Question

If the foot of the perpendicular drawn from the point (2,0,-3) to the plane $\pi$ is (1,-2,0) and the equation of the plane is $ax+by-3z+d=0$ then $a+b+d=$

• A. 0
• B. 1
• C. 6
• D. 2

Hint:

The vector connecting a point to the foot of its perpendicular on a plane is parallel to the plane's normal vector. This allows you to find the direction ratios of the normal. The foot of the perpendicular must also lie on the plane, which helps find the constant term 'd'.

Answer 1

The Correct Option is C

Let the given point be $P(2,0,-3)$ and the foot of the perpendicular on the plane be $Q(1,-2,0)$.
The line segment PQ is perpendicular to the plane.
Therefore, the vector $\vec{PQ}$ is parallel to the normal vector of the plane.
$\vec{PQ} = \vec{Q} - \vec{P} = (1-2)\hat{i} + (-2-0)\hat{j} + (0-(-3))\hat{k} = -1\hat{i} - 2\hat{j} + 3\hat{k}$.
The direction ratios of the normal to the plane are $(-1, -2, 3)$.
The equation of the plane is given as $ax+by-3z+d=0$.
The direction ratios of the normal from the given equation are $(a, b, -3)$.
Since the two normal vectors are parallel, their direction ratios must be proportional.
$\frac{a}{-1} = \frac{b}{-2} = \frac{-3}{3}$.
From the third ratio, we get $\frac{-3}{3} = -1$. Let this be the constant of proportionality.
$\frac{a}{-1} = -1 \implies a = 1$.
$\frac{b}{-2} = -1 \implies b = 2$.
So, the equation of the plane is $1x+2y-3z+d=0$.
Since the point Q(1,-2,0) lies on the plane, its coordinates must satisfy the equation.
Substitute the coordinates of Q to find $d$:
$1(1) + 2(-2) - 3(0) + d = 0$.
$1 - 4 - 0 + d = 0 \implies -3 + d = 0 \implies d = 3$.
The required value is the sum $a+b+d$.
$a+b+d = 1 + 2 + 3 = 6$.