Question

The equation of the line passing through the point $(1,2,-4)$ and perpendicular to the two lines
$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$ will be

• A. $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$
• B. $\frac{x-1}{-2}=\frac{y-2}{3}=\frac{z+4}{8}$
• C. $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+4}{8}$
• D. None of these

Answer 1

Answer: (A)

Let $a$, $b$, $c$, be the direction ratios of the required line.
Then, equation of line passing through $(1,2,-4)$ and having $d$.$r^{\prime }s(a,b,c)$ is
$\frac{x-1}{a}=\frac{y-2}{b}=\frac{z+4}{c}\dots \left(1\right)$
Now, line $\left(1\right)$ is perpendicular to the lines
$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$
$\therefore 3a-16b+7c=0\dots \left(2\right)$
$3a+8b-5c=0\dots \left(3\right)$
On solving $\left(2\right)$ and $\left(3\right)$, we get
$\frac{a}{24}=\frac{b}{36}=\frac{c}{72}$
i.e., $\frac{a}{2}=\frac{b}{3}=\frac{c}{6}$
So, required equation of line $\left(1\right)$ is
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$