JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Vector Algebra

A vector

a

is parallel to the line of intersection of the plane determined by the vectors

\hat{i}, \hat{i} + \hat{j}

and the plane determined by the vectors

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

. The obtuse angle between

a

and the vector

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

Answer: (A)

Solution

The normal to the first plane is

n_{1} = \hat{i} \times (\hat{i} + \hat{j}) = \hat{k}

. The normal to the second plane is

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

. The line of intersection is parallel to

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

. So

a = - \hat{i} + \hat{j}

. Given

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

, we have

a \cdot b = - 1 - 2 = - 3

∣ a ∣ = 2

∣ b ∣ = 3

. Thus

cos θ = \frac{- 3}{2 \cdot 3} = - \frac{1}{2}

, giving

θ = \frac{3 π}{4}

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option