JEE Challenger
More from Three Dimensional Geometry
MathMedAdv 2008 • Paper 1

Intersection of Three Planes and Parallelism of Intersection Lines

Consider three planes

P1:x−y+z=1P2:x+y−z=−1P3:x−3y+3z=2.\begin{array}{l} P_1 : x - y + z = 1 \\ P_2 : x + y - z = -1 \\ P_3 : x - 3y + 3z = 2 . \end{array}

Let L1,L2,L3L_1, L_2, L_3 be the lines of intersection of the planes P2P_2 and P3P_3, P3P_3 and P1P_1, and P1P_1 and P2P_2, respectively.

STATEMENT-1 : At least two of the lines L1,L2L_1, L_2 and L3L_3 are non-parallel.

and

STATEMENT-2 : The three planes do not have a common point.

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 GeometryPlane in 3DLine in 3D

Step-by-Step Solution

To determine the correctness of the given statements, we analyze the geometry of the three given planes:
P1:x−y+z=1P_1 : x - y + z = 1
P2:x+y−z=−1P_2 : x + y - z = -1
P3:x−3y+3z=2P_3 : x - 3y + 3z = 2

The normal vectors to the planes are:
n⃗1=i^−j^+k^\vec{n}_1 = \hat{i} - \hat{j} + \hat{k}
n⃗2=i^+j^−k^\vec{n}_2 = \hat{i} + \hat{j} - \hat{k}
n⃗3=i^−3j^+3k^\vec{n}_3 = \hat{i} - 3\hat{j} + 3\hat{k}

Analysis of STATEMENT-1:

The direction vector d⃗1\vec{d}_1 of line L1L_1 (intersection of P2P_2 and P3P_3) is given by:

d⃗1=n⃗2×n⃗3=∣i^j^k^11−11−33∣=0i^−4j^−4k^=−4(j^+k^)\vec{d}_1 = \vec{n}_2 \times \vec{n}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -3 & 3 \end{vmatrix} = 0\hat{i} - 4\hat{j} - 4\hat{k} = -4(\hat{j} + \hat{k})

The direction vector d⃗2\vec{d}_2 of line L2L_2 (intersection of P3P_3 and P1P_1) is given by:

d⃗2=n⃗3×n⃗1=∣i^j^k^1−331−11∣=0i^+2j^+2k^=2(j^+k^)\vec{d}_2 = \vec{n}_3 \times \vec{n}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 3 \\ 1 & -1 & 1 \end{vmatrix} = 0\hat{i} + 2\hat{j} + 2\hat{k} = 2(\hat{j} + \hat{k})

The direction vector d⃗3\vec{d}_3 of line L3L_3 (intersection of P1P_1 and P2P_2) is given by:

d⃗3=n⃗1×n⃗2=∣i^j^k^1−1111−1∣=0i^+2j^+2k^=2(j^+k^)\vec{d}_3 = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 0\hat{i} + 2\hat{j} + 2\hat{k} = 2(\hat{j} + \hat{k})

Since d⃗1,d⃗2,\vec{d}_1, \vec{d}_2, and d⃗3\vec{d}_3 are all scalar multiples of j^+k^\hat{j} + \hat{k}, all three lines L1,L2,L_1, L_2, and L3L_3 are parallel to one another.

Therefore, STATEMENT-1 is False.

Analysis of STATEMENT-2:

To check if the three planes have a common point, we solve the system of equations. Consider the linear combination 2P1−P22P_1 - P_2:
2(x−y+z)−(x+y−z)=2(1)−(−1)2(x - y + z) - (x + y - z) = 2(1) - (-1)
x−3y+3z=3x - 3y + 3z = 3

Comparing this with P3P_3:
x−3y+3z=2x - 3y + 3z = 2

Since 3≠23 \neq 2, the system of equations is inconsistent and has no solution. Hence, the three planes do not intersect at a common point.

Therefore, STATEMENT-2 is True.

Conclusion:

  • STATEMENT-1 is False
  • STATEMENT-2 is True

This corresponds to Option (D).

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