Question

The line which passes through the origin and intersect the two lines
$\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{3},\frac{x-4}{2}=\frac{y+3}{3}=\frac{z-14}{4},$ is

• A. $\frac{x}{1}=\frac{y}{-3}=\frac{z}{5}$
• B. $\frac{x}{-1}=\frac{y}{3}=\frac{z}{5}$
• C. $\frac{x}{1}=\frac{y}{3}=\frac{z}{-5}$
• D. $\frac{x}{1}=\frac{y}{4}=\frac{z}{-5}$

Answer 1

Answer: (A)

Let the line be $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}...(i)$
If line (i) intersects with the line
$\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{3}$,
then $\left|\begin{array}{ccc}a& b& c\\ 2& 4& 3\\ 4& -3& 14\end{array}\right|=0$
$\Rightarrow 9a-7b-10c=0$
from (i) and (ii), we have
$\frac{a}{1}=\frac{b}{-3}=\frac{c}{5}$
$\therefore$ The line is $\frac{x}{1}=\frac{y}{-3}=\frac{z}{5}$