Q72 / 160 · AP EAMCET 2013 · Mathematics
$P, Q, R$ and $S$ are four points with the position vectors $3 \mathbf{i}-4 \mathbf{j}+5 \mathbf{k},-4 \mathbf{i}+5 \mathbf{j}+\mathbf{k}$ and $-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}$, respectively. Then, the line $P Q$ meets the line $R S$ at the point
- A$3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}$
- Correct answerB$-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}$
- C$-\mathbf{i}+4 \mathbf{j}+\mathbf{k}$
- D$\mathbf{i}+\mathbf{j}+\mathbf{k}$
Solution
Let the coordinates of four points $P, Q, R$ and $S$ be $(3,-4,5), \quad(0,0,4), \quad(-4,5,1) \quad$ and $\quad(-3,4,3)$ respectively.
Now, equation of line $P Q$ is
$$
\begin{aligned}
& \frac{x-3}{0-3}=\frac{y+4}{0+4}=\frac{z-5}{4-5} \\
\Rightarrow \quad & \frac{x-3}{-3}=\frac{y+4}{4}=\frac{z-5}{-1}=r_1
\end{aligned}
$$
Equation of line $R S$ is
$$
\frac{x+4}{-3+4}=\frac{y-5}{4-5}=\frac{z-1}{3-1}
$$
$$
\Rightarrow \quad \frac{x+4}{1}=\frac{y-5}{-1}=\frac{z-1}{2}=r_2
$$
Let $\left(-3 r_1+3,4 r_1-4,-r_1+5\right)$ and $\left(r_2-4,-r_2+5\right.$, $\left.2 r_2+1\right)$ be the points on line (i) and (ii), respectively. Since, both lines intersect at a common point, then
$$
\begin{aligned}
& & -3 r_1+3 & =r_2-4 \\
\Rightarrow & & 3 r_1+r_2 & =7 \\
\text { and } & & -r_2+5 & =4 r_1-4 \\
\Rightarrow & & 4 r_1+r_2 & =9
\end{aligned}
$$
On subtracting Eq. (iv) from Eq. (iii), we get
$$
r_1=2
$$
On putting the value of $r_1$ in Eq. (iii), we get
$$
3(2)+r_2=7 \Rightarrow r_2=1
$$
So, required point of intersection is $(-3,4,3)$ i.e.,
$$
-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}
$$
Hence, the correct answer is option: 'B'.