Question

The length intercepted by a line with direction ratios 2, 7, -5 between the lines
$\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}$ and $\frac{x+3}{-3}=\frac{y-3}{-3}=\frac{z-6}{4}$ is

• A. $\sqrt{75}$
• B. $\sqrt{78}$
• C. $\sqrt{83}$
• D. None of these

Answer 1

Answer: (B)

The general points on the given lines are respectively $P(5+3t,7-t,-2+t)$ and $Q(-3-3s,3+2s,6+4s)$. Direction ratios of PQ are
$<3-3s-5-3t,3+2s-7+t,6+4s+2-t>$
i.e., $<-8-3s-3t,-4+2s+t,+8+4s-t>$
If PQ is the desired line then direction ratios of PQ should be proportional to $<2,7,-5>$, therefore,
$\frac{-8-3s-3t}{2}=\frac{-4+2s+t}{7}=\frac{8+4s-t}{-5}$
Taking first and second numbers, we get
$-56-21s-21t=-8+4s+2t$
$\Rightarrow 25s+23t=-48\dots \left(i\right)$
Taking second and third members, we get
$20-10s-5t=56+28s-7t$
$\Rightarrow 38s-2t=-36\dots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$ for t and s, we get
$s=-1$ and $t=-1$.
The coordinates of P and Q are respectively
$\left(5+3\left(-1\right),7-\left(-1\right),-2-1\right)=\left(2,8,-3\right)$
and $\left(-3-3\left(-1\right),3+2\left(-1\right),6+4\left(-1\right)\right)=\left(0,1,2\right)$
$\therefore$ The required line intersects the given lines in the points $\left(2,8,-3\right)$ and $\left(0,1,2\right)$ respectively.
Length of the line intercepted between the given lines
$=|PQ|=\sqrt{{\left(0-2\right)}^2+{\left(1-8\right)}^2+{\left(2+3\right)}^2}=\sqrt{78}.$