Question

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

• A. $3i+4j+3k$
• B. $-3i+4j+3k$
• C. $-i+4j+k$
• 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 $PQ$ 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 $\left(-3r_1+3,4r_1-4,-r_1+5\right)$ and $(r_2-4,-r_2+5,$,
$2r_2+1)$ be the points on line (i) and (ii), respectively. Since, both lines intersect at a common point, then
$-3r_1+3=r_2-4$
$\Rightarrow 3r_1+r_2=7$ ...(iii)
and $-r_2+5=4r_1-4$
$\Rightarrow 4r_1+r_2=9$ ...(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., $-3i+4j+3k$