Question

If the foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$, then the equation of the plane is

• A. $2x + y + 3z = 14$
• B. $x + 2y + 3z = 14$
• C. $x + 2y + 3z + 14 = 0$
• D. $x + 2y - 3z = 14$

Hint:

Plane from Foot of Perpendicular.
Vector from origin to foot gives normal. Use point-normal form: $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.

Answer 1

The Correct Option is B

Direction vector of perpendicular = $(1,2,3)$ $\rightarrow$ normal to the plane. Using point-normal form: \[ (x-1)(1) + (y-2)(2) + (z-3)(3) = 0 \Rightarrow x + 2y + 3z = 14 \]

Answer 2

Step 1: Understand the problem
The foot of the perpendicular from the point \((0,0,0)\) to a plane is given as \((1,2,3)\). We need to find the equation of the plane.

Step 2: Use the foot of perpendicular property
If the foot of the perpendicular from point \(P_0 (x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is \(P_1 (x_1, y_1, z_1)\), then:
- The vector from \(P_0\) to \(P_1\) is perpendicular to the plane.
- The normal vector to the plane is \(\vec{n} = (A, B, C)\).
- The point \(P_1\) lies on the plane.

Step 3: Write the vector from origin to foot and the normal vector
The vector \(\overrightarrow{OP_1} = (1, 2, 3)\) is along the normal vector \(\vec{n} = (A, B, C)\), so:
\[ (A, B, C) = \lambda (1, 2, 3) \] for some scalar \(\lambda\).

Step 4: Write the plane equation using the normal vector and foot point
The plane passes through \((1, 2, 3)\), so:
\[ A(x - 1) + B(y - 2) + C(z - 3) = 0 \] Substitute \(A = \lambda\), \(B = 2\lambda\), \(C = 3\lambda\):
\[ \lambda(x - 1) + 2\lambda(y - 2) + 3\lambda(z - 3) = 0 \] Divide both sides by \(\lambda\) (assuming \(\lambda \neq 0\)):
\[ (x - 1) + 2(y - 2) + 3(z - 3) = 0 \] Simplify:
\[ x - 1 + 2y - 4 + 3z - 9 = 0 \] \[ x + 2y + 3z - 14 = 0 \] or \[ x + 2y + 3z = 14 \]

Final answer:
\[ \boxed{x + 2y + 3z = 14} \]