The plane ell x+my=0 is rotated about its lin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the equation of the plane after rotation be $P_{3}: \ell x+my+nz=0$.The line of intersection of the planes $P_{1}: \ell x+my=0$ and $P_{2}: z=

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the equation of the plane after rotation be $P_{3}: \ell x+my+nz=0$.The line of intersection of the planes $P_{1}: \ell x+my=0$ and $P_{2}: z=0$ is the line where $\ell x+my=0$ and $z=0$.The normal vectors are $\vec{n}_{1} = (\ell, m, 0)$ and $\vec{n}_{3} = (\ell, m, n)$.The angle $\alpha$ between the planes $P_{1}$ and $P_{3}$ is given by $\cos \alpha = \frac{|\vec{n}_{1} \cdot \vec{n}_{3}|}{|\vec{n}_{1}| |\vec{n}_{3}|}$.$\cos \alpha = \frac{|\ell^{2}+m^{2}|}{\sqrt{\ell^{2}+m^{2}} \sqrt{\ell^{2}+m^{2}+n^{2}}} = \sqrt{\frac{\ell^{2}+m^{2}}{\ell^{2}+m^{2}+n^{2}}}$.Squaring both sides, $\cos^{2} \alpha = \frac{\ell^{2}+m^{2}}{\ell^{2}+m^{2}+n^{2}}$.$\Rightarrow \cos^{2} \alpha (\ell^{2}+m^{2}+n^{2}) = \ell^{2}+m^{2}$.$\Rightarrow n^{2} \cos^{2} \alpha = (\ell^{2}+m^{2})(1 - \cos^{2} \alpha) = (\ell^{2}+m^{2}) \sin^{2} \alpha$.$\Rightarrow n^{2} = (\ell^{2}+m^{2}) 	an^{2} \alpha$.$\Rightarrow n = \pm \sqrt{\ell^{2}+m^{2}} 	an \alpha$.Substituting $n$ into the equation of $P_{3}$, we get $\ell x+my \pm z \sqrt{\ell^{2}+m^{2}} 	an \alpha = 0$.

The plane $\ell x+my=0$ is rotated about its line of intersection with the plane $z=0$ through an angle $\alpha$. The equation of the new plane is

Similar Questions

In the following case,determine the direction cosines of the normal to the plane and the distance from the origin: $x+y+z=1$

$A$ variable plane is at a constant distance $h$ from the origin and meets the coordinate axes in $A, B, C$. The locus of the centroid of $\triangle ABC$ is

Find the equation of a plane that makes equal intercepts of unit length on the axes.

Let $\pi$ be the plane that passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$. Then the foot of the perpendicular drawn from the point $(1, 2, 1)$ to the plane $\pi$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module