# Find unit vectors that satisfy the given conditions: a. The unit vector in R^3 oppositely...

## Question:

Find unit vectors that satisfy the given conditions:

a. The unit vector in {eq}R^3 {/eq} oppositely directed to -j + k is_____. (You can represent that vector in any form - either as a list of components enclosed in angular brackets or as a linear combination of the standard basis vectors i, j, k.)

b. The unit vector in {eq}R^2 {/eq} that has the same direction as the vector from the point A=(-4, -2) to the point B= (-4, 2) is _____

## Vectors:

It is important to keep in mind that vectors are not fixed at any particular location: they are simply directed line segments. When we add them, we add components. Two vectors are parallel if they are scalar multiples of each other since all a scalar multiple does is affect the length of a vector (it can't possible change its direction).

## Answer and Explanation:

**Part A**

Bracket notation is preferable as it makes explicit all ones and zeroes and ensures that everything is in proper order. So we have

{eq}\begin{align*} \vec v &= - \hat j + \hat k = \left< 0, -1, 1 \right> \end{align*} {/eq}

We want the vector that zeroes this out:

{eq}\begin{align*} \vec v + \vec u &= 0 \\ \left< 0, -1, 1 \right> + \vec u &= 0 \\ \end{align*} {/eq}

And so we must have

{eq}\begin{align*} \vec u &= \left< 0, 1, -1 \right> \end{align*} {/eq}

All we really need to do is flip the sign to get the opposite directed vector. It has magnitude

{eq}\begin{align*} | \vec u | &= \sqrt{\left( 0 \right)^2+\left( 1 \right)^2+\left( -1 \right)^2} \\ &= \sqrt2 \end{align*} {/eq}

And so the vector we seek is

{eq}\begin{align*} \hat u &= \frac{\vec u }{| \vec u |} \\ &= \frac{\left< 0, 1, -1 \right>}{\sqrt2} \end{align*} {/eq}

**Part B**

Next, what we are really looking for is the unit vector that points from A to B. The resultant vector, i.e. direction vector, from A to B is

{eq}\begin{align*} \vec u &= \left< -4,2 \right> - \left< -4, -2 \right> \\ &= \left< 0, 4 \right> \end{align*} {/eq}

This obviously has magnitude 4, so we have

{eq}\begin{align*} \vec u &= \left< 0,1 \right> \end{align*} {/eq}

This is clearly true since to get from point A to point B all we have to do is head in the positive {eq}y {/eq} direction.

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from Common Entrance Test (CET): Study Guide & Syllabus

Chapter 57 / Lesson 3