3D Geometry 3 Question 66
####66. (i) Find the equation of the plane passing through the points $(2,1,0),(5,0,1)$ and $(4,1,1)$.
(ii) If $P$ is the point $(2,1,6)$, then the point $Q$ such that $P Q$ is perpendicular to the plane in (a) and the mid-point of $P Q$ lies on it.
$(2003,4$ M)
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Answer:
(i) ($x+y-2_{z}$)
(ii) $Q(6,5,-2)$
Solution:
- (i) Equation of plane passing through $(2,1,0)$ is
$a(x-2)+b(y-1)+c(z-0)=0$
It also passes through $(5,0,1)$ and $(4,1,1)$.
$\Rightarrow 3 a-b+c=0$ and $2 a-0 b+c=0$
On solving, we get $\frac{a}{-1}=\frac{b}{-1}=\frac{c}{2}$
$\therefore$ Equation of plane is
$ \begin{aligned} -(x-2)-(y-1)+2(z-0) & =0 \\ -(x-2)-y+1+2 z & =0 \\ x & \end{aligned} $
(ii) Let the coordinates of $Q$ be $(\alpha, \beta, \gamma)$.
Equation of line $P Q \Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}$
Since, mid-point of $P$ and $Q$
$ \frac{\alpha+2}{2}, \frac{\beta+1}{2}, \frac{\gamma+6}{2} $
which lies in line $P Q$.
$ \Rightarrow \quad \frac{\frac{\alpha-2}{2}-2}{2}=\frac{\frac{\beta+1}{2}-1}{1}=\frac{\frac{\gamma+6}{2}-6}{-2} $
$ \begin{gathered} =\frac{1 \frac{\alpha+2}{2}-2+1 \frac{\beta+1}{2}-1-2 \frac{\gamma+6}{2}-6}{1 \cdot 1+1 \cdot 1+(-2)(-2)}=2 \\ \because \frac{\alpha+2}{2}-1 \frac{\beta+1}{2}-2 \frac{\gamma+6}{2}=3 \\ \Rightarrow \quad \alpha=6, \beta=5, \gamma=-2 \Rightarrow Q(6,5,-2) \end{gathered} $
