#### The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove the equation of the plane in its new position is $ax+by\pm \left ( \sqrt{a^{2}+b^{2}} \tan \alpha \right )z=0$

Given, the plane ax + by = 0 is rotated about its line of intersection with  z = 0 by an angle $\alpha$

To prove: equation of the plane in its new position is

$ax+by\pm z\sqrt{a^{2}+b^{2}}\tan \alpha=0$

Proof: Two planes are given, ax + by = 0 …(i) and z = 0 …(ii)

We know, the equation of the plane passing through the line of intersection of the planes (i) and (ii) is

$ax + by + \lambda z = 0...(iii)$

where, $\lambda \epsilon \mathbb{R}$

The angle between the new plane and plane (i) is given as $\alpha$

Since the angle between two planes is equivalent to the angle between their normals, the direction ratio of normal to ax + by = 0 or ax + by +0z = 0 is (a, b, 0).

$\Rightarrow \vec{l}=a\hat{i}+b\hat{j}$

And, the direction ratio of normal to $ax + by + \lambda z = 0$ is (a, b, λ).

$\Rightarrow \vec{m}=a\hat{i}+b\hat{j}+\lambda \hat{k}$

Also, we know, angle between 2 normal vectors of the two given planes can be given as;

$\cos \alpha=\frac{\vec{l}\vec{m}}{\left |\vec{l} \right |\left |\vec{m} \right |}$

If we substitute the values of these vectors, we get

$\cos \alpha=\frac{\left (a\hat{i}+b\hat{j} \right )\left ( a\hat{i}+b\hat{j}+\lambda \hat{k} \right )}{\left |\left (a\hat{i}+b\hat{j} \right ) \right |\left |\left ( a\hat{i}+b\hat{j}+\lambda \hat{k} \right ) \right |}\\$

$\Rightarrow \cos \alpha=\frac{a.a+b.b+0.\lambda}{\sqrt{a^{2}+b^{2}}\sqrt{a^{2}+b^{2}+\lambda^{2}}}\\ \Rightarrow \cos \alpha=\frac{a^{2}+b^{2}}{\sqrt{a^{2}+b^{2}}\sqrt{a^{2}+b^{2}+\lambda^{2}}}$

We then multiply $\sqrt{a^{2}+b^{2}}$ by the numerator and denominator on the right hand side of the equation to get

$\Rightarrow \cos \alpha=\frac{a^{2}+b^{2}}{\sqrt{a^{2}+b^{2}}\sqrt{a^{2}+b^{2}+\lambda^{2}}}\times \frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}}}\\ \\ \Rightarrow \cos \alpha=\frac{\left (a^{2}+b^{2} \right )\sqrt{a^{2}+b^{2}}}{\left (a^{2}+b^{2} \right )\sqrt{a^{2}+b^{2}+\lambda^{2}}}\\ \\ \Rightarrow \cos \alpha=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}+\lambda^{2}}}\\ \\$

Applying square on both sides,

$\Rightarrow \cos^{2} \alpha=\left (\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}+\lambda^{2}}} \right )^{2}\\ \\ \Rightarrow \cos^{2} \alpha=\frac{a^{2}+b^{2}}{a^{2}+b^{2}+\lambda ^{2}}$
$\Rightarrow (a^{2} + b^{2} + \lambda^{2}) cos^{2} \alpha = a^{2} + b^{2}\\ \Rightarrow a^{2} \cos^{2} \alpha + b^{2} \cos^{2} \alpha + \lambda^{2} \cos^{2} \alpha = a^{2} + b^{2}\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = a^{2} + b^{2} - a^{2} \cos^{2} \alpha -b^{2}\cos^{2} \alpha\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = a^{2} - a^{2} \cos^{2} \alpha + b^{2} -b^{2} \cos^{2} \alpha\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = a^{2}(1 -\cos^{2} \alpha) + b^{2}(1 - \cos^{2} \alpha)\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = (a^{2} + b^{2})(1 - \cos^{2} \alpha)\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = (a^{2} + b^{2}) \sin^{2} \alpha [since, \sin^{2} \alpha + \cos^{2} \alpha = 1]$

$\Rightarrow \lambda^{2} = \frac{(a^{2} + b^{2}) \sin^{2} \alpha }{\cos^{2} \alpha}\\ Since \frac{sin^{2}\alpha}{\cos^{2}\alpha}=\tan^{2}\alpha\\ \Rightarrow \lambda^{2}=\left ( a^{2}+b^{2} \right )tan^{2}\alpha\\ \Rightarrow \lambda =\pm \sqrt{\left ( a^{2}+b^{2} \right )tan^{2}\alpha}\\ \Rightarrow \lambda =\pm \sqrt{\left ( a^{2}+b^{2} \right )}tan^{2}\alpha\\$

Substituting the value of $\lambda$ in equation (iii) to find the plane equation,

ax + by + λz = 0

$\lambda =\pm \sqrt{\left ( a^{2}+b^{2} \right )}tan^{2}\alpha\\$

Hence proved.