Mathematics - Vector Question with Solution | TestHub

Given $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$

Since the vector makes equal angles with the coordinate axes$OX,OY$and$OZ$, so

$\overrightarrow{\mathit{a}}=\lambda \left(\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\right)=\frac{\lambda }{\sqrt{3}}\left(\hat{i}+\hat{j}+\hat{k}\right)$

Now projection of $\overrightarrow{a}$on $\overrightarrow{\mathit{b}}=7$

$\Rightarrow \frac{\lambda }{\sqrt{3}}\frac{\left(\hat{i}+\hat{j}+\hat{k}\right)·\left(3\hat{i}+4\hat{j}\right)}{5}=7$

$\Rightarrow \lambda =5\sqrt{3}$

So $\overrightarrow{a}=5\left(\hat{\hat{i}}+\hat{j}+\hat{k}\right)$

Let $\overrightarrow{b}=5\alpha \left(\hat{i}+\hat{j}+\hat{\mathit{k}}\right)+\beta \left(\hat{i}\right)$

Since both vectors are perpendicular $\overrightarrow{\mathit{a}}·\overrightarrow{b}=0$

$\Rightarrow 25\alpha \left(3\right)+5\beta =0$

$\Rightarrow 15\alpha +\beta =0\Rightarrow \beta =-15\alpha$

i.e. $\overrightarrow{b}=5\alpha \left(-2\hat{i}+\hat{j}+\hat{k}\right)$

$\left|\overrightarrow{b}\right|=5\sqrt{3}$

$\overrightarrow{b}=\pm \frac{5}{\sqrt{2}}\left(-2\hat{i}+\hat{j}+\hat{k}\right)$

Projection of $\overrightarrow{\mathit{b}}$ on $3\hat{i}+4\hat{j}$ is

$\frac{\overrightarrow{b}·\left(3\hat{i}+4\hat{j}\right)}{5}=\pm \frac{5}{\sqrt{2}}\left(\frac{-6+4}{5}\right)=\pm \sqrt{2}$