Chapter 10: Q 42. (page 801)

In Exercises 37–42, find $||v||$ and find the unit vector in the direction of v.
$v = <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$

Short Answer

Expert verified

$||v|| = 1$ and the unit vector in the direction of vis $<\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$ .

Step by step solution

Step 1. Given information.

The given vector is:

$v = <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$

Step 2. Find the norm of the vector.

$v = <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}> ||v|| = \sqrt{{(\frac{\sqrt{2}}{2})}^{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2}} = \sqrt{\frac{2}{4} + \frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{4}{4}} = 1$

Therefore, the norm of the given vector is 1.

Step 3. Find the unit vector in the direction of v.

The unit vector in the direction of vis the vector:

$\frac{1}{||v||} v = \frac{1}{1} <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}> = <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$

Therefore, the unit vector in the direction of $v = <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$ isrole="math" localid="1649497503883" $<\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$ .

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In Exercises 37–42, find $||v||$ and find the unit vector in the direction of v.
$v = <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the norm of the vector.

Step 3. Find the unit vector in the direction of v.

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In Exercises 37–42, find v and find the unit vector in the direction of v.v=22,12,12

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the norm of the vector.

Step 3. Find the unit vector in the direction of v.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Geometry

Probability and Statistics

Statistics

Applied Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

In Exercises 37–42, find $||v||$ and find the unit vector in the direction of v.
$v = <\frac{\sqrt{2}}{2}, \frac{1}{2}, \frac{1}{2}>$