# Suppose a vector \vec v in \mathbb{R}^3 is given by \vec v = \langle 3,2,-2 \rangle. FInd a...

## Question:

Suppose a vector {eq}\vec v {/eq} in {eq}\mathbb{R}^3 {/eq} is given by {eq}\vec v = \langle 3,2,-2 \rangle {/eq}. Find a vector {eq}\vec w {/eq} that is parallel to (and in the same direction as) {eq}\vec v {/eq} with magnitude {eq}6. {/eq}

## The Magnitude of a Vector:

If {eq}\vec{v}=\left<a,b,c\right> {/eq} is a vector over {eq}\mathbb{R}^3 {/eq}, then the magnitude of {eq}\vec{v} {/eq} is the real number

{eq}\|\vec{v}\|=\sqrt{a^2+b^2+c^2} \, . {/eq}

Geometrically, the magnitude of {eq}\vec{v} {/eq} can be interpreted as the length of {eq}\vec{v} {/eq}. If {eq}c {/eq} is any real number, then

{eq}\|c\vec{v}\|=|c|\|\vec{v}\| \, . {/eq}

We'll start by computing the magnitude of {eq}\vec{v}=\left<3,2,-2\right> {/eq}:

{eq}\begin{align*} \|\vec{v}\|&=\sqrt{3^2+2^2+(-2)^2}\\ &=\sqrt{9+4+4}\\ &=\sqrt{17} \, . \end{align*} {/eq}

So if we want to find a vector which has magnitude 6 and points in the same direction as {eq}\vec{v} {/eq}, we need to multiply {eq}\vec{v} {/eq} by {eq}\dfrac{6}{\sqrt{17}} {/eq}. That is, the vector we're looking for is:

{eq}\begin{align*} \frac{6}{\sqrt{17}}\vec{v}&=\frac{6}{\sqrt{17}}\left<3,2,-2\right>\\ &=\left<\frac{18}{\sqrt{17}},\frac{12}{\sqrt{17}},-\frac{12}{\sqrt{17}}\right> \, . \end{align*} {/eq}

In summary, the vector {eq}\boxed{\vec{w}=\left<\frac{18}{\sqrt{17}},\frac{12}{\sqrt{17}},-\frac{12}{\sqrt{17}}\right>} {/eq} points in the same direction as {eq}\vec{v} {/eq} and has magnitude {eq}6 {/eq}.