Question

If $(2,3, - 1)$ is the foot of the perpendicular from $(4,2,1)$ to a plane , then the equation of that plane is $ax + by + cz = d$ then $a + d$ is ,
A) $3$
B) $1$
C) $- 2$
D) $2$

Answer 1

In this question try to find out the equation of plane for this $(2,3, - 1)$ is the foot of the perpendicular from $(4,2,1)$ to a plane so the Direction Ratio of normal to the plane is equal to $(2, - 1,2)$ hence equation of plane $2x - y + 2z = d$ . As we know that the point $(2,3, - 1)$ is lying on the plane, put the value get d from here .

Complete step-by-step answer:
In this equation we have to find the equation of plane that is $ax + by + cz = d$ ,
As it is given that the $(2,3, - 1)$ is the foot of the perpendicular from $(4,2,1)$ to a plane .
Hence the Direction Ratio of normal to the plane is $(4 - 2,2 - 3,1 + 1)$ or it is equal to $(2, - 1,2)$ .
So direction ratio of normal to the plane is $(2, - 1,2)$
Hence the equation of plane whose direction ratio of normal is $(2, - 1,2)$
$2x - y + 2z = d$ .
Hence the plane $2x - y + 2z = d$ is nothing but the equation of plane $ax + by + cz = d$ means that both planes are equal .
So on comparing we get $a = 2,b = - 1,c = 2$ .
Now we have to find the value of $d$ for this we know that the point $(2,3, - 1)$ is lie on the plane as it is the foot of perpendicular,
Hence it is on the plane $2x - y + 2z = d$
So on putting the value of $(2,3, - 1)$ on the plane $2x - y + 2z = d$
$2(2) - 3 + 2( - 1) = d$
$4 - 3 - 2 = d$
Hence $d = - 1$
So $a + d$ = $2 + ( - 1)$ = $1$

So, the correct answer is “Option B”.

Note: The equation of a plane which is parallel to each of the XY YZ, and ZX planes and passes through the $A(a,b,c)$ hence the equation of planes will be,
The equation of the plane which is parallel to the XY -plane is $z = c$ .
The equation of the plane which is parallel to the YZ-plane is $x = a$ .
The equation of the plane which is parallel to the ZX -plane is $y = b$