Question

Show that the equation of the plane through the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+1}{2}$ and parallel to $\frac{x-2}{2}=\frac{y-1}{3}=\frac{z+4}{5}$

Answer 1

Consider the problem

Since, required plane is passes through

$\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+1}{2}...\left(1\right)$

So, it passes through point $(1,-6,-1)$

Since, it is passes through above line $\left(1\right)$ and parallel to

$\frac{x-2}{2}=\frac{y-1}{3}=\frac{z+4}{5}$

So, normal to the plane is perpendicular to vectors

$\overrightarrow{a}=3\hat{i}+4\hat{j}+2\hat{k}and\overrightarrow{b}=2\hat{i}+3\hat{j}+5\hat{k}$

Equation of plane passing through $(1,-6,-1)$ is

$A\left(x-1\right)+B\left(y+6\right)+C\left(z+1\right)=0...\left(1\right)$

since, it is $\perp$ to vectors $\overrightarrow{a}and\overrightarrow{b}$

So,

$\begin{array}{c}3A+4B+2C=0...\left(2\right)\\ 2A+3B+5C=0.....\left(3\right)\end{array}$

Now, Solving $\left(2\right)$ and $\left(3\right)$ we get

$\begin{array}{c}\frac{A}{20-6}=\frac{B}{4-15}=\frac{C}{9-8}\\ \Rightarrow \frac{A}{14}=\frac{B}{-11}=\frac{C}{1}=k\end{array}$

$\Rightarrow A=14k,B=-11k,C=k$

Now putting values of $A$, $B$ and $C$ in $\left(1\right)$ we get,

$\begin{array}{c}14k\left(x-1\right)+\left(-11k\right)\left(y+6\right)+k\left(z+1\right)=0\\ \Rightarrow 14\left(x-1\right)-11\left(y+6\right)+z+1=0\\ \Rightarrow 14x-14-11y-66+z+1=0\\ \Rightarrow 14x-11y+z-79=0\end{array}$