JEE Challenger
More from Three Dimensional Geometry
MathMedAdv 2013 • Paper 2

Equation of Plane Perpendicular to Two Given Planes Passing Through Line Intersection

Given the lines L1:x−12=y−1=z+31L_1 : \frac{x-1}{2} = \frac{y}{-1} = \frac{z+3}{1} and L2:x−41=y+31=z+32L_2 : \frac{x-4}{1} = \frac{y+3}{1} = \frac{z+3}{2}, alongside the planes P1:7x+y+2z=3P_1 : 7x + y + 2z = 3 and P2:3x+5y−6z=4P_2 : 3x + 5y - 6z = 4. Suppose the plane that passes through the intersection point of L1L_1 and L2L_2 and is perpendicular to both P1P_1 and P2P_2 is represented by ax+by+cz=dax + by + cz = d.

Match the entries in List-I with the corresponding values in List-II:

List - IList - IIP. a=1. 13Q. b=2. −3R. c=3. 1S. d=4. −2\begin{array}{ll} \textbf{List - I} & \textbf{List - II} \\ \text{P. } a = & \text{1. } 13 \\ \text{Q. } b = & \text{2. } -3 \\ \text{R. } c = & \text{3. } 1 \\ \text{S. } d = & \text{4. } -2 \end{array}

Options

A

P →\to 3, Q →\to 2, R →\to 4, S →\to 1

Correct
B

P →\to 1, Q →\to 3, R →\to 4, S →\to 2

C

P →\to 3, Q →\to 2, R →\to 1, S →\to 4

D

P →\to 2, Q →\to 4, R →\to 1, S →\to 3

Topics & Concepts

Three Dimensional GeometryPlane in 3DLine in 3DDirection Cosines and Ratios

Step-by-Step Solution

To find the equation of the plane ax+by+cz=dax + by + cz = d, we first determine the point of intersection of the lines L1L_1 and L2L_2, and then find the normal vector to the plane.

Step 1: Find the point of intersection of lines L1L_1 and L2L_2

Let a general point on line L1L_1 in terms of parameter λ\lambda be: (x,y,z)=(2λ+1,−λ,λ−3)(x, y, z) = (2\lambda + 1, -\lambda, \lambda - 3)

Let a general point on line L2L_2 in terms of parameter μ\mu be: (x,y,z)=(μ+4,μ−3,2μ−3)(x, y, z) = (\mu + 4, \mu - 3, 2\mu - 3)

Equating the xx and yy coordinates: 2λ+1=μ+4  ⟹  2λ−μ=3— (1)2\lambda + 1 = \mu + 4 \implies 2\lambda - \mu = 3 \quad \text{--- (1)} −λ=μ−3  ⟹  λ+μ=3— (2)-\lambda = \mu - 3 \implies \lambda + \mu = 3 \quad \text{--- (2)}

Adding equations (1)(1) and (2)(2): 3λ=6  ⟹  λ=23\lambda = 6 \implies \lambda = 2 Substituting λ=2\lambda = 2 into equation (2)(2): μ=3−2=1\mu = 3 - 2 = 1

Checking the zz-coordinate for consistency:

  • From L1L_1: z=2−3=−1z = 2 - 3 = -1
  • From L2L_2: z=2(1)−3=−1z = 2(1) - 3 = -1

Thus, the lines intersect at the point: (x0,y0,z0)=(2(2)+1,−2,2−3)=(5,−2,−1)(x_0, y_0, z_0) = (2(2) + 1, -2, 2 - 3) = (5, -2, -1)

Step 2: Find the normal vector to the required plane

The plane is perpendicular to both planes:

  • P1:7x+y+2z=3  ⟹  n⃗1=7i^+j^+2k^P_1: 7x + y + 2z = 3 \implies \vec{n}_1 = 7\hat{i} + \hat{j} + 2\hat{k}
  • P2:3x+5y−6z=4  ⟹  n⃗2=3i^+5j^−6k^P_2: 3x + 5y - 6z = 4 \implies \vec{n}_2 = 3\hat{i} + 5\hat{j} - 6\hat{k}

The normal vector n⃗\vec{n} to the required plane is parallel to n⃗1×n⃗2\vec{n}_1 \times \vec{n}_2:

n⃗=∣i^j^k^71235−6∣\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 1 & 2 \\ 3 & 5 & -6 \end{vmatrix}

n⃗=i^(−6−10)−j^(−42−6)+k^(35−3)\vec{n} = \hat{i}(-6 - 10) - \hat{j}(-42 - 6) + \hat{k}(35 - 3) n⃗=−16i^+48j^+32k^\vec{n} = -16\hat{i} + 48\hat{j} + 32\hat{k}

Dividing by −16-16 gives the proportional direction ratios: n⃗=1i^−3j^−2k^\vec{n} = 1\hat{i} - 3\hat{j} - 2\hat{k}

Therefore, the coefficients are: a=1,b=−3,c=−2a = 1, \quad b = -3, \quad c = -2

Step 3: Find the value of dd

The equation of the plane passing through the point (5,−2,−1)(5, -2, -1) is: 1(x−5)−3(y+2)−2(z+1)=01(x - 5) - 3(y + 2) - 2(z + 1) = 0 x−3y−2z−5−6−2=0x - 3y - 2z - 5 - 6 - 2 = 0 x−3y−2z=13x - 3y - 2z = 13

Comparing this with ax+by+cz=dax + by + cz = d, we have: a=1,b=−3,c=−2,d=13a = 1, \quad b = -3, \quad c = -2, \quad d = 13

Step 4: Matching with List-II

  • P. a=1  ⟹  3\text{P. } a = 1 \implies 3
  • Q. b=−3  ⟹  2\text{Q. } b = -3 \implies 2
  • R. c=−2  ⟹  4\text{R. } c = -2 \implies 4
  • S. d=13  ⟹  1\text{S. } d = 13 \implies 1

Hence, the correct matching is: P→3,Q→2,R→4,S→1\text{P} \to 3, \quad \text{Q} \to 2, \quad \text{R} \to 4, \quad \text{S} \to 1

Correct Option: (A)

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