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 vetor $\overrightarrow{TA}$ is $\sqrt{5}$ units , then the modulus of a position vector of $A$ is:

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

Answer 1

The correct option is B $\sqrt{171}$

$\overrightarrow{p}=3\hat{i}-\hat{j}+2\hat{k}\&\overrightarrow{q}=\hat{i}+2\hat{j}-4\hat{k}\overrightarrow{V_{PR}}=(4,-1,2)\&\overrightarrow{V_{QS}}=(-2,1-2)$
$L_{PR}:\overrightarrow{r}=(3\hat{i}-\hat{j}+2\hat{k})+\lambda (4\hat{i}-1\hat{j}+2\hat{k})L_{QS}:\overrightarrow{r}=(\hat{i}+2\hat{j}-4\hat{k})+\mu (-2\hat{i}+1\hat{j}-2\hat{k})$
Now $T$ on $PR$
$(3+4\lambda ,-1-\lambda ,2+2\lambda )$
SImilarly $T$ on $QS$
$(1-2\mu ,2+\mu ,-4-2\mu )$
For $\lambda \&\mu :\begin{array}{c}3+4\lambda =1-2\mu \Rightarrow \mu +2\lambda =-1\\ -1-\lambda =2+\mu \Rightarrow \mu +\lambda =-3\end{array}\}\begin{array}{c}\lambda =2\\ \mu =-5\end{array}$
And
$2+2\lambda =-4-2\mu \Rightarrow T\equiv (11,-3,6)$
D.R. of $TA=\overrightarrow{V_{QS}}\times \overrightarrow{V_{PR}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 1& -2\\ 4& -1& 2\end{array}\right|=0\hat{i}-4\hat{j}-2\hat{k}{L}_{TA}:\overrightarrow{r}(11\hat{i}-3\hat{j}+6\hat{k})+\lambda (-4\hat{j}-2\hat{k})$
Now
$A=(11,-3-4\lambda ,6-2\lambda )TA=\sqrt{5}\Rightarrow (4\lambda {)}^2+(2\lambda {)}^2=5\Rightarrow 16{\lambda }^2+4{\lambda }^2=5\times \lambda =\pm \frac{1}{2}$
$A:(11,-5,5)\text{or}A:(11,-1,7)\Rightarrow |A|=\sqrt{121+25+25}\text{or}|A|=\sqrt{121+1+49}=\sqrt{171}\text{or}\sqrt{171}$