JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Three Dimensional Geometry

Consider three planes

P_{1} : x - y + z = 1 P_{2} : x + y - z = - 1 P_{3} : x - 3 y + 3 z = 2.

Let

L_{1}, L_{2}, L_{3}

be the lines of intersection of the planes

P_{2}

and

P_{3}, P_{3}

and

P_{1}

, and

P_{1}

and

P_{2}

, respectively.

STATEMENT-1 : At least two of the lines

L_{1}, L_{2}

and

L_{3}

are non-parallel.

and

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

(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

Answer: (D)

Solution

Finding direction vectors of the lines:

For

L_{1}

(intersection of

P_{2}

and

P_{3}

d_{1} = n_{2} \times n_{3} = (1, 1, - 1) \times (1, - 3, 3) = (0, - 4, - 4) ∥ (0, 1, 1)

For

L_{2}

(intersection of

P_{3}

and

P_{1}

d_{2} = n_{3} \times n_{1} = (1, - 3, 3) \times (1, - 1, 1) = (0, 2, 2) ∥ (0, 1, 1)

For

L_{3}

(intersection of

P_{1}

and

P_{2}

d_{3} = n_{1} \times n_{2} = (1, - 1, 1) \times (1, 1, - 1) = (0, 2, 2) ∥ (0, 1, 1)

All three lines have direction vectors proportional to

(0, 1, 1)

, so all three lines are parallel. Therefore, Statement 1 is FALSE.

Checking for common point:
The determinant of coefficients:

111 - 1 1 - 3 1 - 1 3 = 1 (3 - 3) + 1 (3 + 1) + 1 (- 3 - 1) = 0 + 4 - 4 = 0

Since the determinant is zero, the planes either have a common line or no common point. Since the three lines of intersection are parallel (not coincident), the planes do not have a common point. Statement 2 is TRUE.

Answer: (D)

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option

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JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Three Dimensional Geometry

Solution

More JEE Advanced questions on Three Dimensional Geometry