3D Geometry 3 Question 23
####23. On which of the following lines lies the point of intersection of the line, $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane, $x+y+z=2$ ?
(2019 Main, 10 Jan II)
(a) $\frac{x-4}{1}=\frac{y-5}{1}=\frac{z-5}{-1}$
(b) $\frac{x+3}{3}=\frac{4-y}{3}=\frac{z+1}{-2}$
(c) $\frac{x-2}{2}=\frac{y-3}{2}=\frac{z+3}{3}$
(d) $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$
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Answer:
(d)
Solution:
- Given equation of line is
$ \begin{aligned} & \frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=r(\text { let }) \\ & \Rightarrow x=2 r+4 ; y=2 r+5 \text { and } z=r+3 \end{aligned} $
$\therefore$ General point on the line (i) is
$ P(2 r+4,2 r+5, r+3) $
So, the point of intersection of line (i) and plane $x+y+z=2$ will be of the form $P(2 r+4,2 r+5, r+3)$ for some $r \in R$.
$\Rightarrow(2 r+4)+(2 r+5)+(r+3)=2$
$[\because$ the point will lie on the plane]
$\Rightarrow 5 r=-10 \Rightarrow r=-2$
So, the point of intersection is $P(0,1,1)$
[putting $r=-2$ in $(2 r+4,2 r+5, r+3)$ ]
Now, on checking the options, we get
$\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$ contain the point $(0,1,1)$
