Mathematics - Vector Question with Solution | TestHub

Projection of $\overrightarrow{a} along \overrightarrow{b}=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}·\hat{b}$

$\overrightarrow{u}=\left\{\left(\hat{i}+\hat{j}\right)·\frac{a\hat{i}+b\hat{j}}{\sqrt{a^2+b^2}}\right\}\frac{a\hat{i}+b\hat{j}}{\sqrt{a^2+b^2}}=\frac{\left(a+b\right)}{\sqrt{a^2+b^2}}·\frac{a\hat{i}+b\hat{j}}{\sqrt{a^2+b^2}}$

$\overrightarrow{\mathrm{v}}=\left\{\left(\hat{i}+\hat{j}\right)·\frac{a\hat{i}-b\hat{j}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right\}\frac{a\hat{i}-b\hat{j}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=\frac{\left(\mathrm{a}-\mathrm{b}\right)}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}·\frac{a\hat{i}-b\hat{j}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$

given $\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ \mathrm{a}& \mathrm{b}& 0\\ \mathrm{a}& -\mathrm{b}& 0\end{array}\right|=8$

$\left|\hat{k}\left(-ab-ab\right)\right|=8$

$ab=4$

Given $\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|=\overrightarrow{w}$

$\frac{a+b}{\sqrt{a^2+b^2}}\frac{\left|a-b\right|}{\sqrt{a^2+b^2}}=\sqrt{2}$

For $a>b$,

$2a=\sqrt{2}\sqrt{a^2+b^2}$

$2a^2=a^2+b^2$

${\mathrm{a}}^2={\mathrm{b}}^2\Rightarrow \mathrm{a}=\mathrm{b}=2$

Similarly, for $b>a, a=b=2$

$a+b=4$

$a-b=0$

Length of diagonal $\overrightarrow{\left|PR\right|}=\left|\overrightarrow{\mathrm{PS}}+\overrightarrow{\mathrm{PQ}}\right|=\left|2\mathrm{a}\hat{\mathrm{i}}\right|=2a=4$.