Q 42.

Question

In Exercises 37–42, find $||v||$ and find the unit vector in the direction of v.

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

Step-by-Step Solution

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Answer

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

1Step 1. Given information.

The given vector is:

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

2Step 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.

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

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

Q 41.

Q 43.

Other exercises in this chapter

Q 40.

In Exercises 37–42, find v and find the unit vector in the direction of v.v=2,1,-5

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Q 41.

In Exercises 37–42, find v and find the unit vector in the direction of v.v=1,1,1

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Q 43.

Find a vector in the direction of3,1,2 and with magnitude 5.

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Q 44.

Find a vector in the direction of -1,2,3 with magnitude 3.

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