The lines \(\displaystyle\frac{1-x}{2}=\frac{y-1}{3}=\frac{z}{1}\) and \(\displaystyle\frac{2x-3}{2p}=\frac{y}{-1}=\frac{z-4}{7}\) are perpendicular to each other for p equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
The direction ratios can be read only when each line is in the standard form \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\), with the coefficient of x, y and z equal to 1.
First line:
\( \displaystyle \frac{1-x}{2} = \frac{x-1}{-2} \)
so its direction ratios are \(-2, 3, 1\).
Second line:
\( \displaystyle \frac{2x-3}{2p} = \frac{x-\frac{3}{2}}{p} \)
so its direction ratios are \(p, -1, 7\).
The lines are perpendicular when \(a_1a_2+b_1b_2+c_1c_2 = 0\):
\((-2)(p)+(3)(-1)+(1)(7) = 0\)
\(-2p+4 = 0\)
\(p = 2\)
Why the other options are incorrect
Option (1) −2 — Incorrect. This comes from reading the first direction ratio as 2 instead of −2: \(2p-3+7 = 0\) gives \(p = -2\).
Option (2) \(\displaystyle\frac{1}{2}\) — Incorrect. With \(p = \frac{1}{2}\), \(-2p+4 = 3 \ne 0\).
Option (4) 3 — Incorrect. With \(p = 3\), \(-2p+4 = -2 \ne 0\).