Question

Let $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} \quad a_{i}>0, i=1,2,3$ 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
$\frac{\overrightarrow{ b } \cdot(3 \hat{ i }+4 \hat{ j })}{5}=\pm \frac{5}{\sqrt{2}}\left(\frac{-6+4}{5}\right)=\pm \sqrt{2}$