Question

Foot of the perpendicular drawn from the point $(1,\frac{3}{2},2)$ to the plane $2x-2y+4z+5=0$ is

• A. $(\frac{-1}{2},3,-1)$
  
  
• B. $(0,\frac{5}{2},0)$
  
  
• C. $(\frac{5}{2},\frac{1}{2},5)$
  
  
• D. $(\frac{-1}{2},\frac{3}{2},-1)$
  
  

Answer 1

The correct option is B $(0,\frac{5}{2},0)$




$\text{Let}Q(a,b,c)$ be the foot of the perpendicular.
Direction ratio of $PQ$ is
$(a-1,b-\frac{3}{2},c-2)$
Direction ratios of the normal to the plane is $2,-2,4.$
$\because$the two direction ratios are parallel
$\therefore a-1=2\lambda ;b-\frac{3}{2}=-2\lambda ;c-2=4\lambda$
$\Rightarrow a=2\lambda +1;b=-2\lambda +\frac{3}{2};c=4\lambda +2$
$\because Q(a,b,c)$ lies on the plane $2x-2y+4z+5=0$
$\therefore 2(2\lambda +1)-2(-2\lambda +\frac{3}{2})+4(4\lambda +2)+5=0$
$\Rightarrow \lambda =-\frac{1}{2}$
$\therefore$ Coordinates of $Q$ are $(0,\frac{5}{2},0).$