Q 44.

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

Verified

Answer

The required vector is $\frac{3}{\sqrt{14}} <- 1, 2, 3>$ .

1Step 1. Given information.

The given vector is:

$v = <- 1, 2, 3>$ with magnitude 3.

2Step 2. Find the norm of the vector.

$v = <- 1, 2, 3> ||v|| = \sqrt{{(- 1)}^{2} + 2^{2} + 3^{2}} = \sqrt{1 + 4 + 9} = \sqrt{14}$

Therefore, the norm of the given vector is $\sqrt{14}$ .

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.

The required vector is:

$\frac{a}{||v||} v = \frac{3}{\sqrt{14}} <- 1, 2, 3>$

Therefore, the vector in the direction of $v = <- 1, 2, 3>$ and with a magnitude of 3 is $\frac{3}{\sqrt{14}} <- 1, 2, 3>$ .