JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Three Dimensional Geometry

Let

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

and

P_{2} : x + 2 y + z = 2

be two planes. Then, which of the following statement(s) is (are) TRUE?

(A)The line of intersection of $P_{1}$ and $P_{2}$ has direction ratios 1, 2, -1
(B)The line $\frac{3 x - 4}{9} = \frac{1 - 3 y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_{1}$ and $P_{2}$
(C)The acute angle between $P_{1}$ and $P_{2}$ is $6 0^{\circ}$ ✓ Correct
(D)If $P_{3}$ is the plane passing through the point (4, 2, -2) and perpendicular to the line of intersection of $P_{1}$ and $P_{2}$ , then the distance of the point (2, 1, 1) from the plane $P_{3}$ is $\frac{2}{3}$ ✓ Correct

Answer: (C), (D)

Solution

Direction ratios of the line of intersection:

n_{1} \times n_{2} = \hat{i} 21 \hat{j} 12 \hat{k} - 1 1 = 3 \hat{i} - 3 \hat{j} + 3 \hat{k}

, i.e., direction ratios

1, - 1, 1

. (A) is FALSE.

The given line has direction ratios

3, - 3, 3

1, - 1, 1

, which is parallel (not perpendicular) to the line of intersection. (B) is FALSE.

Angle between planes:

cos θ = \frac{∣2 \cdot 1 + 1 \cdot 2 + ( - 1 ) \cdot 1∣}{6 6} = \frac{3}{6} = \frac{1}{2}

, so

θ = 6 0^{\circ}

. (C) is TRUE.

Plane

P_{3}

perpendicular to

(1, - 1, 1)

through

(4, 2, - 2)

1 (x - 4) - 1 (y - 2) + 1 (z + 2) = 0

gives

x - y + z = 0

. Distance from

(2, 1, 1)

\frac{∣2 - 1 + 1∣}{3} = \frac{2}{3}

. (D) is TRUE.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · One or more correct options