Let the plane containing the line of intersection of the planes P_1: x+(+4)y+z=1 and P_2: …

Question

Let the plane containing the line of intersection of the planes $P_1:\ x+(\lambda+4)y+z=1$ and $P_2:\ 2x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2\lambda,\lambda,-\lambda)$ from the plane $P_2$ is:

Accepted Answer

Equation of plane passing through point of intersection of $\mathrm{P} 1$ and $\mathrm{P} 2$

$$
\begin{aligned}
& \mathrm{P}_1+\mathrm{kP}_2 = 0 \\
& (\mathrm{x}+(\lambda+4) \mathrm{y}+\mathrm{z}-1)+\mathrm{k}(2 \mathrm{x}+\mathrm{y}+\mathrm{z}-2)=0
\end{aligned}
$$

Passing through $(0,1,0)$ and $(1,0,1)$

$$
\begin{aligned}
& (\lambda+4-1)+\mathrm{k}(1-2)=0 \\
& (\lambda+3)-\mathrm{k}=0\quad...(1)
\end{aligned}
$$

Also passing $(1,0,1)$

$$
\begin{aligned}
& (1+1-1)+\mathrm{k}(2+1-2)=0 \\
& 1+\mathrm{k}=0 \\
& \mathrm{k}=-1
\end{aligned}
$$

put in (1)

$$
\begin{aligned}
& \lambda+3+1=0 \\
& \lambda=-4
\end{aligned}
$$

Then point $(2 \lambda, \lambda,-\lambda)$

$$
\begin{aligned}
& (-8,-4,4) \\
& d=\left|\frac{-16-4+4-2}{\sqrt{6}}\right| \\
& d=\frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=3 \sqrt{6}
\end{aligned}
$$

Solutions · 1