Question

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 $\vec{r} = \hat{i} + \hat{j} + \hat{k} + λ (\hat{i} - \hat{j} - \hat{k}) .$ π₂ denotes the equation of the plane containing the line L and a point with position vector $⟨ 0, 1, 0 ⟩ .$

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Linked Question 1

The acute angle between π₁ and π₂, is

$\tan^{- 1} 1 - \tan^{- 1} (\sqrt{2} + 1)$

$\tan^{- 1} 1 - \tan^{- 1} (2 - \sqrt{3})$

$\tan^{- 1} 1 - \tan^{- 1} (\sqrt{2} - 1)$

${tan}^{- 1} 1 - \tan^{- 1} (2 + \sqrt{3})$

$\vec{r} \cdot {\vec{n}}_{1} = \vec{a} \cdot \vec{n}$

$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j}) = 2$

x + y = 2 ............ plane π₁

Now, $\vec{AB} = \hat{i} + \hat{k}$ ; A(1, 1, 1); B(0, 1, 0)

Now, $\vec{AB} \times \vec{V} = {\vec{n}}_{2} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & - 1 & - 1 \end{matrix} |$

${\vec{n}}_{2} = \hat{i} (0 + 1) - \hat{j} (- 1 - 1) + \hat{k} (- 1)$

$= {\vec{n}}_{2} = \hat{i} + 2 \hat{j} - \hat{k}$

Hence, equation of plane π₂ is $(\vec{r} - \hat{j}) \cdot (\hat{i} + 2 \hat{j} - \hat{k})$ = 0 or x + 2y – z = 2

If θ is the acute angle between π₁ and π₂ then

$\Rightarrow \cos θ = \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{| {\vec{n}}_{1} | | {\vec{n}}_{2} |}$

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

Linked Question 2

Vector of magnitude $\sqrt{6}$ along the line of intersection of planes π₁ and π₂ and perpendicular to normal vector of plane π₁, is

$\pm \sqrt{2} (1, - 1, 1)$

$\pm \sqrt{2} (1, 1, - 1)$

$\pm \sqrt{2} (1, 1, 1)$

$\pm \sqrt{2} (- 1, 1, 1)$

$\vec{r} \cdot {\vec{n}}_{1} = \vec{a} \cdot \vec{n}$

$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j}) = 2$

x + y = 2 ............ plane π₁

Now, $\vec{AB} = \hat{i} + \hat{k}$ ; A(1, 1, 1); B(0, 1, 0)

Now,

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

${\vec{n}}_{2} = \hat{i} (0 + 1) - \hat{j} (- 1 - 1) + \hat{k} (- 1)$

$= {\vec{n}}_{2} = \hat{i} + 2 \hat{j} - \hat{k}$

Hence, equation of plane π₂ is $(\vec{r} - \hat{j}) \cdot (\hat{i} + 2 \hat{j} - \hat{k})$ = 0 or x + 2y – z = 2

Vector along the line of intersection of planes π₁ and π₂ is ${\vec{n}}_{1} \times {\vec{n}}_{2}$ which is also perpendicular to the normal vector of plane π₁.

$∴ {\vec{n}}_{1} \times {\vec{n}}_{2} = - \hat{i} + \hat{j} + \hat{k}$

So, required vector $= \pm \sqrt{6} (\frac{- \hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})$

$= \pm \sqrt{2} (- \hat{i} + \hat{j} + \hat{k}) or \pm \sqrt{2} (- 1, 1, 1)$