Question

Let the position vectors of two points $P$ and $Q$ be $3\hat{i}-\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}-4\hat{k}$, respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $PR$ and $QS$ are $(4,-1,2)$ and $(-2,1,-2)$, respectively. Let lines $PR$ and QS intersect at $T$. If the vector $\overrightarrow{TA}$ is perpendicular to both $\overrightarrow{PR}$ and $\overrightarrow{QS}$ and the length of vector $\overrightarrow{TA}$ is $\sqrt{5}$ units, then the modulus of a position vector of $A$ is :

• A. $\sqrt{482}$
• B. $\sqrt{171}$
• C. $\sqrt{5}$
• D. $\sqrt{227}$

Answer 1

Answer: (B)

$P(3,-1,2)$
$Q(1,2,-4)$
$\overrightarrow{PR}\Vert 4\hat{i}-\hat{j}+2\hat{k}$
$\overrightarrow{QS}\Vert -2\hat{i}+\hat{j}-2\hat{k}$
dr's of normal to the plane containing $P,TQ$ will be proportional to :
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -1& 2\\ -2& 1& -2\end{array}\right|$

$\therefore \frac{\ell }{0}=\frac{m}{4}=\frac{n}{2}$
For point, $T:\overrightarrow{PT}=\frac{x-3}{4}=\frac{y+1}{-1}=\frac{z-2}{2}=\lambda$
$\overrightarrow{QT}=\frac{x-1}{-2}=\frac{y-1}{1}=\frac{z+4}{-2}=\mu$
$T:(4\lambda +3,-\lambda -1,2\lambda +2)$
$\cong (2\mu +1,\mu +2,-2\mu -4)$
$4\lambda +3=-2\mu +1$
$\Rightarrow 2\lambda +\mu =-1$
$\lambda +\mu =-3$
$\Rightarrow \lambda =2$
$\&\mu =-5\lambda +\mu =-3$
$\Rightarrow \lambda =2$
So point $T:(11,-3,6)$
$\overrightarrow{OA}=(11\hat{i}-3\hat{j}+6\hat{k})\pm \left(\frac{2\hat{j}+\hat{k}}{\sqrt{5}}\right)\sqrt{5}$
$\overrightarrow{OA}=(11\hat{i}-3\hat{j}+6\hat{k})\pm (2\hat{j}+\hat{k})$
$\overrightarrow{OA}=11\hat{i}-\hat{j}+7\hat{k}$
or
$9\hat{i}-5\hat{j}+5\hat{k}$
$|\overrightarrow{OA}|=\sqrt{121+1+49}=\sqrt{171}$
or
$\sqrt{81+25+25}=\sqrt{131}$