If the distance between the plane 23x-10y-2z+48=0 and the plane containing the lines x+12=…

Question

If the distance between the plane $23x-10y-2z+48=0$ and the plane containing the lines $\dfrac{x+1}{2}=\dfrac{y-3}{4}=\dfrac{z+1}{3}$ and $\dfrac{x+3}{2}=\dfrac{y+2}{6}=\dfrac{z-1}{\lambda}$ ($\lambda\in\mathbb R$) is equal to $\dfrac{k}{\sqrt{633}}$, then $k$ is equal to \_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

Since both lines lie in the plane containing them, and $(-1,3,-1)$ is a point of that plane (on the first line), the distance between the two parallel planes equals the distance from $(-1,3,-1)$ to $23x-10y-2z+48=0$:
$$\frac{|23(-1)-10(3)-2(-1)+48|}{\sqrt{23^2+10^2+2^2}}=\frac{|-23-30+2+48|}{\sqrt{633}}=\frac{3}{\sqrt{633}}\implies k=3.$$

Solutions · 1