Scalar and Vector Examples

Example 1 : Find the vector of magnitude 5 which are perpendicular to the vectors $\vec{a}$ = $2\hat{i} + \hat{j} – 3\hat{k}$ and $\vec{b}$ = $\hat{i} – 2\hat{j} + \hat{k}$.

Solution : Unit vectors perpendicular to $\vec{a}$ & $\vec{b}$ = $\pm$$\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|$

$\therefore$   $\vec{a}\times\vec{b}$ = $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 2 & -2 \\ \end{vmatrix}$ = $-5\hat{i} – 5\hat{j} – 5\hat{k}$

$\therefore$   Unit Vectors = $\pm$ $-5\hat{i} – 5\hat{j} – 5\hat{k}\over 5\sqrt{3}$

Hence the required vectors are $\pm$ $5\sqrt{3}\over 3$($\hat{i} + \hat{j} + \hat{k}$)

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Example 2 : If $\vec{a}$, $\vec{b}$, $\vec{c}$ are three non zero vectors such that $\vec{a}\times\vec{b}$ = $\vec{c}$ and $\vec{b}\times\vec{c}$ = $\vec{a}$, prove that $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually at right angles and |$\vec{b}$| = 1 and |$\vec{c}$| = |$\vec{a}$|

Solution : $\vec{a}\times\vec{b}$ = $\vec{c}$ and $\vec{b}\times\vec{c}$ = $\vec{a}$

$\implies$   $\vec{c}\perp\vec{a}$ , $\vec{c}\perp\vec{b}$ and $\vec{a}\perp\vec{b}$, $\vec{a}\perp\vec{c}$

$\implies$   $\vec{a}\perp\vec{b}$, $\vec{b}\perp\vec{c}$ and $\vec{c}\perp\vec{a}$

$\implies$   $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular vectors.

Again, $\vec{a}\times\vec{b}$ = $\vec{c}$ and $\vec{b}\times\vec{c}$ = $\vec{a}$

$\implies$   |$\vec{a}\times\vec{b}$| = |$\vec{c}$| and |$\vec{b}\times\vec{c}$| = |$\vec{a}$|

$\implies$   $|\vec{a}||\vec{b}|sin{\pi\over 2}$ = |$\vec{c}$| and $|\vec{b}||\vec{c}|sin{\pi\over 2}$ = |$\vec{a}$|     ($\because$ $\vec{a}\perp\vec{b}$ and $\vec{b}\perp\vec{c}$)

$\implies$   $|\vec{a}||\vec{b}|$ = |$\vec{c}$| and $|\vec{b}||\vec{c}|$ = |$\vec{a}$|

$\implies$   ${|\vec{b}|}^2$ |$\vec{c}$| = |$\vec{c}$|

$\implies$   ${|\vec{b}|}^2$ = 1

$\implies$   $|\vec{b}|$ = 1

putting in $|\vec{a}||\vec{b}|$ = |$\vec{c}$|

$\implies$   $|\vec{a}|$ = |$\vec{c}$|

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Example 3 : For any three vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ prove that [$\vec{a}$ + $\vec{b}$ $\vec{b}$ + $\vec{c}$ $\vec{c}$ + $\vec{a}$] = 2[$\vec{a}$ $\vec{b}$ $\vec{c}$]

Solution : We have [$\vec{a}$ + $\vec{b}$ $\vec{b}$ + $\vec{c}$ $\vec{c}$ + $\vec{a}$]

= {($\vec{a}$ + $\vec{b}$)$\times$($\vec{b}$ + $\vec{c}$)}.($\vec{c}$ + $\vec{a}$)

= {$\vec{a}$$\times$$\vec{b}$ + $\vec{a}$$\times$$\vec{c}$ + $\vec{b}$$\times$$\vec{b}$ + $\vec{b}$$\times$$\vec{c}$}.($\vec{c}$ + $\vec{a}$)       {$\vec{b}$$\times$$\vec{b}$ = 0}

= {$\vec{a}$$\times$$\vec{b}$ + $\vec{a}$$\times$$\vec{c}$ + $\vec{b}$$\times$$\vec{c}$}.($\vec{c}$ + $\vec{a}$)

= ($\vec{a}\times\vec{b}$).$\vec{c}$ + ($\vec{a}\times\vec{c}$).$\vec{c}$ + ($\vec{b}\times\vec{c}$).$\vec{c}$ + ($\vec{a}\times\vec{b}$).$\vec{a}$ + ($\vec{a}\times\vec{c}$).$\vec{a}$ + ($\vec{b}\times\vec{c}$).$\vec{a}$

= [$\vec{a}$ $\vec{b}$ $\vec{c}$] + 0 + 0 + 0 + 0 + [$\vec{b}$ $\vec{c}$ $\vec{a}$]

= [$\vec{a}$ $\vec{b}$ $\vec{c}$] + [$\vec{a}$ $\vec{b}$ $\vec{c}$] = 2[$\vec{a}$ $\vec{b}$ $\vec{c}$]

Practice these given scalar and vector examples to test your knowledge on concepts of scalar and vectors.