3D Geometry 3 Question 41
####41. The equation of a plane passing through the line of intersection of the planes $x+2 y+3 z=2$ and $x-y+z=3$ and at a distance $2 / \sqrt{3}$ from the point $(3,1,-1)$ is (2012)
(a) $5 x-11 y+z=17$
(c) $x+y+z=\sqrt{3}$
(b) $\sqrt{2} x+y=3 \sqrt{2}-1$
(d) $x-\sqrt{2} y=1-\sqrt{2}$
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Answer:
Correct Answer: 41. (a)
Solution:
- Key Idea
(i) Equation of plane through intersection of two planes,
$ \begin{aligned} & \text { i.e. }\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+\lambda \ & \qquad\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0 \end{aligned} $
(ii) Distance of a point $\left(x_{1}, y_{1}, z_{1}\right)$ from
$ \begin{aligned} a x+b y+c z+d & =0 \ & =\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} \end{aligned} $
Equation of plane passing through intersection of two planes $x+2 y+3 z=2$ and $x-y+z=3$ is
$ (x+2 y+3 z-2)+\lambda(x-y+z-3)=0 $
$\Rightarrow(1+\lambda) x+(2-\lambda) y+(3+\lambda) z-(2+3 \lambda)=0$ whose distance from $(3,1,-1)$ is $\frac{2}{\sqrt{3}}$.
$\Rightarrow \frac{|3(1+\lambda)+1 \cdot(2-\lambda)-1(3+\lambda)-(2+3 \lambda)|}{\sqrt{(1+\lambda)^{2}+(2-\lambda)^{2}+(3+\lambda)^{2}}}=\frac{2}{\sqrt{3}}$
$\Rightarrow \quad \frac{|-2 \lambda|}{\sqrt{3 \lambda^{2}+4 \lambda+14}}=\frac{2}{\sqrt{3}}$
$\Rightarrow \quad 3 \lambda^{2}=3 \lambda^{2}+4 \lambda+14$
$\Rightarrow \quad \lambda=-\frac{7}{2}$
$\therefore 1-\frac{7}{2} x+2+\frac{7}{2} \quad y+3-\frac{7}{2} z-2-\frac{21}{2}=0$
$\Rightarrow \quad-\frac{5 x}{2}+\frac{11}{2} y-\frac{1}{2} z+\frac{17}{2}=0$
or$ \quad 5 x-11 y+z-17=0 $
