JEE Advanced · 2006 · 09 Apr 2006 · Vector Algebra

Let

A

be vector parallel to line of intersection of planes

P_{1}

and

P_{2}

through origin.

P_{1}

is parallel to the vectors

2 \hat{j} + 3 \hat{k}

and

4 \hat{j} - 3 \hat{k}

and

P_{2}

is parallel to

\hat{j} - \hat{k}

and

3 \hat{i} + 3 \hat{j}

, then the angle between vectors

A

and

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

Answer: (B), (D)

Solution

The normal to plane

P_{1}

is:

n_{1} = (2 \hat{j} + 3 \hat{k}) \times (4 \hat{j} - 3 \hat{k}) = \hat{i} 00 \hat{j} 24 \hat{k} 3 - 3 = - 18 \hat{i}

The normal to plane

P_{2}

is:

n_{2} = (\hat{j} - \hat{k}) \times (3 \hat{i} + 3 \hat{j}) = \hat{i} 03 \hat{j} 13 \hat{k} - 1 0 = 3 \hat{i} - 3 \hat{j} - 3 \hat{k}

The line of intersection is parallel to:

A = n_{1} \times n_{2} = (- 18 \hat{i}) \times (3 \hat{i} - 3 \hat{j} - 3 \hat{k}) = 54 \hat{j} - 54 \hat{k} = 54 (\hat{j} - \hat{k})

A

is parallel to

\hat{j} - \hat{k}

.

Let

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

cos θ = \frac{A \cdot B}{∣ A ∣∣ B ∣} = \frac{54 ( 1 + 2 )}{54 2 \cdot 3} = \frac{162}{162 2} = \frac{1}{2}

Since

cos θ = \pm \frac{1}{2}

, we have

θ = \frac{π}{4}

\frac{3 π}{4}

JEE Advanced · 2006 · 09 Apr 2006 · One or more correct options