---
title: "Plane through three points and distance"
subject: "Computer Aided Design"
chapter: 4
key: "semester_4/computer_aided_design:plane-surface-three-points"
source: https://www.bctengineer.com/q/semester_4/computer_aided_design/plane-surface-three-points
site: BCT Engineering
---

# Plane through three points and distance

Computer Aided Design · Chapter 4: Representation of Surfaces and Solids

- Asked in: year not given, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/computer_aided_design/plane-surface-three-points
- Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-4

## Question

Find the equation of the plane that passes through the points A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3). Also find the unit normal, the parametric form of the plane, and the perpendicular distance of the point Q(2, 2, 2) from the plane.

## Short answer

The plane is 6x+3y+2z=6, unit normal (0.857, 0.429, 0.286), parametric form P(u,w)=(1-u-w, 2u, 3w); distance of Q(2,2,2) is 2.286 units.

## Answer

#### Equation of plane
Two vectors in the plane:

$$
\mathbf{AB}=(-1,2,0),\qquad \mathbf{AC}=(-1,0,3)
$$

Normal:

$$
\mathbf n=\mathbf{AB}\times\mathbf{AC}=
\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-1&2&0\\-1&0&3\end{vmatrix}
=(6-0)\mathbf i-(-3-0)\mathbf j+(0+2)\mathbf k=(6,\ 3,\ 2)
$$

Plane: $6x+3y+2z=d$. Using A: $d=6$.

$$
6x+3y+2z=6\quad\Longleftrightarrow\quad \frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1
$$

Check B: $6(0)+3(2)+0=6$. Check C: $0+0+2(3)=6$. 

#### Unit normal
$|\mathbf n|=\sqrt{36+9+4}=7$, so $\hat{\mathbf n}=(0.857,\ 0.429,\ 0.286)$.

#### Parametric form
$$
\mathbf P(u,w)=\mathbf A+u\,\mathbf{AB}+w\,\mathbf{AC}=(1-u-w,\ 2u,\ 3w)
$$

where $u,w$ in $[0,1]$ give the part of the plane bounded by the parallelogram on $\mathbf{AB}$ and $\mathbf{AC}$.

#### Distance of Q(2, 2, 2)
$$
D=\frac{|6(2)+3(2)+2(2)-6|}{7}=\frac{16}{7}=2.286
$$

**Answer: plane 6x + 3y + 2z = 6; unit normal (0.857, 0.429, 0.286); distance of Q = 2.286 units.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the plane normal found?
   A. AB x AC
   B. AB . AC
   C. A x B
   D. AB + AC
   Answer: A. The cross product of two in-plane vectors gives the normal (6,3,2). (Key point: Normal)

2. What is the plane equation?
   A. x + 2y + 3z = 6
   B. 6x + 3y + 2z = 6
   C. 6x + 3y + 2z = 1
   D. 3x + 6y + 2z = 6
   Answer: B. Intercept form x/1 + y/2 + z/3 = 1 gives 6x + 3y + 2z = 6. (Key point: Plane equation)

3. What is the distance of Q(2,2,2)?
   A. 22/7
   B. 16
   C. 16/7 = 2.286
   D. 2.0
   Answer: C. |12 + 6 + 4 - 6|/7 = 16/7. (Key point: Distance)

4. What is the parametric form?
   A. P = u A + w B
   B. P = AB + AC
   C. P = A x B
   D. P = A + u AB + w AC
   Answer: D. P(u,w) = (1-u-w, 2u, 3w). (Key point: Parametric form)

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