MCQ
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
    $\sqrt{7}$
  • $\sqrt{2}$
  • C
    $2$
  • D
    $7$

Answer

Correct option: B.
$\sqrt{2}$
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 $\vec{b}=7$

$\Rightarrow \frac{\overrightarrow{ a } \cdot \overrightarrow{ b }}{|\overrightarrow{ b }|}=7$

$\frac{\lambda}{\sqrt{3}} \frac{(\hat{ i }+\hat{ j }+\hat{ k }) \cdot(3 \hat{ i }+4 \hat{ j })}{5}=7$

$\lambda=5 \sqrt{3}$

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

now $\overrightarrow{ b }=5 \alpha(\hat{ i }+\hat{ j }+\hat{ k })+\beta(\hat{ i })$

$\overrightarrow{ a } \cdot \overrightarrow{ b }=0$

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

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

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

$|\vec{b}|=5 \sqrt{3}$

$\Rightarrow \alpha=\pm \frac{1}{\sqrt{2}}$

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

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