Question

Find a vector of magnitude 3 in the direction of v=4i-3k

Answer 1

Answer:

$\frac{12i}{5} - \frac{9k}{5}$

Step-by-step explanation:

The magnitude of a vector v = ai + bk is

$|v| = \sqrt{a^{2} + b^{2}$

In this problem, we have that:

Find a vector of magnitude 3 in the direction of v=4i-3k:

The first step is finding the unit vector of v, $v_{u}$.

So

$|v| = \sqrt{4^{2} + (-3)^{2}} = 5$

$v_{u} = \frac{4i}{5} - \frac{3k}{5}$

Magnitude 3, same direction.

We multiply the unit vector by +3, since it is in the same direction. If it was in the opposite direction, we would have multiplied by -3.

The answer is:

$\frac{12i}{5} - \frac{9k}{5}$