Q 48.

Question

Find a vector in the direction opposite to $<- 4, 5, - 1>$ and with magnitude 3.

Step-by-Step Solution

Verified

Answer

The required vector is $\frac{3}{\sqrt{42}} <- 4, 5, - 1>$ .

1Step 1. Given information.

The given vector is:

$v = <- 4, 5, - 1>$ with magnitude 3.

2Step 2. Find the norm of the vector.

$v = <- 4, 5, - 1> ||v|| = \sqrt{{(- 4)}^{2} + 5^{2} + {(- 1)}^{2}} = \sqrt{16 + 25 + 1} = \sqrt{42}$

Therefore, the norm of the given vector is $\sqrt{42}$ .

3Step 3. Find the vector.

We know that the vector in the direction of v is $\frac{a}{||v||} (- v)$ . Here a is the magnitude of the given vector and the negative sign shows the direction opposite to $v = <- 4, 5, - 1>$ .

The required vector is:

$\frac{a}{||v||} v = \frac{3}{\sqrt{42}} (- <- 4, 5, - 1>) = \frac{3}{\sqrt{42}} <4, - 5, 1>$

Therefore the vector in the direction opposite to $v = <- 4, 5, - 1>$ and with a magnitude of is $\frac{3}{\sqrt{42}} <4, - 5, 1>$ .

Q 47.

Q 49.

Other exercises in this chapter

Q 46.

Find a vector in the direction of 9,0,-6 and with magnitude 7.

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Q 47.

Find a vector in the direction opposite to-1,-4,-6and with magnitude 7.

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Q 49.

Find a vector in the direction opposite to 0,-3,4 and with magnitude 10.

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Q 50.

Find a unit vector in the direction opposite to 3,4,5.

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