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Dot Product Vectors Worksheet

Dot Product Vectors Worksheet
  • Page 1
 1.  
Find < 5a, 8b > · < 5a + 8b, 5a - 8b >.
a.
25a2 - 64b2 + 80ab
b.
25a2 - 64b2 + 40ab
c.
25a2 - b2 + 80ab
d.
25a2 - b2 + 40ab


Solution:

< 5a, 8b > · < 5a + 8b, 5a - 8b > = 5a(5a + 8b) + 8b(5a - 8b)
[Use the definition of dot product.]

= 25a2 + 40ab + 40ba - 64b2

= 25a2 - 64b2 + 80ab
[As ab = ba.]


Correct answer : (1)
 2.  
Let u = < 6, 9 > and v = < 3, 5 >. Which of the following is equal to projvu?
a.
<189 34, 315 34>
b.
<315 , 189 >
c.
<3 34, 5 34>
d.
<189 , 315 >


Solution:

u = < 6, 9 > and v = < 3, 5 > are the two vectors.

|v| = 32+52 = 34
[Find the magnitude of v.]

u · v = < 6, 9 > · < 3, 5 >
[Find the dot product of u, v.]

= 6(3) + 9(5) = 63
[Use the definition of dot product.]

projvu = (uv|v |2) v
[Use the definition of Projvu.]

= (63 / 34) < 3, 5 >
[Substitute the values of u · v, |u| and |v|.]

= < 189 / 34, 315 / 34>
[Simplify.]


Correct answer : (1)
 3.  
Let u, v be two nonzero vectors. Which of the following is correct?
a.
projvu = (u · v)
b.
projvu = (uv|v |2)v
c.
Projection of u in the perpendicular direction of v = u - (uv|v |2)v
d.
Both B and C


Solution:

u, v are two nonzero vectors.

Projection of u on v = (uv|v |2)v
[Use the definition of Projvu.]

Projection of u in the perpendicular direction of v = u - (uv|v |2)v.
[Use the definition of projection of u in the perpendicular direction of v.]


Correct answer : (4)
 4.  
Let u = (cos 70°) i + (sin 70°) j and v = (cos 10°) i + (sin 10°) j. Which of the following is the angle between the vectors u, v?
a.
10°
b.
60°
c.
80°
d.
70°


Solution:

u = (cos 70°) i + (sin 70°) j and v = (cos 10°) i + (sin 10°) j are the two vectors.

|u| = cos2 70° + sin2 70° = 1
[Use the formula to find the magnitude of u.]

|v| = cos2 10° + sin2 10° = 1
[Use the formula to find the magnitude of v.]

u · v = (cos 70°)(cos 10°) + (sin 70°)(sin 10°)
[Use the definition of the dot product to find u · v.]

= cos (70° - 10°)
[Use cos A cos B + sin A sin B = cos (A - B).]

= cos 60° = 12

Let θ be the angle between u and v.

So, cos θ = u  v|u||v|.
[Write the formula to find cos θ.]

= 12(1)(1) = 12
[Substitute the values of u · v, |u|, |v|.]

θ = Cos-1(12) = 60°
[Solve for θ.]


Correct answer : (2)
 5.  
Which of the following is correct for a vector u of a plane?
a.
u = (u · i) + (u · j)
b.
u = (u · i)j + (u · j)i
c.
u = (u · j) i
d.
u = (u · i)i + (u · j)j


Solution:

Let u = <x, y>

i = <1, 0> and j = <0, 1> are the standard unit vectors.

u · i = <x, y> · <1, 0> = x

u · j = <x, y> · <0, 1> = y

(u · i)i + (u · j)j = xi + yj = u

(u · i)j + (u · j)i = xj + yiu

u · i + u · j = x + yu

(u · j)i = yiu

So, (u · i)i + (u · j)j = u for any vector u of a plane.


Correct answer : (4)
 6.  
Let u = < sin 3 α, cos 3 α >, v = < cos 6β, sin 6β >. Which of the following is the value of u · v?
a.
sin (3α + 6β)
b.
cos (3α - 6β)
c.
sin (3α - 6β)
d.
cos (3α + 6β)


Solution:

u = < sin 3 α, cos 3 α >, v = < cos 6β, sin 6β > are the two vectors.

u · v = < sin 3 α, cos 3 α> · < cos 6β, sin 6β >

= (sin 3 α)(cos 6β) + (cos 3 α)(sin 6β)
[Use the definition of dot product.]

= sin (3 α + 6β)
[Use sin A cos B + cos A sin B = sin(A + B).


Correct answer : (1)
 7.  
Let u, v be two vectors. Which of the following is the value of |(3u + 6v)|2?
a.
|3u|2 + |6v|2 + 2(3u · 6v)
b.
|3u|2 - |6v|2
c.
|3u|2 + |6v|2 + 2|3u| |6v|
d.
|3u|2 + |6v|2


Solution:

u, v are the two vectors.

|(3u + 6v)|2 = (3u + 6v) · (3u + 6v)
[For any vector x, |x|2 = x · x.]

= (3u + 6v) · 3u + (3u + 6v) · 6v
[For any three vectors x, y, z, x · (y + z) = x · y + x · z.]

= 3u · 3u + 6v · 3u + 3u · 6v + 6v · 6v

= |3u|2 + 3u · 6v + 3u · 6v + |6v|2
[For any vectors x, y, x · y = y · x.]

= |3u|2 + |6v|2 + 2(3u · 6v)


Correct answer : (1)
 8.  
Let u, v be two vectors. Which of the following is the value of (5u + 10v) · (5u - 10v)?
a.
|5u|2 - |10v|2
b.
|5u + 10v|2
c.
|5u - 10v|2
d.
|5u|2 + |10v|2


Solution:

u, v are the two vectors.

(5u + 10v) · (5u - 10v) = 5u · (5u - 10v) + 10v · (5u - 10v)
[For any three vectors x, y, z, (x + y) · z = x · z + y · z.]

= 5u · 5u - 5u · 10v + 10v · 5u - 10v · 10v

= |5u|2 - 5u · 10v + 5u · 10v - |10v|2
[For any vectors x, y, x · y = y · x.]

= |5u|2 - |10v|2
[For a vector x, x - x = 0.]


Correct answer : (1)
 9.  
Let u, v be two vectors. Which of the following is correct?
a.
- |u| |v| ≤ u · v < |u| |v|
b.
- |u| |v| ≤ u · v ≤ |u| |v|
c.
- |u| |v| ≥ u · v ≥ |u| |v|
d.
- |u| |v| < u · v < |u| |v|


Solution:

u, v are two vectors. Let θ be the angle between u and v.

u · v = |u| |v| cos θ
[Use the definition of dot product.]

-1 ≤ cos θ ≤ 1
[Range of cos θ.]

- |u| |v| ≤ |u| |v| cos θ ≤ |u| |v|
[Multiply the inequality by |u| |v|, which is positive.]

- |u| |v| ≤ u · v ≤ |u| |v|
[Use u · v = |u| |v| cos θ.]


Correct answer : (2)
 10.  
Let u, v be any two nonzero vectors. Which of the following is correct?
a.
u · v is maximum if the angle between u, v is 180°.
b.
u · v is minimum if the angle between u, v is 0°.
c.
u · v is independent of the angle between u, v.
d.
u · v is maximum if the angle between u, v is 0°.


Solution:

u, v are two nonzero vectors. Let θ be the angle between them.

u · v = |u| |v| cos θ
[Use the definition of dot product.]

If θ = 0°, u · v = |u| |v| cos 0° = |u| |v|

If θ = 180°, u · v = |u| |v| cos 180° = - |u| |v|

So, u · v is maximum if the angle between them is 0°.


Correct answer : (4)

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