Question

If the points \((1, 1, \lambda)\) and \((-3, 0, 1)\) are equidistant from the plane \(3x + 4y - 12z + 13 = 0\), then the values of \(\lambda\) are

• A. \(-1, \frac{7}{3}\)
• B. \(1, \frac{-7}{3}\)
• C. \(-1, \frac{-7}{3}\)
• D. \(1, \frac{7}{3}\)

Hint:

When points are equidistant from a plane, they lie on planes parallel to the given plane at the same distance on either side. The equation \(|P_1| = |P_2|\) covers both "same side" and "opposite side" cases.

Answer 1

The Correct Option is D

Step 1: Distance Formula:
Distance of a point \((x_1, y_1, z_1)\) from plane \(ax+by+cz+d=0\) is \(\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}\). Normalizing factor (denominator): \(\sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\).
Step 2: Calculate Distances:
Distance \(d_1\) for \(P(1, 1, \lambda)\): \[ d_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{13} = \frac{|20 - 12\lambda|}{13} \] Distance \(d_2\) for \(Q(-3, 0, 1)\): \[ d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{13} = \frac{|-9 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13} \]
Step 3: Equate Distances and Solve:
\[ \frac{|20 - 12\lambda|}{13} = \frac{8}{13} \] \[ |20 - 12\lambda| = 8 \] This gives two cases: Case 1: \(20 - 12\lambda = 8\) \[ 12\lambda = 12 \implies \lambda = 1 \] Case 2: \(20 - 12\lambda = -8\) \[ 12\lambda = 28 \implies \lambda = \frac{28}{12} = \frac{7}{3} \] Values of \(\lambda\) are \(1\) and \(\frac{7}{3}\).