Let P be the point of intersection of the line (x+3)/(3)=(y+2)/(1)=(1-z)/(2) and the plane…

Question

Let P be the point of intersection of the line $\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}$ and the plane $x+y+z=2$. If the distance of the point P from the plane $3x-4y+12z=32$ is q, then q and 2q are the roots of the equation :

Accepted Answer

Given, equation of line is

$$
\begin{aligned}
& \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}=k \\
& \therefore x=3 k-3, y=k-2, z=1-2 k
\end{aligned}
$$

Since, given that $P \equiv(3 k-3, k-2,1-2 k)$ be the point of intersection of the given line and the plane $x+y+z=2$

$$
\begin{aligned}
&\text { So, }(3 k-3)+(k-2)+(1-2 k)  =2 \\
&\Rightarrow 2 k-4=2 \Rightarrow k =3
\end{aligned}
$$

Thus, $P=(6,1,-5)$

Now, distance of point $P$ from the plane $3 x-4 y+12 z=32$ is

$$
\[\begin{gathered} q =\left|\frac{18-4-60-32}{\sqrt{9+16+144}}\right|=\left|\frac{-78}{\sqrt{169}}\right| \ \Rightarrow  q  =\left|\frac{-78}{13}\right|=\frac{78}{13}=6 \ \therefore  q  =6 \Rightarrow 2 q=12 \end{gathered}$$
\]

Thus, $q$ and $2 q$ are roots of the equation $x^2-18 x+72=0$

Solutions · 1