f8659e…: “As soon as a summer mountain”

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Suppose that $\vec {u} = \vector {-1,2,2}$ and $\vec {v} = \vector {-2,0,1}$ . Find a vector $\vec {w}$ of magnitude $4$ that is parallel to $2\vec {u}-3\vec {v}$ . $\vec {w} = \vector {\answer {\frac {16}{\sqrt {33}} }, \answer { \frac {16}{\sqrt {33}} } , \answer { \frac {4}{\sqrt {33}} }}$

To begin, let’s find a vector in the same direction as $2\vec {u}-3\vec {v}$ . Using the rules of addition and scalar multiplication, we find:

$2\vec {u}-3\vec {v} = \vector {\answer {4} , \answer {4} , \answer {1}}$

How should we proceed?

Multiply this result by $4$ ; that is, $\vec {w} = \vector {16 , 16, 4}$ . Find the magnitude of $\vector {4, 4, 1}$ and scale it appropriately if necessary.

We compute:

$|2\vec {u}-3\vec {v}| = \sqrt {\left ( \answer {4} \right )^2+\left ( \answer {4} \right )^2+\left ( \answer {1}\right )^2 } = \sqrt {\answer {33}}$ (type the components in the order of $2\vec {u}-3\vec {v}$ )

A unit vector in the direction of $\vec {w}$ is thus $\frac {2\vec {u}-3\vec {v}}{\left |2\vec {u}-3\vec {v}\right |}$ , so:

$\uvec {w} = \vector {\answer {\frac {4}{\sqrt {33}} }, \answer { \frac {4}{\sqrt {33}} } , \answer { \frac {1}{\sqrt {33}} }}$

and $\vec {w} = \answer {4} \uvec {w}$ .