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
$-56 - 21s - 21t = -8+ 4s + 2t$
$\Rightarrow 25s + 23t = -48\quad \dots \left(i\right)$
Taking second and third members, we get
$20 -10s - 5t = 56 + 28s - 7t$
$\Rightarrow 38s - 2t = -36\quad \dots \left(ii\right)$
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
$= | PQ | =\sqrt{\left(0-2\right)^{2}+\left(1-8\right)^{2}+\left(2+3\right)^{2}} = \sqrt{78}.$