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MathematicsAssertion and reason+3 / -1Question 12 of 69

IIT-JEE 2008 · Paper 1, Mathematics question 12

Consider three planes
P1:x−y+z=1,P2:x+y−z=−1,P3:x−3y+3z=2.P_1:x-y+z=1,\qquad P_2:x+y-z=-1,\qquad P_3:x-3y+3z=2.
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 L1L_1, L2L_2 and L3L_3 are non-parallel.

and

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

Choose one option
Answer and worked solution

Official answer: D

Directions of the lines. With normals n1=(1,−1,1)n_1=(1,-1,1), n2=(1,1,−1)n_2=(1,1,-1), n3=(1,−3,3)n_3=(1,-3,3):
n2×n3=(0,−4,−4),n3×n1=(0,2,2),n1×n2=(0,2,2).n_2\times n_3=(0,-4,-4),\quad n_3\times n_1=(0,2,2),\quad n_1\times n_2=(0,2,2).
All three are parallel to (0,1,1)(0,1,1). Statement 1 is false.

Common point. P1+P2P_1+P_2 gives x=0x=0; then P1P_1 gives z−y=1z-y=1 but P3P_3 gives 3(z−y)=23(z-y)=2. Contradiction: no common point. Statement 2 is true.

(D)

Topics:Equation of a planeEquation of a line in space
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