Let A→ be vector parallel to line of intersection of planes P1 and P2. Plane P1 is parallel to the vectors 2j^+3k^ and 4j^−3k^ and that P2 is parallel to j^−k^ and 3i^+3j^ , then the angle between vector A→ and a given vector 2i^+j^−2k^ is

Normal to plane P 1 is n → 1 = ( 2 j ^ + 3 k ^ ) × ( 4 j ^ − 3 k ^ ) = − 18 i ^ . Normal to plane P 2 is n → 2 = ( j ^ − k ^ ) × ( 3 i ^ + 3 j ^ ) = 3 i ^ − 3 j ^ − 3 k ^ ∴ A → is parallel to ± n → 1 × n → 2 = ± ( − 54 j ^ + 54 k ^ ) ∴ Angle between A → and 2 i ^ + j ^ − 2 k ^ is cos ⁡ θ = ± ( − 54 j ^ + 54 k ^ ) ⋅ ( 2 i ^ + j ^ − 2 k ^ ) 54 2 ⋅ 3 = ± 1 2 ⇒ θ = π 4 (or) 3 π 4

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Let $\vec{A}$ be vector parallel to line of intersection of planes P₁ and P₂. Plane P₁ is parallel to the vectors $2 \hat{j} + 3 \hat{k} and 4 \hat{j} - 3 \hat{k}$ and that P₂ is parallel to $\hat{j} - \hat{k} and 3 \hat{i} + 3 \hat{j}$ , then the angle between vector $\vec{A}$ and a given vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ is

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$\frac{3 π}{4}$

$\frac{π}{2}$

$\frac{π}{6}$

$\frac{π}{4}$

answer is B, D.

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Detailed Solution

Normal to plane P₁ is
${\vec{n}}_{1} = (2 \hat{j} + 3 \hat{k}) \times (4 \hat{j} - 3 \hat{k}) = - 18 \hat{i} .$
Normal to plane P₂ is
${\vec{n}}_{2} = (\hat{j} - \hat{k}) \times (3 \hat{i} + 3 \hat{j}) = 3 \hat{i} - 3 \hat{j} - 3 \hat{k}$
$∴ \vec{A} is parallel to \pm ({\vec{n}}_{1} \times {\vec{n}}_{2}) = \pm (- 54 \hat{j} + 54 \hat{k})$
$∴ Angle between \vec{A} and 2 \hat{i} + \hat{j} - 2 \hat{k} is$
$\begin{array}{l} \cos θ = \pm \frac{(- 54 \hat{j} + 54 \hat{k}) \cdot (2 \hat{i} + \hat{j} - 2 \hat{k})}{54 \sqrt{2} \cdot 3} = \pm \frac{1}{\sqrt{2}} \\ \Rightarrow θ = \frac{π}{4} (or) \frac{3 π}{4} \end{array}$

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