Foot of perpendicular from point to a plane in 3D — find coordinates

Question

Find the coordinates of the foot of the perpendicular drawn from the point $A(1, 2, 3)$ to the plane $2x + 3y - z + 4 = 0$ . Also find the perpendicular distance.

(CBSE 2024, similar pattern)

Solution — Step by Step

The perpendicular from $A$ to the plane is along the normal to the plane. The normal direction from the equation $2x + 3y - z + 4 = 0$ is $\vec{n} = (2, 3, -1)$ .

The perpendicular line through $A(1, 2, 3)$ along direction $(2, 3, -1)$ :

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{-1} = t

So any point on this line is $(1 + 2t,\ 2 + 3t,\ 3 - t)$ .

The foot of perpendicular $F$ lies on the plane. Substitute into $2x + 3y - z + 4 = 0$ :

2(1 + 2t) + 3(2 + 3t) - (3 - t) + 4 = 0

2 + 4t + 6 + 9t - 3 + t + 4 = 0

14t + 9 = 0

t = -\frac{9}{14}

x = 1 + 2\left(-\frac{9}{14}\right) = 1 - \frac{9}{7} = -\frac{2}{7}

y = 2 + 3\left(-\frac{9}{14}\right) = 2 - \frac{27}{14} = \frac{1}{14}

z = 3 - \left(-\frac{9}{14}\right) = 3 + \frac{9}{14} = \frac{51}{14}

Foot of perpendicular: $F = \left(-\dfrac{2}{7},\ \dfrac{1}{14},\ \dfrac{51}{14}\right)$

Using the distance formula:

d = \frac{|2(1) + 3(2) - 1(3) + 4|}{\sqrt{4 + 9 + 1}} = \frac{|2 + 6 - 3 + 4|}{\sqrt{14}} = \frac{9}{\sqrt{14}} = \mathbf{\frac{9\sqrt{14}}{14}}

Why This Works

The foot of perpendicular is the point on the plane closest to $A$ . The shortest path from a point to a plane is always along the normal direction. So we travel from $A$ in the direction of $\vec{n}$ until we hit the plane.

The parametric approach is general — it works for perpendicular from a point to any plane or even to a line. The direct distance formula $d = |ax_1 + by_1 + cz_1 + d|/\sqrt{a^2 + b^2 + c^2}$ gives only the distance, not the foot coordinates.

Alternative Method

You can use the projection formula directly. The foot $F = A - \frac{(\vec{n} \cdot \vec{OA} + d)}{|\vec{n}|^2}\vec{n}$ , where $d$ is the constant term. This is a one-step vector formula that gives $F$ without the parametric approach.

For CBSE 12th boards, this is a standard 5-mark question. The parametric line method is cleaner and easier to follow. For JEE MCQs, the direct formula is faster. Know both approaches.

Common Mistake

Students sometimes use the direction $(a, b, c)$ from the general equation but forget the sign of $c$ . In the plane $2x + 3y - z + 4 = 0$ , the coefficient of $z$ is $-1$ , not $+1$ . The normal direction is $(2, 3, -1)$ . Using $(2, 3, 1)$ gives a completely wrong foot of perpendicular. Copy the coefficients directly from the equation, signs included.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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