JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Three Dimensional Geometry

Consider the planes

3 x - 6 y - 2 z = 15

and

2 x + y - 2 z = 5

.

STATEMENT-1 : The parametric equations of the line of intersection of the given planes are

x = 3 + 14 t, y = 1 + 2 t, z = 15 t

.

because

STATEMENT-2 : The vector

14 \hat{i} + 2 \hat{j} + 15 \hat{k}

is parallel to the line of intersection of given planes.

(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

The direction vector of the line of intersection is perpendicular to both normal vectors of the planes.

Normal vectors:

n_{1} = 3 \hat{i} - 6 \hat{j} - 2 \hat{k}

and

n_{2} = 2 \hat{i} + \hat{j} - 2 \hat{k}

Direction vector:

d = n_{1} \times n_{2} = \hat{i} 32 \hat{j} - 6 1 \hat{k} - 2 - 2 = 14 \hat{i} + 2 \hat{j} + 15 \hat{k}

So Statement-2 is TRUE.

For

z = 0

, solving the two plane equations:

3 x - 6 y = 15

and

2 x + y = 5

gives

x = 3, y = - 1

.

The parametric equations should be:

x = 3 + 14 t, y = - 1 + 2 t, z = 15 t

Since the given equations have

y = 1 + 2 t

instead of

y = - 1 + 2 t

, Statement-1 is FALSE.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option