Question

The point of intersection of the lines
$\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}$; $\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}$ is

• A. $\left(2,10,4\right)$
• B. $\left(21,\frac{5}{3},\frac{10}{3}\right)$
• C. $\left(5,7,-2\right)$
• D. $\left(-3,3,6\right)$

Answer 1

Answer: (B)

The given equation of lines are
$\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=k\text{(say)}\dots (1)$
and $\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}\dots \left(2\right)$
$\therefore$ Any point on line $(1)$ is $(3k+5,7-k,k-2)$.
It should lie on line $(2)$
$\therefore \frac{3k+5+3}{-36}=\frac{7-k-3}{2}=\frac{k-2-6}{4}$
On solving, we get $k=\frac{16}{3}$
$\therefore$ Coordinates of intersection point are $\left(21,\frac{5}{3},\frac{10}{3}\right)$.