The plane lx + my = 0 is rotated by an angle | vedclass

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(A) The equation of any plane passing through the line of intersection of the planes $lx + my = 0$ and $z = 0$ is given by $lx + my + \lambda z = 0$.T

Vedclass · Accepted Answer

$lx + my \pm z\sqrt{l^2 + m^2} 	an \alpha = 0$

Vedclass · Answer

(A) The equation of any plane passing through the line of intersection of the planes $lx + my = 0$ and $z = 0$ is given by $lx + my + \lambda z = 0$.The normal to the first plane $lx + my = 0$ is $\vec{n_1} = (l, m, 0)$.The normal to the new plane $lx + my + \lambda z = 0$ is $\vec{n_2} = (l, m, \lambda)$.The angle between the two planes is $\alpha$,which is the same as the angle between their normals. The cosine of the angle between the normals is given by:$\cos \alpha = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|l^2 + m^2|}{\sqrt{l^2 + m^2} \sqrt{l^2 + m^2 + \lambda^2}}$$\cos \alpha = \frac{l^2 + m^2}{\sqrt{l^2 + m^2} \sqrt{l^2 + m^2 + \lambda^2}} = \sqrt{\frac{l^2 + m^2}{l^2 + m^2 + \lambda^2}}$Squaring both sides:$\cos^2 \alpha = \frac{l^2 + m^2}{l^2 + m^2 + \lambda^2}$$(l^2 + m^2 + \lambda^2) \cos^2 \alpha = l^2 + m^2$$\lambda^2 \cos^2 \alpha = (l^2 + m^2)(1 - \cos^2 \alpha) = (l^2 + m^2) \sin^2 \alpha$$\lambda^2 = (l^2 + m^2) 	an^2 \alpha$$\lambda = \pm \sqrt{l^2 + m^2} 	an \alpha$Substituting $\lambda$ back into the equation,we get $lx + my \pm z\sqrt{l^2 + m^2} 	an \alpha = 0$.

The plane $lx + my = 0$ is rotated by an angle $\alpha$ about its line of intersection with the plane $z = 0$. Find the equation of the plane in its new position.

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