# Find vectors that satisfy the given conditions: 1. The vector in the opposite direction to \vec u...

## Question:

Find vectors that satisfy the given conditions:

1. The vector in the opposite direction to {eq}\vec u = \langle -1,-5 \rangle {/eq} and determine the half of its length.

2. What is the vector of length {eq}2 {/eq} and in the same direction as {eq}\vec v = \langle -1,-4,-4, \rangle {/eq}

## Vector:

Let {eq}\vec v {/eq} be any vector.

Then the vector that is in the same direction of {eq}\vec v {/eq} and having length {eq}m {/eq} is obtained by using the formula {eq}\ \displaystyle m\frac{\vec v}{\left \| \vec v \right \|}. {/eq}

The vector that is opposite in the direction of {eq}\vec v {/eq} and having length {eq}n {/eq} is obtained by the formula {eq}\ \displaystyle -n\frac{\vec v}{\left \| \vec v \right \|}. {/eq}

1. Let {eq}\vec v {/eq} be the vector that lies in the opposite direction of {eq}\vec u = \langle -1,-5 \rangle {/eq} and has half-length of {eq}\vec u. {/eq}

To find {eq}\vec v, {/eq} first we find the unit vector that is opposite to {eq}\vec u {/eq} and multiply the unit vector with the half of length of {eq}\vec u. {/eq}

Length of {eq}\vec u {/eq} is {eq}\left \| \vec u \right \|=\sqrt{(-1)^2+(-5)^2}=\sqrt{26}. {/eq}

Now {eq}\displaystyle \frac{1}{2}\left \| \vec u \right \|=\frac{\sqrt{26}}{2}. {/eq}

The unit vector that is opposite to {eq}\vec u = \langle -1,-5 \rangle {/eq} is:

{eq}\begin{align*} \frac{-\vec u}{\left \| \vec u \right \|} &= \frac{-\langle -1,-5 \rangle}{\sqrt{26}}. \end{align*} {/eq}

Then the required vector is:

{eq}\begin{align*} \vec v &= \frac{-\langle -1,-5 \rangle}{\sqrt{26}}*\frac{\sqrt{26}}{2}\\ \\ &=\frac{\langle 1,5 \rangle}{2}\\ \\ &=\left \langle \frac{1}{2},\frac{5}{2} \right \rangle. \end{align*} {/eq}

2. Let {eq}\vec w {/eq} be the vector that lies in the same direction of {eq}\vec v = \langle -1,-4,-4 \rangle {/eq} and has length {eq}2. {/eq}

To find {eq}\vec w, {/eq} we find the unit vectorin the direction of {eq}\vec v {/eq} and multiply the unit vector with {eq}2. {/eq}

Now the unit vector in the direction of {eq}\vec v {/eq} is:

{eq}\begin{align*} \frac{\vec v}{\left \| \vec v \right \|} &=\frac{\langle -1,-4,-4 \rangle}{\left \| \langle -1,-4,-4 \rangle \right \|}\\ \\ &=\frac{\langle -1,-4,-4 \rangle}{\sqrt{(-1)^2+(-4)^2+(-4)^2}}\\ \\ &=\frac{\langle -1,-4,-4 \rangle}{\sqrt{33}}. \end{align*} {/eq}

Then the required vector is:

{eq}\begin{align*} \vec w &=2*\frac{\langle -1,-4,-4 \rangle}{\sqrt{33}}\\ \\ &=\left \langle \frac{-2}{\sqrt{33}},\frac{-8}{\sqrt{33}},\frac{-8}{\sqrt{33}} \right \rangle. \end{align*} {/eq}