JEE Challenger
More from Three Dimensional Geometry
MathHardAdv 2015 • Paper 1

Points on Locus of Feet of Perpendiculars from Line to Plane

In R3\mathbb{R}^3, let LL be a straight line passing through the origin. Suppose that all the points on LL are equidistant from the two planes P1:x+2y−z+1=0P_1 : x + 2y - z + 1 = 0 and P2:2x−y+z−1=0P_2 : 2x - y + z - 1 = 0. Let MM denote the locus of the feet of the perpendiculars dropped from the points on LL onto the plane P1P_1. Which of the following points lie(s) on MM?

Options

A

(0,−56,−23)\left(0, -\frac{5}{6}, -\frac{2}{3}\right)

Correct
B

(−16,−13,16)\left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)

Correct
C

(−56,0,16)\left(-\frac{5}{6}, 0, \frac{1}{6}\right)

D

(−13,0,23)\left(-\frac{1}{3}, 0, \frac{2}{3}\right)

Topics & Concepts

Three Dimensional GeometryLine in 3DPlane in 3D

Step-by-Step Solution

To determine which of the given points lie on the locus MM, we proceed step-by-step:

Step 1: Find the equation of the line LL

Let the line LL pass through the origin (0,0,0)(0, 0, 0) with direction vector v⃗=(a,b,c)\vec{v} = (a, b, c). The parametric equation of LL is given by: (x,y,z)=(at,bt,ct),t∈R(x, y, z) = (at, bt, ct), \quad t \in \mathbb{R}

The equations of the two planes are: P1:x+2y−z+1=0P_1 : x + 2y - z + 1 = 0 P2:2x−y+z−1=0P_2 : 2x - y + z - 1 = 0

The normal vectors to these planes are: n⃗1=(1,2,−1)andn⃗2=(2,−1,1)\vec{n}_1 = (1, 2, -1) \quad \text{and} \quad \vec{n}_2 = (2, -1, 1)

For all points on the line LL to be at a constant distance from the plane P1P_1, the line LL must be parallel to P1P_1. This requires: v⃗⋅n⃗1=0  ⟹  a+2b−c=0\vec{v} \cdot \vec{n}_1 = 0 \implies a + 2b - c = 0

Similarly, for all points on LL to be at a constant distance from the plane P2P_2, the line LL must be parallel to P2P_2: v⃗⋅n⃗2=0  ⟹  2a−b+c=0\vec{v} \cdot \vec{n}_2 = 0 \implies 2a - b + c = 0

Hence, the direction vector v⃗\vec{v} must be parallel to the cross product n⃗1×n⃗2\vec{n}_1 \times \vec{n}_2:

v⃗=n⃗1×n⃗2=∣i^j^k^12−12−11∣=i^(2−1)−j^(1−(−2))+k^(−1−4)=(1,−3,−5)\vec{v} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix} = \hat{i}(2 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - 4) = (1, -3, -5)

Thus, any point on LL can be parametrized by: (x,y,z)=(t,−3t,−5t),t∈R(x, y, z) = (t, -3t, -5t), \quad t \in \mathbb{R}

Step 2: Find the locus MM of the feet of the perpendiculars on P1P_1

Let Q(t)Q(t) be the foot of the perpendicular from a point A(t)=(t,−3t,−5t)A(t) = (t, -3t, -5t) on LL onto the plane P1P_1. The vector from A(t)A(t) to Q(t)Q(t) is parallel to the normal vector n⃗1=(1,2,−1)\vec{n}_1 = (1, 2, -1), so we can write: Q(t)=(t+k,−3t+2k,−5t−k)Q(t) = (t + k, -3t + 2k, -5t - k)

Since Q(t)Q(t) lies on the plane P1:x+2y−z+1=0P_1 : x + 2y - z + 1 = 0, substituting its coordinates into the plane equation gives: (t+k)+2(−3t+2k)−(−5t−k)+1=0(t + k) + 2(-3t + 2k) - (-5t - k) + 1 = 0 (t−6t+5t)+k(1+4+1)+1=0(t - 6t + 5t) + k(1 + 4 + 1) + 1 = 0 0+6k+1=0  ⟹  k=−160 + 6k + 1 = 0 \implies k = -\frac{1}{6}

Therefore, the coordinates of the foot of the perpendicular Q(t)Q(t) are: Q(t)=(t−16,−3t−13,−5t+16)Q(t) = \left(t - \frac{1}{6}, -3t - \frac{1}{3}, -5t + \frac{1}{6}\right) This parametric equation represents the line MM.

Step 3: Check the given options

  • For Option (B): (−16,−13,16)\left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right) Setting t=0t = 0: Q(0)=(0−16,−0−13,−0+16)=(−16,−13,16)Q(0) = \left(0 - \frac{1}{6}, -0 - \frac{1}{3}, -0 + \frac{1}{6}\right) = \left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right) Hence, this point lies on MM.

  • For Option (A): (0,−56,−23)\left(0, -\frac{5}{6}, -\frac{2}{3}\right) Setting the xx-coordinate to 00: t−16=0  ⟹  t=16t - \frac{1}{6} = 0 \implies t = \frac{1}{6} Substituting t=16t = \frac{1}{6} into yy and zz: y=−3(16)−13=−12−13=−56y = -3\left(\frac{1}{6}\right) - \frac{1}{3} = -\frac{1}{2} - \frac{1}{3} = -\frac{5}{6} z=−5(16)+16=−46=−23z = -5\left(\frac{1}{6}\right) + \frac{1}{6} = -\frac{4}{6} = -\frac{2}{3} Thus, Q(16)=(0,−56,−23)Q\left(\frac{1}{6}\right) = \left(0, -\frac{5}{6}, -\frac{2}{3}\right), so this point also lies on MM.

  • For Options (C) and (D): Setting the yy-coordinate to 00: −3t−13=0  ⟹  t=−19-3t - \frac{1}{3} = 0 \implies t = -\frac{1}{9} Then the xx-coordinate must be: x=−19−16=−518x = -\frac{1}{9} - \frac{1}{6} = -\frac{5}{18} Neither Option (C) (x=−56)\left(x = -\frac{5}{6}\right) nor Option (D) (x=−13)\left(x = -\frac{1}{3}\right) matches x=−518x = -\frac{5}{18}. Thus, points (C) and (D) do not lie on MM.

Therefore, the correct options are A and B.

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