3D Geometry Ques 3
- A perpendicular is drawn from a point on the line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}$ to the plane $x+y+z=3$ such that the foot of the perpendicular $Q$ also lies on the plane $x-y+z=3$. Then, the coordinates of $Q$ are
(2019 Main, 10 April II)
(a) $(-1,0,4)$
(b) $(4,0,-1)$
(c) $(2,0,1)$
(d) $(1,0,2)$
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Answer:
Correct Answer: 3.(c)
Solution: (c) Key Idea Use the foot of perpendicular $Q\left(x_2, y_2, z_2\right)$ drawn from point $P\left(x_1, y_1, z_1\right)$ to the plane $a x+b y+c z+d=0$, is given by
$ \begin{gathered} \frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{z_2-z_1}{c} \\ =-\frac{a x_1+b y_1+c z_1+d}{a^2+b^2+c^2} \end{gathered} $
Let a general point on the given line
$ \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}=r $
is $P(2 r+1,-1-r, r)$.
Now, let foot of perpendicular $Q\left(x_1, y_1, z_1\right)$ to be drawn from point $P(2 r+1,-1-r, r)$ to the plane $x+y+z=3$, then
$ \begin{aligned} \frac{x_1-2 r-1}{1} & =\frac{y+1+r}{1}=\frac{z-r}{1} \\ & =-\left(\frac{2 r+1-1-r+r-3}{1+1+1}\right) \\ \Rightarrow \quad x_1-2 r-1 & =y+r+1 \\ & =z-r=\frac{1}{3}(3-2 r)=1-\frac{2}{3} r \\ \Rightarrow \quad x_1 & =2+\frac{4 r}{3}, y=-\frac{5 r}{3} \text { and } z=1+\frac{r}{3} \end{aligned} $
So, point $Q=\left(2+\frac{4 r}{3},-\frac{5 r}{3}, 1+\frac{r}{3}\right)$, lies on the plane $x-y+z=3$ also.
So, $\quad 2+\frac{4 r}{3}+\frac{5 r}{3}+1+\frac{r}{3}=3 \Rightarrow r=0$
Therefore, the coordinates of point $Q$ are $(2,0,1)$.
BETA
