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 $\left(4,-1,2\right)$ and $\left(-2,1,-2\right)$ respectively. Let lines $PR$ and $QS$ intersect at $T$. If the vector $TA$ is perpendicular to both $PR$ and $QS$ and the length of vector $TA$ 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

$\sqrt{171}$

Explanation for the correct option:

Finding the modulus of a position vector of $A$:

Given the direction ratios of $PR$ is $\left(4,-1,2\right)$ and $QS$ is $\left(-2,1,-2\right)$ also the position vectors of the points $P\left(3,-1,2\right)$ and $Q\left(1,2-4\right)$.

Now the equation of the line $PR$ is obtained as,

$\Rightarrow \frac{x-3}{4}=\frac{y+1}{-1}=\frac{z-2}{2}=\lambda \dots \dots \left(1\right)$

Also the equation of the line $QS$ is,

$\Rightarrow \frac{x-1}{-2}=\frac{y-2}{1}=\frac{z+4}{-2}=\mu \dots \dots \left(2\right)$

Since the two lines $PR$ and $QS$ are intesecting at the point $T$ then,

$\Rightarrow x=4\lambda +3;y=-\lambda -1;z=2\lambda +2\left[\text{from equation}\left(1\right)\right]$

Also,

$\Rightarrow x=-2\mu +1;y=\lambda +2;z=-2\lambda -4\left[\text{from equation}\left(2\right)\right]$

Since the $y$ and $z$ coordinate are the same, we have

$$\begin{gathered} \Rightarrow -\lambda -1=\mu +2 \\ \Rightarrow \mu =-\lambda -3 \end{gathered}$$

And the $x$ coordinate is obtained as,

$\Rightarrow 4\lambda +3=-2\mu +1$

Substituting $\mu$ as $-\lambda -3$ in the above equation to find the value of $\lambda$.

$$\begin{gathered} \Rightarrow 4\lambda +3=-2\left(-\lambda -3\right)+1 \\ \Rightarrow 4\lambda +3=2\lambda +6+1 \\ \Rightarrow 2\lambda =4 \\ \Rightarrow \lambda =2 \end{gathered}$$

Using the value of $\lambda$ we find the coordinates of $T$,

$\Rightarrow T=\left(11,-3,6\right)$

Now find the cross-product of $\overrightarrow{PR}$ and $\overrightarrow{QS}$.

$$\begin{gathered} \Rightarrow \overrightarrow{PR}\times \overrightarrow{QS}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -1& 2\\ -2& 1& -2\end{array}\right| \\ \Rightarrow \overrightarrow{PR}\times \overrightarrow{QS}=\left(2-2\right)\hat{i}-\left(-8+4\right)\hat{j}+\left(4-2\right)\hat{k} \\ \Rightarrow \overrightarrow{PR}\times \overrightarrow{QS}=0\hat{i}+4\hat{j}+2\hat{k} \end{gathered}$$

Therefore, the coordinate of the normal vector is $\left(0,4,2\right)$ and the direction ratio of $\overrightarrow{n}$ is $\left(0,2,1\right)$.

To find the coordinate of $A$

$$\begin{gathered} \Rightarrow x-11=0;y+3=2;z-2=5 \\ \Rightarrow x=11;y=-1;z=7 \end{gathered}$$

The coordinate of $A$ is $\left(11,-1,7\right)$.

Using the distance formula find the distance of $TA$.

$$\begin{gathered} \Rightarrow TA=\sqrt{{\left(11-11\right)}^2+{\left(-3+11\right)}^2+{\left(6-7\right)}^2} \\ \Rightarrow TA=\sqrt{0+4+1} \\ \Rightarrow TA=\sqrt{5} \end{gathered}$$

And the position vector of $A$ is,

$$\begin{gathered} \Rightarrow \left|\overrightarrow{OA}\right|=\sqrt{121+1+49} \\ \Rightarrow \left|\overrightarrow{OA}\right|=\sqrt{171} \end{gathered}$$

The modulus of a position vector of $A$ is $\sqrt{171}$.

Hence, option (B) is the correct answer.