A vector $\vec{a}$ 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 $\vec{a}$ and the vector $\vec{b} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ is

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$\frac{3 π}{4}$

$\frac{4 π}{5}$

$\frac{5 π}{6}$

$\frac{2 π}{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} {\vec{n}}_{1} = \hat{i} \times (\hat{i} + \hat{j}) = \hat{k} \\ {\vec{n}}_{2} = (\hat{i} + \hat{k}) \times (\hat{i} - \hat{j}) \\ = \hat{i} + \hat{j} - \hat{k} \end{array}$

Line of intersection along ${\vec{n}}_{1} \times {\vec{n}}_{2}$

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

D.R of $\vec{a} = - \hat{i} + \hat{j}$

D.R of $\vec{b} = \hat{i} - 2 \hat{j} + 2 \hat{k}$

$\vec{a} \cdot \vec{b} = - 3 and (\vec{a}, \vec{b}) = θ$

$\Rightarrow \cos θ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{- 3}{\sqrt{2} \times 3} = \frac{- 1}{\sqrt{2}} ∴ θ = \frac{3 π}{4}$

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