Question

Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}{a}_i>0,i=1,2,3$ be a vector which makes equal angles with the coordinates axes $OX,OY$ and $OZ$. Also, let the projection of $\vec{a}$ on the vector $3\hat{i}+4\hat{j}$ be $7$ . Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ with $90^{\circ }$. If $\vec{a},\vec{b}$ and $x$-axis are coplanar, then projection of a vector $\vec{b}$ on $3\hat{i}+4\hat{j}$ is equal to

• A. $\sqrt{7}$
• B. $\sqrt{2}$
• C. $2$
• D. $7$

Answer 1

Answer: (B)

$\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$
$\overrightarrow{a}=\lambda \left(\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\right)=\frac{\lambda }{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
Now projection of $\vec{a}$ on $\vec{b}=7$
$\Rightarrow \frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{b}|}=7$
$\frac{\lambda }{\sqrt{3}}\frac{(\hat{i}+\hat{j}+\hat{k})\cdot (3\hat{i}+4\hat{j})}{5}=7$
$\lambda =5\sqrt{3}$
$\vec{a}=5(\hat{i}+\hat{j}+\hat{k})$
now $\vec{b}=5\alpha (\hat{i}+\hat{j}+\hat{k})+\beta (\hat{i})$
$\vec{a}\cdot \vec{b}=0$
$\Rightarrow 25\alpha (3)+5\beta =0$
$\Rightarrow 15\alpha +\beta =0\Rightarrow \beta =-15\alpha$
$\vec{b}=5\alpha (-2\hat{i}+\hat{j}+\hat{k})$
$|\vec{b}|=5\sqrt{3}$
$\Rightarrow \alpha =\pm \frac{1}{\sqrt{2}}$
$\vec{b}=\pm \frac{5}{\sqrt{2}}(-2\hat{i}+\hat{j}+\hat{k})$
Projection of $\vec{b}$ on $3\hat{i}+4\hat{j}$ is
$\frac{\overrightarrow{b}\cdot (3\hat{i}+4\hat{j})}{5}=\pm \frac{5}{\sqrt{2}}\left(\frac{-6+4}{5}\right)=\pm \sqrt{2}$