10DQ - University Physics with Modern Physics

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.

Step-by-Step Solution

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 1: 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.

Step 2: 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.