Mathematics - Vector Question with Solution | TestHub

Given,

Equation of line $x=y, z=1$

$\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r$

Now, let any point on the above line be $Q\left(r, r, 1\right)$

Similarly, for line $x=-y, z=-1$ we get,

$\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k$

So, any point on the above line will be $R\left(k, -k, -1\right)$

Now, the direction ratio of $\overrightarrow{PQ}=\left(a-r\right)\hat{i}+\left(a-r\right)\hat{j}+\left(a-1\right)\hat{k}$

Now, finding foot of perpendicular from $P$ on line $\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}$ using perpendicular condition we get,

$a-r+a-r+0=0$

$\Rightarrow 2a=2r$

$\Rightarrow a=r$

Hence, point $Q\left(a,a,1\right)$

And similarly, the direction ratio of 

$\overrightarrow{PR}=\left(a-k\right)i+\left(a+k\right)\hat{j}+\left(a+1\right)\hat{k}$

And again, finding foot of perpendicular from $P$ on line $\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}$ using perpendicular condition we get,

$a-k-a-k=0\Rightarrow k=0$

Hence, point $R\left(0,0,-1\right)$

Also, given $\overrightarrow{PQ}\perp \overrightarrow{PR}$

Now, using perpendicular condition we get,

$\left(a-r\right)\left(a-k\right)+\left(a-r\right)\left(a+k\right)+\left(a-1\right)\left(a+1\right)=0$

$\Rightarrow \left(a-a\right)\left(a-0\right)+\left(a-a\right)\left(a+0\right)+\left(a-1\right)\left(a+1\right)=0$

$\Rightarrow \left(a-1\right)\left(a+1\right)=0$

$\Rightarrow a=1$ or $-1$

Hence, $12a^2=12$