The Cartesian equation of the line passing through the point $(-1,3,-2)$ and perpendicular to the lines…
The Cartesian equation of the line passing through the point $(-1,3,-2)$ and perpendicular to the lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and
$\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$ is
$\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}$
$\frac{x-1}{-2}=\frac{y+3}{-7}=\frac{z-2}{-4}$
$\frac{x+1}{2}=\frac{y+3}{7}=\frac{z+2}{4}$
$\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$
Solution
$P=(-1,3,-2)$
Let $\mathrm{dr}$ 's or required line be $(a, b, c)$
Dr's of line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are $(1,2,3)$
Given that required line perpendicular to above line
$\Rightarrow \quad a_1 a_2+b_1 b_2+c_1 c_2=0$
$\Rightarrow \quad a+2 b+3 c=0$
Similarly required line perpendicular to
$
\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}
$
$\Rightarrow \quad-3 a+2 b+5 c=0$
On solving Eqs. (i) and (ii), we get
$
\begin{aligned}
& \frac{a}{10-6}=\frac{b}{-9-5}=\frac{c}{2+6} \\
& \Rightarrow \quad \frac{a}{2}=\frac{b}{-7}=\frac{c}{4} \\
&
\end{aligned}
$
$\therefore$ Required line is passing through $(-1,3,-2)$ and having dr's $(2,-7,4)$
$
\therefore \quad \frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}
$