Question

A vector $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$ of length $2\sqrt{3}\text{units},$ which makes equal angles with the vectors $\overrightarrow{b}=y\hat{i}-2z\hat{j}+3x\hat{k}$ and $\overrightarrow{c}=2z\hat{i}+3x\hat{j}-y\hat{k}$ and is perpendicular to $\overrightarrow{d}=\hat{i}-\hat{j}+2\hat{k}$ and makes an obtuse angle with $y-$axis is

• A. $2(\hat{i}+\hat{j}+\hat{k})$
• B. $2(\hat{i}-\hat{j}-\hat{k})$
• C. $2(\hat{i}+\hat{j}+\hat{k})$
• D. $2(-\hat{i}+\hat{j}-\hat{k})$

Answer 1

The correct option is B $2(\hat{i}-\hat{j}-\hat{k})$

Given : $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$ is perpendicular to $\overrightarrow{d}=\hat{i}-\hat{j}+2\hat{k}$
$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{d}=0\Rightarrow x-y+2z=0\cdots \left(i\right)$
Moreover, $|\overrightarrow{b}|=|\overrightarrow{c}|$ so $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{a}\cdot \overrightarrow{c}$ as $\overrightarrow{a}$ makes equal angles with vectors $\overrightarrow{b}$ and $\overrightarrow{c}.$
Thus, $xz-2xy-yz=0\cdots \left(ii\right)$
Also $x^2+y^2+z^2=12\cdots \left(iii\right)$ and $y<0,$ substituting the value of $y$ from $\left(i\right)$ in $\left(ii\right)$
we get, $x^2+2xz+z^2=0$
So, $x=-z$ and $y=z.$
Again, substituting these value in $\left(iii\right)$
we get, $z^2=4,i.e.z=\pm 2$ but $y<0$ and $y=z$
So, $z=-2=y$ and $x=2$