Question

If $\left|\overrightarrow{a}\right|=aand\left|\overrightarrow{b}\right|=b,$ prove that ${\left(\frac{\overrightarrow{a}}{a^2}-\frac{\overrightarrow{b}}{b^2}\right)}^2={\left(\frac{\overrightarrow{a}-\overrightarrow{b}}{ab}\right)}^2.$

Answer 1

$$\begin{gathered} {\left(\frac{\overrightarrow{a}}{a^2}-\frac{\overrightarrow{b}}{b^2}\right)}^2 \\ ={\left|\frac{\overrightarrow{a}}{a^2}\right|}^2+{\left|\frac{\overrightarrow{b}}{b^2}\right|}^2-\frac{2\overrightarrow{a}.\overrightarrow{b}}{a^2{b}^2} \\ =\frac{{\left|\overrightarrow{a}\right|}^2}{a^4}+\frac{{\left|\overrightarrow{b}\right|}^2}{b^4}-\frac{2\overrightarrow{a}.\overrightarrow{b}}{a^2{b}^2} \\ =\frac{a^2}{a^4}+\frac{b^2}{b^4}-\frac{2\overrightarrow{a}.\overrightarrow{b}}{a^2{b}^2}\text{(From the given information)} \\ =\frac{1}{a^2}+\frac{1}{b^2}-\frac{2\overrightarrow{a}.\overrightarrow{b}}{a^2{b}^2} \\ =\frac{b^2+a^2-2\overrightarrow{a}.\overrightarrow{b}}{a^2{b}^2} \\ =\frac{a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}}{a^2{b}^2} \\ =\frac{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{a}\right|}^2-2\overrightarrow{a}.\overrightarrow{b}}{a^2{b}^2}\text{(From the given information)} \\ =\frac{{\left(\overrightarrow{a}-\overrightarrow{b}\right)}^2}{a^2{b}^2} \\ ={\left(\frac{\overrightarrow{a}-\overrightarrow{b}}{ab}\right)}^2 \end{gathered}$$