Let [P] be the plane containing the straight line [(x-3)/(9)=(y+4)/(-1)=(z-7)/(-5)] and pe…

Question

Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the point $(2,-5,11)$, then $$\mathrm{d}^{2}$$ is equal to :

Accepted Answer

Let $\langle a, b, c\rangle$ be direction ratios of plane containing

$$
\begin{aligned}
&\text { lines } \frac{x}{2}=\frac{y}{3}=\frac{z}{5} \text { and } \frac{x}{3}=\frac{y}{7}=\frac{z}{8} . \\
&\therefore \quad 2 a+3 b+5 c=0 \quad \ldots \text { (i) } \\
&\text { and } 3 a+7 b+8 c=0 \quad \ldots \text { (ii) } \\
&\text { from eq. (i) and (ii) }: \frac{a}{24-35}=\frac{b}{15-16}=\frac{c}{14-9} \\
&\therefore \text { D. R's}  \text {. of plane are }<11,1,-5>
\end{aligned}
$$

Let $D . R$'s of plane $P$ be $$ then.

$11 a_{1}+b_{1}-5 c_{1}=0$

and $9 a_{1}-b_{1}-5 c_{1}=0$

From eq. (iii) and (iv) :

$$
\begin{aligned}
&\frac{a\_{1}}{-5-5}=\frac{b\_{1}}{-45+55}=\frac{c\_{1}}{-11-9} \\
&\therefore \text { D.R's }  \text { of plane } P \text { are }
\end{aligned}
$$

Equation plane $P$ is : $1(x-3)-1(y+4)+2(z-7)=0$

$$

Solutions · 1