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  • 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 ax+bypm ( sqrt{a^{2}+b^{2}} tan alpha )=0 is z = 0.

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

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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.

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