3D Geometry 3 Question 67
####67. If the distance between the plane $A x-2 y+z=d$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$, then $|d|$ is equal to….
(2010)
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Answer:
$|D|=6$
Solution:
- Equation of the plane containing the lines $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$
$ \text { is } a(x-2)+b(y-3)+c(z-4)=0 $
where, $3 a+4 b+5 c=0$
$ 2 a+3 b+4 c=0 $
and $a(1-2)+b(2-3)+c(2-3)=0$
i.e.
$ a+b+c=0 $
From Eqs. (ii) and (iii), $\frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$, which satisfy
Eq. (iv).
Plane through lines is $x-2 y+z=0$.
Given plane is $A x-2 y+z=d$ is $\sqrt{6}$.
$\therefore$ Planes must be parallel, so $A=1$ and then
$ \frac{|d|}{\sqrt{6}}=\sqrt{6} \Rightarrow|d|=6 $
