Question

If $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$ makes equal angles with $\overrightarrow{b}=y\hat{i}-2z\hat{j}+3x\hat{k}$ and $\overrightarrow{c}=2z\hat{i}+3x\hat{j}-y\hat{k}$ and $\overrightarrow{a}\perp \overrightarrow{d}$, where $\overrightarrow{d}=\hat{i}-\hat{j}+2\hat{k}$.If $\left|\overrightarrow{a}\right|=2\sqrt{3}$, then $\overrightarrow{a}.\overrightarrow{b}=$

• A. $12$
• B. $-12$
• C. $24$
• D. $-24$

Answer 1

The correct option is D $-24$

${\left|\overrightarrow{b}\right|}^2={\left|\overrightarrow{c}\right|}^2=9x^2+y^2+4z^2$
$\angle (\overrightarrow{a}.\overrightarrow{b})=\angle (\overrightarrow{a}.\overrightarrow{c})$
$\Rightarrow \overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}$
$\Rightarrow xy-2yz+3zx=2zx+3xy-yz$
$\Rightarrow 2xy+yz-zx=0$ ...........$\left(1\right)$
$\overrightarrow{a}.\overrightarrow{d}=0$
$\Rightarrow x-y+2z=0$ ...........$\left(2\right)$
Eliminating $y$ from $\left(1\right),\left(2\right)$, we get
From $\left(1\right)2xy=zx-yz=(x-y)z$
From $\left(2\right)x-y=-2z$
$\Rightarrow 2xy=(-2z)z$
$\Rightarrow 2xy=-2z^2$
$\Rightarrow xy=-z^2$
$\Rightarrow xy=-z^2$ ...........$\left(3\right)$
Again we have From $\left(2\right)y=x+2z$
Substituting $y=x+2z$ in $\left(1\right)$ we get
$2xy+yz-zx=0$
$\Rightarrow 2x(x+2z)+(x+2z)z-zx=0$
$(x+2z)(2x+z)-zx=0$
$\Rightarrow 2x^2+4xz+zx+2z^2-zx=0$
$\Rightarrow 2x^2+4xz+2z^2-zx=0$
$\Rightarrow x^2+2xz+zx+z^2=0$
$\Rightarrow {(x+z)}^2=0$
$\Rightarrow z=-x$
From $\left(3\right)xy=-2z^2$
$\Rightarrow xy=-2\left(x^2\right)$
$\Rightarrow xy=-\left(x^2\right)$
$\therefore y=-x$
$\left|\overrightarrow{a}\right|=\sqrt{x^2+y^2+z^2}$
$=\sqrt{x^2+x^2+x^2}=x\sqrt{3}$
Given$\left|\overrightarrow{a}\right|=2\sqrt{3}=x\sqrt{3}$
$\Rightarrow x=2,y=-2,z=-2$
$x^2+y^2+z^2=2^2+2^2+2^2=12$
$\Rightarrow 3x^2=12$
$\Rightarrow x^2=4$
$\therefore x=\pm 2$
$\Rightarrow (x,y,z)=(2,-2,-2)=(-2,2,2)$
$\overrightarrow{a}.\overrightarrow{b}=xy-2yz+3zx$
$=-4-8-12=-24$