JEE Challenger
More from Three Dimensional Geometry
MathMedAdv 2012 • Paper 2

Equation of Plane Through Line of Intersection

The equation of a plane passing through the line of intersection of the planes

x+2y+3z=2x + 2y + 3z = 2 and x−y+z=3x - y + z = 3 and at a distance 23\frac{2}{\sqrt{3}} from the point (3,1,−1)(3, 1, -1) is

Options

A

5x−11y+z=175x - 11y + z = 17

Correct
B

2x+y=32−1\sqrt{2}x + y = 3\sqrt{2} - 1

C

x+y+z=3x + y + z = \sqrt{3}

D

x−2y=1−2x - \sqrt{2}y = 1 - \sqrt{2}

Topics & Concepts

Three Dimensional GeometryPlane in 3D

Step-by-Step Solution

To find the equation of the required plane, we use the concept of the family of planes. Any plane passing through the line of intersection of the given planes x+2y+3z−2=0andx−y+z−3=0x + 2y + 3z - 2 = 0 \quad \text{and} \quad x - y + z - 3 = 0 can be represented as:

(x+2y+3z−2)+λ(x−y+z−3)=0(x + 2y + 3z - 2) + \lambda(x - y + z - 3) = 0

Rearranging the terms in the standard form Ax+By+Cz+D=0Ax + By + Cz + D = 0, we get: (1+λ)x+(2−λ)y+(3+λ)z−(2+3λ)=0— (1)(1 + \lambda)x + (2 - \lambda)y + (3 + \lambda)z - (2 + 3\lambda) = 0 \quad \text{--- (1)}

The perpendicular distance dd from a point (x0,y0,z0)(x_0, y_0, z_0) to a plane Ax+By+Cz+D=0Ax + By + Cz + D = 0 is given by: d=∣Ax0+By0+Cz0+D∣A2+B2+C2d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}

Here, the point is (3,1,−1)(3, 1, -1) and the given distance is d=23d = \frac{2}{\sqrt{3}}.

Substitute the point (3,1,−1)(3, 1, -1) into equation (1): Numerator=∣(1+λ)(3)+(2−λ)(1)+(3+λ)(−1)−(2+3λ)∣\text{Numerator} = |(1 + \lambda)(3) + (2 - \lambda)(1) + (3 + \lambda)(-1) - (2 + 3\lambda)| Numerator=∣3+3λ+2−λ−3−λ−2−3λ∣=∣−2λ∣\text{Numerator} = |3 + 3\lambda + 2 - \lambda - 3 - \lambda - 2 - 3\lambda| = |-2\lambda|

Now, calculate the denominator: A2+B2+C2=(1+λ)2+(2−λ)2+(3+λ)2\sqrt{A^2 + B^2 + C^2} = \sqrt{(1 + \lambda)^2 + (2 - \lambda)^2 + (3 + \lambda)^2} =(1+2λ+λ2)+(4−4λ+λ2)+(9+6λ+λ2)= \sqrt{(1 + 2\lambda + \lambda^2) + (4 - 4\lambda + \lambda^2) + (9 + 6\lambda + \lambda^2)} =3λ2+4λ+14= \sqrt{3\lambda^2 + 4\lambda + 14}

Setting the distance formula equal to 23\frac{2}{\sqrt{3}}: ∣−2λ∣3λ2+4λ+14=23\frac{|-2\lambda|}{\sqrt{3\lambda^2 + 4\lambda + 14}} = \frac{2}{\sqrt{3}}

Canceling 22 from both sides: ∣λ∣3λ2+4λ+14=13\frac{|\lambda|}{\sqrt{3\lambda^2 + 4\lambda + 14}} = \frac{1}{\sqrt{3}}

Squaring both sides: λ23λ2+4λ+14=13\frac{\lambda^2}{3\lambda^2 + 4\lambda + 14} = \frac{1}{3} 3λ2=3λ2+4λ+143\lambda^2 = 3\lambda^2 + 4\lambda + 14 4λ+14=0  ⟹  λ=−724\lambda + 14 = 0 \implies \lambda = -\frac{7}{2}

Substitute λ=−72\lambda = -\frac{7}{2} back into the equation of the family of planes: (x+2y+3z−2)−72(x−y+z−3)=0(x + 2y + 3z - 2) - \frac{7}{2}(x - y + z - 3) = 0

Multiplying the entire equation by 22: 2(x+2y+3z−2)−7(x−y+z−3)=02(x + 2y + 3z - 2) - 7(x - y + z - 3) = 0 (2x+4y+6z−4)−(7x−7y+7z−21)=0(2x + 4y + 6z - 4) - (7x - 7y + 7z - 21) = 0 −5x+11y−z+17=0-5x + 11y - z + 17 = 0

Multiplying by −1-1: 5x−11y+z=175x - 11y + z = 17

Thus, the correct option is A.

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