Question

If the distance between the plane, $23x-10y-2z+48=0$ and the plane containing the lines$\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}$ and $\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda }\left(\lambda ϵR\right)$ is equal to $\frac{k}{\sqrt{633}}$ then $k$ is equal to

Answer 1

lines $\frac{x+1}{2}=\frac{y-3}{4}=\frac{3+1}{3}=t$ and

$\frac{x+3}{2}=\frac{y+2}{6}=\frac{2-1}{\lambda }=S$

So point on line be.

$(2t-1,4t+3,3t-1)$ and $(2s-3,6s-2,\lambda s+1)$

as both point are same,

$\begin{array}{c}2t-1=2s-3\\ 2t=2s-2\\ t=s-\mathrm{1......}\left(1\right)\end{array}|\begin{array}{c}4t+3=6s-2\\ 4t=6s-5\\ .......\left(2\right)\end{array}|\begin{array}{c}3t-1=\lambda s+1\\ 3t=\lambda s+2\\ .........\left(3\right)\end{array}|$

solving (1) and (2)

$4t=4s-4$

$4t=6s-5$

- - +

$\overline{-2s+1=0}$

$s=\frac{1}{2},t=\frac{-1}{2}$

Put in (3) we get

$\frac{-3}{2}=\frac{\lambda }{2}+2$

$-3=\lambda +4$

$\lambda =-7$

So point is.

$(-2,1,\frac{-5}{2})$

distance from plane $23x-10y-2_3+48=0$

$\frac{|23(-2)-10\left(1\right)+2\left(\frac{5}{2}\right)+48|}{\sqrt{{23}^2+{10}^2+2^2}}=\frac{|-46-10+5+48|}{\sqrt{633}}$

$\frac{3}{\sqrt{633}}$

by comparing we get

$K=3$