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

Vectors are very important concepts of mathematics that have many applications, which include calculating trajectories of rockets and making free body diagrams of various objects.

Answer: The vector of magnitude 3 in the direction of v = (4i - 3k) is 12i/5 - 9k/5.

Let's understand the solution in detail.

Explanation:

We have the vector v = (4i - 3k)

First, we have to find the unit vector of v, using the equation u =v / |v| , where |v| is the magnitude of v.

Hence, |v| = √[4² + (-3)²] = 5

Now, we have unit vector u = 4i/5 - 3k/5.

Now, to find a vector of the same direction with a magnitude of 3, we multiply the unit vector with 3.

Hence, the required vector = 3 (4i/5 - 3k/5) = 12i/5 - 9k/5.

Therefore, the vector of magnitude 3 in the direction of v = (4i - 3k) is 12i/5 - 9k/5.