(^.^) practice exercises vector
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Transcript of (^.^) practice exercises vector
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