Q 21.

Question

In Exercises 20-23, find the dot product of the given pairs of vectors and the angle between the two vectors.

$u = <2, 0, - 5>, v = <- 3, 7, - 1>$

Step-by-Step Solution

Verified

Answer

The dot product is $- 1$ and the angle is $\cos^{- 1} (\frac{- 1}{\sqrt{1711}})$ .

1Step 1. Given information.

The given pairs of vectors are:

$u = <2, 0, - 5> and v = <- 3, 7, - 1>$

2Step 2. Find the dot product.

The dot product is: $u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}$

$u = <2, 0, - 5> and v = <- 3, 7, - 1> = 2 (- 3) + 0 (7) + (- 5) (- 1) = - 6 + 0 + 5 = - 1$

Therefore, the dot product of the given two vectors $u = <2, 0, - 5> and v = <- 3, 7, - 1>$ is $- 1$ .

3Step 3. Find the angle between the two vectors.

The formula for the angle between the two vectors is:

$\cos θ = \frac{u \cdot v}{||u|| ||v||}$

$u = <2, 0, - 5> and v = <- 3, 7, - 1> u \cdot v = - 1 ||u|| = \sqrt{2^{2} + 0^{2} + {(- 5)}^{2}} \sqrt{4 + 0 + 25} = \sqrt{29} ||v|| = \sqrt{{(- 3)}^{2} + 7^{2} + {(- 1)}^{2}} \sqrt{9 + 49 + 1} = \sqrt{59}$

Then,

$\cos θ = \frac{u \cdot v}{||u|| ||v||} \cos θ = \frac{- 1}{\sqrt{29} \sqrt{59}} θ = \cos^{- 1} (\frac{- 1}{\sqrt{1711}})$

Therefore, the angle is $\cos^{- 1} (\frac{- 1}{\sqrt{1711}})$ .

Q 20.

Q 22.

Other exercises in this chapter

Q .18.

Let v1 and v2 be two nonzero position vectors in ℝ2 that are not scalar multiples of each other. Explain why, given any vector w in &#

View solution

Q 20.

In Exercises 20-23, find the dot product of the given pairs of vectors and the angle between the two vectors.u=1,2, v=3,5

View solution

Q 22.

In Exercises 20-23, find the dot product of the given pairs of vectors and the angle between the two vectors.u=3,-1,2, v=-4,-6,3

View solution

Q 23.

In Exercises 20-23, find the dot product of the given pairs of vectors and the angle between the two vectors.u=-5,1,3, v=-3,2,7

View solution