Question

$\overline{a}$ & $\overline{b}$ are two non-collinear vectors then find the value of ${\left\{\frac{\overline{a}}{{\left|\overline{a}\right|}^2}\frac{\overline{b}}{{\left|\overline{b}\right|}^2}\right\}}^2$ is equal to

• A. $\frac{\overline{a}-\overline{b}}{ab}$
• B. ${\left(\frac{\overline{a}-\overline{b}}{ab}\right)}^3$
• C. ${\left(\frac{\overline{a}-\overline{b}}{ab}\right)}^2$
• D. ${(\overline{a}-\overline{b})}^2$

Answer 1

The correct option is C ${\left(\frac{\overline{a}-\overline{b}}{ab}\right)}^2$

$\begin{array}{cc}& {\{\frac{\overline{a}}{|\overline{a}{|}^2}-\frac{\overline{b}}{\mid \overline{b}{|}^2}\}}^2\\ =& \frac{|\overrightarrow{a}{|}^2}{|\overline{a}{|}^4}+\frac{|\overline{b}{|}^2}{|\overrightarrow{b}{|}^4}-\frac{2\overrightarrow{a}\cdot \overrightarrow{b}}{{|\overrightarrow{a}{|}^2|\overline{b}|}^2}\end{array}$

$=\frac{1}{|\overrightarrow{a}{|}^2}+\frac{1}{1\overline{b}{|}^2}-\frac{2\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}{|}^2|\overline{b}{|}^2}=\frac{|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-\dot{2}\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}{|}^2|{\overline{b}|}^2}$

$=\frac{(\overrightarrow{a}-\overrightarrow{b}{)}^2}{|\overline{a}{|}^2|\overline{b}{|}^2}={\left(\frac{\overrightarrow{a}-\overrightarrow{b}}{ab}\right)}^2$

option $C$ is correct.