Q 38.

Question

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

$v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$

Step-by-Step Solution

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Answer

$||v|| = 1$ and the unit vector in the direction of v is $<\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$ .

1Step 1. Given information.

The given vector is:

$v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$

2Step 2. Find the norm of the vector.

$v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}> ||v|| = \sqrt{{(\sqrt{\frac{1}{3}})}^{2} + {(- \sqrt{\frac{2}{3}})}^{2}} = \sqrt{\frac{1}{3} + \frac{2}{3}} = \sqrt{\frac{3}{3}} = 1$

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:

$\frac{1}{||v||} v = \frac{1}{1} <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}> = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$

Therefore, the unit vector in the direction of $v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$ is $<\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$ .

Q 37,

Q 39.

Other exercises in this chapter

Q 36.

Find the norm of the vector. v = 0, 0, 6

View solution

Q 37,

Find v and find the unit vector in the direction of v.v = 3, −4

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Q 39.

In Exercises 37–42, find v and find the unit vector in the direction of v.v=15,13

View solution

Q 40.

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

View solution