Question

(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.

Answer 1

(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
$-(x-2)-(y-1)+2(z-0) =0 $
$-(x-2)-y+1+2 z =0 $
$\Rightarrow x+y-2 z =3$
(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 $Q$
$\left(\frac{\alpha+2}{2}, \frac{\beta+1}{2}, \frac{\gamma+6}{2}\right),$
which lies in line $P Q$.
$\Rightarrow \frac{\frac{\alpha-2}{2}-2}{2} =\frac{\frac{\beta+1}{2}-1}{1} $
$=\frac{\frac{\gamma+6}{2}-6}{-2}$
$=\frac{1\left(\frac{\alpha+2}{2}-2\right)+1\left(\frac{\beta+1}{2}-1\right)-2\left(\frac{\gamma+6}{2}-6\right)}{1 \cdot 1+1 \cdot 1+(-2)(-2)}=2$
${\left[\because\left(\frac{\alpha+2}{2}\right)-1\left(\frac{\beta+1}{2}\right)-2\left(\frac{\gamma+6}{2}\right)=3\right] }$
$\Rightarrow \alpha=6, \beta=5, \gamma=-2$
$\Rightarrow Q(6,5,-2)$