Question

Is the vector$\left(\overbrace{i}+\overbrace{j}+\overbrace{k}\right)$ a unit vector? Is the vector$\left(3.0\overbrace{i}-2.0\overbrace{j}\right)$ a unit vector? Justify your answers.

Short answer

The vectors $\left(\overbrace{\mathrm{i}}+\overbrace{\mathrm{j}}+\overbrace{\mathrm{k}}\right)$ and$\left(3.0\overbrace{\mathrm{i}}-2.0\overbrace{\mathrm{j}}\right)$ are not a unit vector.

Step-by-step solution

Introduction

Unit vectors are the vectors whose magnitude is exactly 1 unit. They are very useful for various reasons. Specifically, the unit vectors $\left[0,1\right]$and $[1,0]$ can together form any other vector.

Explanation

Unit Vector – A unit vector is a vector with magnitude 1.

Let the vector $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$.

Magnitude:

$\begin{array}{rcl}\left|\stackrel{\rightharpoonup }{\mathrm{a}}\right|& =& \sqrt{1^2+1^2+1^2}\\ & =& \sqrt{3}\end{array}$

Since magnitude is not 1, $\stackrel{\rightharpoonup }{a}$is not a unit vector.

Likewise, assume the vector $\overrightarrow{\mathrm{b}}=3.0\hat{\mathrm{i}}-2.0\hat{\mathrm{j}}$.

The amplitude of the vector is

$\begin{array}{rcl}\left|\overrightarrow{\mathrm{b}}\right|& =& \sqrt{3^2+2^2}\\ & =& \sqrt{9+4}\\ & =& \sqrt{13}\end{array}$

Which is not equal to 1.

Hence, it is not a unit vector.