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 $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\quad\ldots\left(1\right)$
If line $\left(1\right)$ intersects the line $\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{3}$, then
$\begin{vmatrix}a&b&c\\ 2&4&3\\ 1&-3&5\end{vmatrix}=0$
$\Rightarrow 29a - 7b-10c = 0\quad\ldots\left(i\right)$
Similarly, if the line $\left(1\right)$ intersects with the line
$\frac{x-4}{2}=\frac{y+3}{3}=\frac{z-14}{4}$, then
$\begin{vmatrix}a&b&c\\ 2&3&4\\ 4&-3&14\end{vmatrix}=0$
$\Rightarrow 9a - 2b - 3c = 0 \quad\ldots\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we have $\frac{a}{1}=\frac{b}{-3}=\frac{c}{5}$
$\therefore $ The line is $\frac{x}{1}=\frac{y}{-3}=\frac{z}{5}$