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)

$P (3,-1,2)$
$Q (1,2,-4)$
$\overrightarrow{ PR } \| 4 \hat{ i }-\hat{ j }+2 \hat{ k }$
$\overrightarrow{ QS } \|-2 \hat{ i }+\hat{ j }-2 \hat{ k }$
dr's of normal to the plane containing $P , T \,Q$ will be proportional to :
$\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 4 & -1 & 2 \\ -2 & 1 & -2\end{vmatrix}$

$\therefore \frac{\ell}{0}=\frac{ m }{4}=\frac{ n }{2}$
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
$\sqrt{81+25+25}=\sqrt{131}$