91Ó°ÊÓ

Two vectors$\mathit{A}$ and$\mathit{B}$ are parallel if and only if they are scalar multiples of one another. $\mathit{A}\mathbf{=}\mathit{k}\mathit{B}$, $\mathit{k}$ is a constant not equal to zero. Two vectors$\mathit{A}$and$\mathit{B}$ are perpendicular if and only if their scalar product is equal to zero.

If two vectors are perpendicular, then $\cos \mathrm{θ}=0$.

$A_x{B}_x+A_y{B}_y+A_z{B}_z=0$if $A$and $B$are perpendicular vectors.

Consider the given vector:

$A=2\hat{i}-\hat{j}+2\hat{k}$

To find a unit vector in the direction of a given vector, divide the vector by its magnitude. So let this unit vector be $a$.

$a=\frac{A}{\left|A\right|}$

role="math" localid="1662643303758" $\begin{array}{rcl}a& =& \frac{2\hat{i}-\hat{j}+2\hat{k}}{\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2+{\left(2\right)}^2}}\\ & =& \frac{1}{3}\left(2\hat{i}-\hat{j}+2\hat{k}\right)\end{array}$

$a=\frac{1}{3}A$

Let this vector be $B$, hence

$\begin{array}{rcl}B& =& 12a\\ & =& 12\left(\frac{1}{3}\left(2\hat{i}-\hat{j}+2\hat{k}\right)\right)\\ & =& 4A\end{array}$

Define a vector $C=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$that is perpendicular on $A$.

$\begin{array}{rcl}A×C& =& 0\\ \left(2\hat{i}-\hat{j}+2\hat{k}\right)×\left(c_1\hat{i}+c_2\hat{j}+c_3\hat{k}\right)& =& 0\\ 2c_1-c_2+2c_3& =& 0\end{array}$.

A possible choice is to take

$c_2=0,c_1=-c_3=1$

Hence,

$C=\hat{i}-\hat{k}$

Vector $C$is perpendicular to $A$, so to find a unit vector, divide it by its magnitude.

So let this unit vector be $c$.

$\begin{array}{rcl}c& =& \frac{C}{\left|C\right|}\\ & =& \frac{\hat{i}-\hat{k}}{\sqrt{1^2+{\left(-1\right)}^2}}\\ & =& \frac{1}{\sqrt{2}}\left(\hat{i}-\hat{k}\right)\\ & =& \frac{1}{\sqrt{2}}C\end{array}$