Discussion Forum



1)

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:

Option A

Explanation:

Let the line be $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$ ...(1)

If line (1) intersects with the line $\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{3}$, then

$\begin{vmatrix}a & b & c\\2& 4 & 3\\4&-3&14 \end{vmatrix}=0$

9a - 7b - 10c = 0   ...(2)

from (1) and (2) , we have,

$\frac{a}{1}=\frac{b}{-3}=\frac{c}{5}$

.'. The line  is $\frac{x}{1}=\frac{y}{-3}=\frac{z}{5}$