Question

$P, Q, R$ and $S$ are four points with the position vectors $3 i -4 j +5 k ,-4 i +5 j + k$ and $-3 i +4 j +3 k$, respectively. Then, the line $P Q$ meets the line $R S$ at the point

• A. $3 i +4 j +3 k$
• B. $-3 i +4 j +3 k$
• C. $- i +4 j + k$
• D. $i + j + k$

Answer 1

Answer: (B)

Let the coordinates of four points $P, Q, R$ and $S$ be $(3,-4,5), \,(0,0,4), \,(-4,5,1)$ and $(-3,4,3)$
respectively. Now, equation of line $P Q$ is
$\frac{x-3}{0-3}=\frac{y+4}{0+4}=\frac{z-5}{4-5}$
$\Rightarrow \frac{x-3}{-3}=\frac{y+4}{4}=\frac{z-5}{-1}=r_{1}$ (say) ...(i)
Equation of line $RS$ is
$\frac{x+4}{-3+4}=\frac{y-5}{4-5}=\frac{z-1}{3-1}$
$\Rightarrow \frac{x+4}{1}=\frac{y-5}{-1}=\frac{z-1}{2}=r_{2}$ (say) ...(ii)
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
$-3 r_{1}+3=r_{2}-4$
$\Rightarrow 3 r_{1}+r_{2}=7$ ...(iii)
and $-r_{2}+5=4 r_{1}-4$
$\Rightarrow 4 r_{1}+r_{2}=9$ ...(iv)
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., $-3 i +4 j +3 k$