the internal angles.

rs = 2ax-6y+az, rT = -3ax-4az, rQ=5ax-ay+8azSolution of the length:RST = rT- rs = -3ax-4az-2ax+6ay-az = -5ax+6ay-5azRST = √(−52)+62+(−52)RST = √86 ≈ 9.274 unitsRTQ=rQ-rT =-3ax-4az-5ax+ay-8az =-8ax+ay-12azRTQ = √¿¿RTQ = √209 ≈ 14.457 units

RSQ = rQ-rS = 5ax-ay+8az-2ax+6ay-az =3ax+5ay+7azRSQ = √32+52+72

RSQ = √83 ≈ 9.110 units

Solution for the internal angles:

α =cos−1[ RST · RTQ|RST||RTQ|] , θ=

cos−1[ RTQ · RSQ|RTQ||RSQ|] , β = cos−1[ RST · RSQ|RST||RSQ|] or |RSQ|

sin α = |RST|

sin θ = |RTQ|

sin βRST =-5ax+6ay-5az, │ RST │ =√86RTQ =-8ax+ay-12az, │ RTQ │= √209

RST · RTQ = (-5)(-8) + (6)(1) + (-5)(-12) = 106

α = cos−1[ 106(186 )(√209) ]

α = 37.754°

√85sin 37.754 °

= √209sin θ

, θ =

sin−1 √209 sin 37.754 °√83

θ = 76.308°

β = 180° - α – θ = 180° - 37.754° β = 65.93°

3.) 1. Given that A=ax+kay-2az and B=ax-3ay+kaz are 40degrees apart, find k. 2. Find the component of (4ax-3ay) along (7ay+10az) 3. Find the projection of A=3ax-ay+7az along the z axis.

1.) AxB = |ax ay az1 k −21 −3 k |

AxB=k2ax - 2ay - 3az - kaz - 6ax - kay= (k2-6)ax-(2+k)ay-(3+k)azAxB = √¿¿AxB =√k4−10k2+10k+49A =√12+k2+(−2)2 = √k2+5B =√12+k2+(−3)2 = √k2+10k4 -10k2 + 10k + 49 = (k2 + 5)( k2 + 10)sin40°k4 -10k2 + 10k + 49 = k4 + 15k2 +50 sin40°-10k2 + 10k + 49 = k4 + 15k2 + 50 (sin40°)2 (shift solve for k)

k = -1.067

2.) Let A = 7ay + 10az and B = 4ax - 3ay

Scalar component:

B·ax =cos−1 A·B|A||B| = cos−1 −21

(5 )(√149)B·ax = 69.875B·aA = 5cos69.875°B·aA = 1.720Vector component:

(B·aA)aA = BaA cos θAB

aA=7ay+10az

√149

aA = 0.573ay + 0.819az(B·aA)aA = 5(0.573ay + 0.819az)cos69.875° (B·aA)aA= 0.986ay + 1.409az

3.) B·ax = cos−1 A·B|A||B|=cos−1 7

√59 = 24.311°

Length of projection:B·aA = BaA cos θAB = √59cos24.31° B·aA = 7

Vector projection:(B·aA)aA = BaA cos θAB

aA =az√72

= az

B =√59

(B·aA)aA = √59cos24.311°az (B·aA)aA = 7az

4.) Consider the three vectors A=ax-2ay-az; B=3ax+ay-7az and C=5ax-6az. If the vectors are adjacent sides of a parallelepiped, find:

a) The volume of the solid. b) The base area. c) The perpendicular height. d) The longest diagonal.

a.) V = A.(BxC)

BxC = |ax ay az3 1 −75 0 −6|

0 = -6ax -35ay + 18ay – 5azBxC = -6ax – 17ay – 5azV = (1)(-6) + (-2)(-17) + (-1)(-5)V= 33 cubic units

b.) Abase = BxC BxC = -6ax – 17ay – 5azAbase = √(−6)2+(−17)2+(−5)2

Abase = 5√14 ≈ 18.708 sq. Units

c.) The perpendicular height.

H = A •(BxC )│BxC│

= 33

5√14 H = 1.764 unit

VECTOR EQUATION OF STRAIGHT LINES

1.) Find the equation of the line passing through (-1,3,9) and is parallel to:

a.) The x-axis A(-1, 3, 9)rA=-ax+3ay+9azrB=axr=rA+tBr=-ax+3ay+9az+t(ax)x=-1+t ; y=3 ; z=9

b) The y-axis rA=-ax+3ay+9azrB=ayr=-ax+3ay+9az+t(ay)x=-1; y=3+t ; z=9

c) The z-axis rA=-ax+3ay+9azrB=azr=-ax+3ay+9az+t(az)x=-1; y=3;z=9+t

d) The vector 8ax + 7ay - 5azA(-1, 3, 9)B=8ax+7ay-5azr=-ax+3ay+9az+t(8ax+7ay-5az)x=-1+8t; y=3+7t; z=9-5t

2.) A line passes through (1,2,-3) and parallel to the direction

2√25ax−3√2

10ay+ √2

2az . Find:

a) The parametric equation of the line A(1, 2, -3) rA=ax+2ay-3az

B=2√25ax−3√2

10ay+ √2

2az

r= ax+2ay-3az+t( 2√25ax−3√2

10ay+ √2

2az )

x = (1+ 2√25t)5 = 5+2√2t

y = (2-√210t ¿10 = 20+√2t

z = (-3+√22t)2 =-6+√2t

x = 5+2√2t; y=20+3√2t ; z=20+√2t

b) The standard equation of the line x= 5+2√2t y=20+3√2t z=20+√2t

t=x−52√2

t=y−203√2

t=z+6√2

x−52√2

= y−203√2

= z+6√2

c) The direction ratio

( 2√2:3√2 : √2)1√2

= 2:3:1

3.) Find the equation of the line passing through the given points:

a) A(4,6,-1) and B(3,0,5) x−43−4

= y−60−6

=z−(−1)5−(−1)

=¿ x−4−1

= y−6−6

= z+16

b) A(1,1,0) and B(-2,-4,8) x−1

−2−1= y−1

−4−1= z−0

8−0=¿ x−1

−3= y−1

−5= z

8

c) A(2/3, -5/7, 0) and B(-8/5, 11/3, 8) x−2/3

−8/5−2/3=y−(−5/7)

113

−(−5/7)= z−0

8−0=¿ x−2/3

−34 /15= y+5 /7

92/21= z

8

d) A(2,0,0) and B(2,0,4)x−22−2

= y−00−0

= z−04−0

=¿ z4; x=∞; y=0

4.) A line is defined by the following parametric equations: x=2-3t; y=-1+2t; z=9+t. Find the distance from the point (4,2,1) to the line.

A = 2ax-ay+9azB = -3ax+2ay+azC = 4ax+2ay+az

RAC = 4ax+2ay+az-2az+ay-9azRAC = 2ax+3ay-8az

RAC2 = 2(2)+3(3)+(-8)(-8)=77

RAC2 . rB=2(-3)+3(2)+(-8)1=-8

rB=√ (−3 )2+22+12=√14

d = √77−( −8√14

)2

d = 8.510 units

b.) A line is defined by the standard Cartesian form as (2x-1) = ( y−4 )

2 ; z=1. Find the distance from the point (0,3,-5) to the line.

t=2x-1 t =( y−4 )

2 z=12x=t+1 2t=y-4

x=

12 +

12 t y=4+2t

A (

12 , 4, 1)

= (

12 ax + 4ay + az)

B (

12 , -2, 0)

= (

12 ax - 2ay)

C (0,3,-5) = (3ay-5az)

RAC = -

12 ax+ay+6az

RAC2 =

14 + 1 + 36 = 37.25

RAC.rB =-1/4-2=-2.25

d =√37.25−(−2.252.062 )

2

d = 6.005units

5.) Find the distance between the given set of parallel lines: a) L1 passes through (1,2,-4) and L2 passes through (0,9,7). Both lines are

parallel with the vector (8ax-3az)

r1(1,2,-4) r2(0,9,7)

rB = 8ax-3az

r12 = -ax+7ay+11azr12

2 = (-1)(-1)+(7)(7)+(11)(11) = 171

r12.rB = (-1)(8)+(7)(0)+(11)(-3) =-41

d =√171−(−41

√82+32 )2

d =12.164units

b) L1 is given as x−1

3= y+4

2=2 z−1and L2 which passes through the point (3,-

9,1)

x−13

= y+42

=2 z−1

A1 (1, 4,

12 ) rB (3, 2,

12 ) A2 (3,-9,1)

rA1A2 = 2ax-5ay+1/2az

rA1A22 = (2)(2)+(-5)(-5)+(

12 )(

12 )=

592

rA1A2.rB=(2)(3)+(-5)(2)+(

12 )(

12 )=-

154

d =√ 592

−(

−154

√32+22+(1/2)2)

2

d =5.33units6.) Find the distance between the following skew lines:

L1 passes through (1,-2,3) and parallel to vector (7ax-2az) and L2 passes through (0,4,-4) and parallel with the z-axis.

A1 (1,-2,3)A2 (0,4,-4)B1 = 7ax-2azB2 parallel to z-axisrA1-rA2=-ax+6ay-7az

B1x B2 ¿ax ay az7 0 −20 0 1

= 7ay

B1x B2 = √72=7

(rA1-rA2) . (B1x B2) = 42

d =(r A1−r A2)▪(B1 x B2)

│B1 x B2│d = 6 units

b) L1 is given by the equation (2x+1)= ( y−3 )

2 =(3 z+4 )

5

and L2 is given by the equation ( x+2)

3 = (2 y−3 )

4 ; z = 1.

( x+2)3 =

(2 y−3 )4 ; z=1.

t = 2x+1 t=( y−3 )

2 t = (3 z+4 )

5t = 2x+1 2t=y-3 5t = 3z+42x = 1-t y=3+2t 3z = 4+5t

x = 12−1

2t , z =

−43

+ 53t

t = ( x+2)

3 = (2 y−3 )

4 ; z=13t = x+2 4t=2y-3 x =3t-2 2y=3+4t

y=32+2 t

rA1 = 12ax + 3ay -

43 az

rA2=-2ax+32ay+az

B1 = 12ax+2ay+5

3az

B2 = 3ax+2ay

rA1-rA2 = −52ax−3

2ay+7

3az

B1x B2 ¿

ax ay az12

2 53

3 2 0

= 5ay + az + 103ax+6az=10

3ax + 5ay + 7az

(rA1-rA2) . (B1x B2) = (-52 )(

103 ) + (

−32 )(5) + (

73 )(7) =

12

B1x B2=√( 103

)2

+52+72 =

√7663

d = (r A1−r A2)▪(B1 x B2)

│B1 x B2│

d=

12

√7663 = 0.172 units

VECTOR EQUATIONS OF PLANES

1.) Find the equations of the planes described:

a) Passing through the point (3,4,1) and perpendicular to the vector (ax-ay+3az). b) Passing through the point (1,0,0) and perpendicular to the x-axis.

a) rA =3ax+4ay+azB = ax-ay+3azd = B . rA = [(3)(1)+(4)(-1)+(1)(3)] = 2

Equation of planeAx – ay + 3az = 2

b.) x-axis = (x, 0, 0)rA = axB = axd = (1)(1)d =1x =1

2.) Find the equation of the following planes:

a) Passing through the points (3,0,-3), (4,8,7) and (-2,-6,-5). At what point on the x-axis does it pass through? b) Passing through the origin and the points (2,4,-5) and (-3,-6,7)

a.) A (3,0,-3) B (4,8,7) C (-2,-6,-5)

x−3 y z+34−3 8 7+3

−2−3 −6 −5+3 = 0

x−3 y z+31 8 10

−5 −6 −2 = 0

0 = -[6(x-3)+(-50y)-6(z+3)} - [-60(x-3)-2y-40(z+3)]0 = 44(x - 3) - 48y + 34(z + 3)44x - 48y + 34z = 30 is the equation of the plane

B=44ax - 48ay + 34az is the perpendicular vector of planeD (44,0,0)

b.) A (0,0,0) B (2,4,-5) C (-3,-6,7)

x y z2 4 −5

−3 −6 7

0=28x + 15y - 12z - [30x+14y-12z]0= -2x + y2x – y = 0

3.)Find the point of intersection between the lines and the planes described:

a ) Line: passes through (6, 1, -2); parallel with (ax - 3ay + 2az).

Plane: passes through (0,9,6) and perpendicular with (-2ay + 7az).

b) Line equation: ( x−3 )

2 = (2y + 3) = ( z−5 )

4 ; plane equation (3x+4y-9z=5).

a.) Line: rA = 6ax = ay-2azB = ax - 3ay + 2az

Equation of linerL = (6ax + ay - 2az) + t(ax - 3ay + 2az)

PlanerA = 9ay + 6azB =-2ay+7azB . r = B. rA = dd = (9)(-2) + (6)(7)d = 24(-2ay+7az) . [(6ax+ay-2az) + t(9ax-3ay+2az)] = 24

0(6+9t)+ -2(1-3t)+ 7(-2+2t) 24

-2 + 6t – 14 + 14t = 2420t = 40t = 2(0, 10, 14)

b.) B . r = B . rA

Linex = 3 + 2ty= -3/2 + t/2z= 5 + 4tRp = 3ax – 3/2ay + 5az + t(2ax + 1/2ay + 4az)

Planed = 5B = 3ax + 4ay - 9azB . r = B.rA

= (2ax+4ay-9az) . [(3ax-

32 ay +50z)

+ t(2ax +

12 ay + 4az)] = 5

x = 3(3+2t) = 9 + 6t

y = 4(-

32 +

12 t) = -6 + 2t

z = -9(5+4t) = -45 + 36t -42 + 14t = 5

t =

4744

x =

33422

y = -

8522

z = -

7211

(

33422 , -

8522 , -

7211 )

4.) Find the line intersection between two planes described as follows:

a) Plane 1: (2ax-5ay+az).r=2 and Plane 2: (-9ax+3ay-8az).r=5 b) Plane 1 passes through (2,5,1) and perpendicular to (4ax-2ay+7az);

Plane 2 passes through (7,2,-5) and perpendicular to the vector (-6ay+9az)

a.) A (2, -5, 1)B (-9, 3, -8)

A x B =

x y z2 −5 1

−9 3 −8

= [(40 - 3)ax + (-9 + 16)ay + (6 - 45)az)]= 37ax + 7ay – 39az

Select xy plane (z = 0)2x -5y = 2

-9x + 3y = 5

x= -

3139

y= -

2839

b.) A1 (2, 5, 1) B1 (4, -2, 7)A2 (7, 2, -5) B2 (0, -6, 9)

B . r = d

(4ax – 2ay + 7az) . r = [(4)(2) + (-2)(5) + (7)(1)](4ax – 2ay + 7az) . r = 5

B . r = d (4ax – 2ay + 7az).r = [(0)(7) + (-6)(2) + (4)(-5)](-6ay + 9az) . r = -57

Select xy plane (z=0)4x – 2y = 5-6y = -57

x = 6

y =

192

5.) Find the distance from the point to the plane described as follows:

a) Point (6,0,-2) and the plane (4ax-9az).r=6. b) Point (2,5,1) and the plane which passes through (5,1,-1) and

is perpendicular to the vector (5ax-2ay+5az).

a.) A (6, 0, -2)B(4, 0, -9) . r = 6

d = 6

d =

d - (rA . B)B

d=

{[(6)( 4 )+( -2 )( -9 )] - 6}√(4 )2 +( -9 )2

d =

36√97

d = 3.655 units

b.) A (5, 1, -1)B (2, 5, 1)Po (2, 5, 1)

d = Po . B = [(2)(5) + (5)(-2) + (1)(5)]d = 5

rA . B = [(5)(5) + (1)(-2) + (-1)(5)] = 18

d =

d - (rA . B)B

d =

(18−5 )√54

d = 1.769 unit