Let π1 denotes the equation of the plane to which the vector (1, 1, 0) is normal and which contains the line L whose equation is r→=i^+j^+k^+λ(i^−j^−k^) and π2 denotes the equation of the plane containing the line L and a point whose position vector (0, 1, 0).

Let π1 denotes the equation of the plane to which the vector (1, 1, 0) is normal and which contains the line L whose equation is r→=i^+j^+k^+λ(i^−j^−k^) and π2 denotes the equation of the plane containing the line L and a point whose position vector (0, 1, 0).

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Let $π_{1}$ denotes the equation of the plane to which the vector (1, 1, 0) is normal and which contains the line L whose equation is $\vec{r} = \hat{i} + \hat{j} + \hat{k} + λ (\hat{i} - \hat{j} - \hat{k})$ and $π_{2}$ denotes the equation of the plane containing the line L and a point whose position vector (0, 1, 0).

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The acute angle between $π_{1}$ and $π_{2}$ is $\tan^{- 1} 1 - \tan^{- 1} (\sqrt{2} - 1)$

The acute angle between $π_{1}$ and $π_{2}$ can be $\tan^{- 1} 1 - \tan^{- 1} (2 - \sqrt{3})$

The vector along a normal to the plane can be $π_{2}$ is $\hat{i} + 2 \hat{j} - \hat{k}$

The vector along a normal to the plane $π_{2}$ is $\hat{i} - 2 \hat{j} + \hat{k}$

answer is B, C.

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x + y = 2 …….. plane $π_{1}$

$\bar{A B} \times \vec{V} = {\vec{n}}_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & - 1 & - 1 \end{matrix} |$

${\vec{n}}_{2} = \hat{i} + 1 \hat{j} - \hat{k} \cos θ = \frac{{\vec{n}}_{1} . {\vec{n}}_{2}}{| {\vec{n}}_{1} | | {\vec{n}}_{2} |}$

$\cos θ = | \frac{(\hat{i} + \hat{j}) . (\hat{i} + 1 \hat{j} - \hat{k})}{\sqrt{2} \sqrt{2}} | = \frac{3}{\sqrt{2} . \sqrt{6}} = \frac{\sqrt{3}}{2} \Rightarrow θ = \cos^{- 1} \frac{\sqrt{3}}{2} \Rightarrow θ = \frac{π}{6}$

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Let π1 denotes the equation of the plane to which the vector (1, 1, 0) is normal and which contains the line L whose equation is r→=i^+j^+k^+λ(i^−j^−k^) and π2 denotes the equation of the plane containing the line L and a point whose position vector (0, 1, 0).

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