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

Question

Let the plane P: $4 x-y+z=10$ 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 $$\mathrm{P}$$, then $$35 \alpha$$ is equal to :

Accepted Answer

Equation of plane after rotation :

$$
\begin{aligned}
& (4 x-y+z-10)+\lambda(x+y-z-y)=0 \\
\Rightarrow & (4+\lambda) x+y(\lambda-1)+z(1-\lambda)-4 \lambda-10=0 \\
& \overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 \\
\Rightarrow & (4+\lambda) 4+(\lambda-1)(-1)+(1-\lambda) 1=0 \\
\Rightarrow & 16+4 \lambda-\lambda+1+1-\lambda=0 \\
\Rightarrow & 2 \lambda=-18 \\
\Rightarrow & \lambda=-9
\end{aligned}
$$

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

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

$$
\begin{aligned}
& =\left|\frac{-10-30-40+26}{\sqrt{100+100+26}}\right|=\frac{54}{15}=\alpha \\
 \therefore 35 \alpha & =35 \cdot \frac{54}{15}=7 \times \frac{54}{3}=7 \times 18=126
\end{aligned}
$$

Solutions · 1