Let the plane P: 4x-y+z=10 be rotated by an angle ( )/(2) about its line of intersection w…

Question

Let the plane P: $4x-y+z=10$ be rotated by an angle $\frac{\pi }{2}$ about its line of intersection with the plane $x+y-z=4$. If $\alpha $ is the distance of the point $(2,3,-4)$ from the new position of the plane $P$, then $35\alpha $ is equal to :

Accepted Answer

Equation of plane after rotation :

$\begin{gathered} (4x-y+z-10)+\lambda (x+y-z-y)=0 \ \ \Rightarrow (4+\lambda )x+y(\lambda -1)+z(1-\lambda )-4\lambda -10=0 \ \ \overset{	o }{{n}_{1}}\cdot \overset{	o }{{n}_{2}}=0 \ \ \Rightarrow (4+\lambda )4+(\lambda -1)(-1)+(1-\lambda )1=0 \ \ \Rightarrow 16+4\lambda -\lambda +1+1-\lambda =0 \ \ \Rightarrow 2\lambda =-18 \ \ \Rightarrow \lambda =-9 $

$	herefore $ equation of plane : $-5x-10y+10z+26=0$

Distance of plane from $(2,3,-4)$

$ =|\frac{-10-30-40+26}{\sqrt{100+100+26}}|=\frac{54}{15}=\alpha \ \ 	herefore 35\alpha =35\cdot \frac{54}{15}=7	imes \frac{54}{3}=7	imes 18=126 \end{gathered}$

Solutions · 1