JEE Challenger
More from Three Dimensional Geometry
MathMedAdv 2007 • Paper 2

Intersection Line of Two Planes in 3D Space

This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Consider the planes 3x−6y−2z=153x - 6y - 2z = 15 and 2x+y−2z=52x + y - 2z = 5.

STATEMENT-1 :\textbf{STATEMENT-1 :} The parametric equations of the line of intersection of the given planes are x=3+14tx = 3 + 14t, y=1+2ty = 1 + 2t, z=15tz = 15t.

because\textbf{because}

STATEMENT-2 :\textbf{STATEMENT-2 :} The vector 14i^+2j^+15k^14\hat{i} + 2\hat{j} + 15\hat{k} is parallel to the line of intersection of given planes.

Options

A

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True

Correct

Topics & Concepts

Three Dimensional GeometryPlanesLine of IntersectionDirection Ratios

Step-by-Step Solution

  1. Check Statement-2:
    The normal vectors to the two planes are n⃗1=3i^−6j^−2k^\vec{n}_1 = 3\hat{i} - 6\hat{j} - 2\hat{k} and n⃗2=2i^+j^−2k^\vec{n}_2 = 2\hat{i} + \hat{j} - 2\hat{k}.
    The line of intersection is parallel to the cross product n⃗1×n⃗2\vec{n}_1 \times \vec{n}_2:
n⃗1×n⃗2=∣i^j^k^3−6−221−2∣=i^(12−(−2))−j^(−6−(−4))+k^(3−(−12))=14i^+2j^+15k^\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & -2 \\ 2 & 1 & -2 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(-6 - (-4)) + \hat{k}(3 - (-12)) = 14\hat{i} + 2\hat{j} + 15\hat{k}

Thus, the vector 14i^+2j^+15k^14\hat{i} + 2\hat{j} + 15\hat{k} is indeed parallel to the line of intersection. Statement-2 is True.

  1. Check Statement-1:
    For the parametric equation to represent the line of intersection, the base point (x,y,z)=(3,1,0)(x, y, z) = (3, 1, 0) (corresponding to t=0t = 0) must satisfy both plane equations:
  • In Plane 1: 3(3)−6(1)−2(0)=9−6=3≠153(3) - 6(1) - 2(0) = 9 - 6 = 3 \neq 15.
    Since (3,1,0)(3, 1, 0) does not lie on the first plane, the given parametric equations do not represent the line of intersection. Statement-1 is False.

Correct Answer: D

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