Question

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

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

Answer 1

The correct option is A $(21,\frac{5}{3},\frac{10}{3})$

$\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=\lambda .............\left(1\right)$

$\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}=\mu .............\left(2\right)$

general point on line (1) is

$(3\lambda +5,-\lambda +7,\lambda -2)$

general point on line (2) is

$(-36\mu -3,2\mu +3,4\mu +6)$

Equating x coordinate

$3\lambda +5=-36\mu -3$

$3\lambda +36\mu +8=\mathrm{0................}\left(iii\right)$

Equating y coordinate

$-\lambda +7=2\mu +3$

$\lambda +2\mu -4=\mathrm{0................}\left(iv\right)$

Doing $1\times \left(iii\right)-3\times \left(iv\right)$

By solving equations $\left(iii\right)$ and $\left(iv\right)$ , we get

$\mu =\frac{-2}{3}$

and $\lambda =\frac{16}{3}$

Now by putting values of $\lambda$ in general point of line (1) or

$\mu$ in general point of line (2)

point of intersection of line (1) and (2) is $(21,\frac{5}{3},\frac{10}{3})$

Therefore option (A) is Correct.