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)

$\left(21, \frac{5}{3}, \frac{10}{3}\right)$

Answer 2

The given equation of lines are $\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=k\,\text{(say)}\,\quad \ldots (1)$ and $\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}\quad\ldots\left(2\right)$ $\therefore$ 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 $\left(21, \frac{5}{3}, \frac{10}{3}\right)$.