Question

Im not sure how to answer this question.

u= 5i-2j+3k v= -2i+3k

Find a vector that is in the direction of the vector v , whose magnitude is twice that of vector u

Answer 1

9514 1404 393

Answer:

(-4√494)/13i +(6√494)/13k ≈ -6.8388i +0j +10.2585k

Explanation:

To answer this question, you need to know two things:

1) the direction of vector v

2) the magnitude of vector u

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direction of v

A direction is specified by a "unit vector", one with the proper ratio of components, and a magnitude of 1. It is found from a given vector by dividing that vector by its magnitude.

The unit vector in the v direction is ...

v/|v| = (-2i +3k)/(√((-2)² +3²) = (-2i +3k)/√13

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magnitude of u

The magnitude of vector u is ...

|u| = √(5² +(-2)² +3²) = √38

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Then the desired vector is ...

(2|u|)(v/|v|) = 2√38(-2i+3k)/√13 = (-4√494)/13i +(6√494)/13k

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Additional comment

We have chosen to "rationalize the denominator" by writing √(38/13) as (√494)/13.