Chapter 9: Q. 55 (page 631)

Find the direction angles of the vector $v = 3 i - 4 j + 2 k .$

Short Answer

Expert verified

The direction angles of the vector are $α \approx 56.1 °, β \approx 138 °, a n d γ \approx 68.2 ° .$

Step by step solution

Step 1. Given Information

The given vector is $v = 3 i - 4 j + 2 k .$

Step 2. Finding the direction angles

To find the direction angles $α, β, a n d γ$ use the formula

$\cos α = \frac{a}{||v||}, \cos β = \frac{b}{||v||}, \cos γ = \frac{c}{||v||} w h e r e ||v|| = \sqrt{a^{2} + b^{2} + c^{2}}$

Let's first find $||v||$

So, $||v|| = \sqrt{3^{2} + {(- 4)}^{2} + 2^{2}} ||v|| = \sqrt{29}$

Step 3. Solve

Proceed with the calculation further by using the formula.

Thus,

$\cos α = \frac{3}{\sqrt{29}} Rationalize the denominator cosα = \frac{3 \sqrt{29}}{29} α = \cos^{- 1} (\frac{3 \sqrt{29}}{29}) α \approx 56.1 °$

And

$\cos β = \frac{- 4}{\sqrt{29}} Rationalize the denominator cosβ = - \frac{4 \sqrt{29}}{29} β = \cos^{- 1} (- \frac{4 \sqrt{29}}{29}) β \approx 138 °$

And

$\cos γ = \frac{2}{\sqrt{29}} Rationalize the denominator cosγ = \frac{2 \sqrt{29}}{29} γ = \cos^{- 1} (\frac{2 \sqrt{29}}{29}) γ \approx 68.2 °$

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Find the direction angles of the vector $v = 3 i - 4 j + 2 k .$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the direction angles

Step 3. Solve

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Find the direction angles of the vectorv=3i-4j+2k.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the direction angles

Step 3. Solve

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Theoretical and Mathematical Physics

Probability and Statistics

Decision Maths

Applied Mathematics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

Find the direction angles of the vector $v = 3 i - 4 j + 2 k .$