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 $PQ$ is perpendicular to the plane in (a) and the mid-point of $PQ$ 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 $(4,1,1)$.
$\Rightarrow 3a-b+c=0$ and $2a-0b+c=0$
On solving, we get
$\frac{a}{-1}=\frac{b}{-1}=\frac{c}{2}$
$\therefore$ Equation of plane is
$-(x-2)-(y-1)+2(z-0)=0$
$-(x-2)-y+1+2z=0$
$\Rightarrow x+y-2z=3$
(ii) Let the coordinates of $Q$ be $(\alpha ,\beta ,\gamma )$.
Equation of line $PQ\Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}$
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 $PQ$.
$\Rightarrow \frac{\frac{\alpha -2}{2}-2}{2}=\frac{\frac{\beta +1}{2}-1}{1}$
$=\frac{\frac{\gamma +6}{2}-6}{-2}$
$=\frac{1\left(\frac{\alpha +2}{2}-2\right)+1\left(\frac{\beta +1}{2}-1\right)-2\left(\frac{\gamma +6}{2}-6\right)}{1\cdot 1+1\cdot 1+(-2)(-2)}=2$
$\left[\because \left(\frac{\alpha +2}{2}\right)-1\left(\frac{\beta +1}{2}\right)-2\left(\frac{\gamma +6}{2}\right)=3\right]$
$\Rightarrow \alpha =6,\beta =5,\gamma =-2$
$\Rightarrow Q(6,5,-2)$