3D Geometry 3 Question 17
####17. The perpendicular distance from the origin to the plane containing the two lines, $\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$, is
(2019 Main, 12 Jan I)
(a) $11 \sqrt{6}$
(b) $\frac{11}{\sqrt{6}}$
(c) 11
(d) $6 \sqrt{11}$
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Answer:
(b)
Solution:
- Let the equation of plane, containing the two lines
$ \frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7} \text { and } \frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7} \text { is } $
$ a(x+2)+b(y-2)+c(z+5)=0 $
$\because$ Plane (i) contain lines, so
$ 3 a+5 b+7 c=0 $
and $\quad a+4 b+7 c=0$
From Eqs. (ii) and (iii), we get
$ \begin{aligned} \frac{a}{35-28}=\frac{b}{7-21} & =\frac{c}{12-5} \\ \Rightarrow \quad \frac{a}{7} & =\frac{b}{-14}=\frac{c}{7} \Rightarrow \frac{a}{1}=\frac{b}{-2}=\frac{c}{1} \end{aligned} $
So, equation of plane will be
$ \begin{array}{rlrl} & 1(x+2)-2(y-2)+1(z+5) & =0 \\ \Rightarrow & x-2 y+z+11 =0 \end{array} $
Now, perpendicular distance from origin to plane is
$ =\frac{11}{\sqrt{1+4+1}}=\frac{11}{\sqrt{6}} $
$[\because$ perpendicular distance from origin to the plane
$ \left.a x+b y+c z+d=0, \text { is } \frac{|d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\right] $
