Question

Find the equation of the plane through the following three points:
(1,2,3) (2,4,5) and (4,5,7)

Answer 1

Answer on Question #52767 – Math – Analytic Geometry

Question

Find the equation of the plane through the following three points:

(1,2,3) (2,4,5) and (4,5,7)

Solution

A(1,2,3), B(2,4,5) and C(4,5,7). Let X(x, y, z).

$$\overrightarrow{AB} = \langle 2, 1, 4, 2, 5, 3 \rangle = \langle 1, 2, 2 \rangle$$ $$\overrightarrow{AC} = \langle 4, 1, 5, 2, 7, 3 \rangle = \langle 3, 3, 4 \rangle$$ $$\overrightarrow{AX} = \langle x, 1, y, 2, z, 3 \rangle$$

Method 1

The cross product (or vector product) of $\overrightarrow{AC}$ and $\overrightarrow{AB}$ is given by

$$\begin{array}{l} \overrightarrow{AC} \times \overrightarrow{AB} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 3 & 3 & 4 \\ 1 & 2 & 2 \end{array} \right| = \vec{i} \left| \begin{array}{cc} 3 & 4 \\ 2 & 2 \end{array} \right| - \vec{j} \left| \begin{array}{cc} 3 & 4 \\ 1 & 2 \end{array} \right| + \vec{k} \left| \begin{array}{cc} 3 & 3 \\ 1 & 2 \end{array} \right| = (3 \cdot 2 - 2 \cdot 4) \vec{i} - (3 \cdot 2 - 1 \cdot 4) \vec{j} + \\ + (3 \cdot 2 - 1 \cdot 3) \vec{k} = -2 \vec{i} - 2 \vec{j} + 3 \vec{k}. \end{array}$$

Since $\overrightarrow{AX} \perp (\overrightarrow{AC} \times \overrightarrow{AB})$, therefore the dot product (or scalar product) of $\overrightarrow{AX}$ and $\overrightarrow{AC} \times \overrightarrow{AB}$ is zero.

Thus,

$$\begin{array}{l} < x - 1, y - 2, z - 3 > \cdot < -2, -2, 3 > = -2(x - 1) - 2(y - 2) + 3(z - 3) = 0 \\ -2x + 2 - 2y + 4 + 3z - 9 = 0 \\ -2x - 2y + 3z - 3 = 0 \\ \end{array}$$

Multiply by (-1) and obtain

$$2x + 2y - 3z + 3 = 0$$

Method 2

Vectors $\overrightarrow{AX}, \overrightarrow{AB}, \overrightarrow{AC}$ lie in one plane, therefore their scalar triple product is zero, therefore

$$\begin{array}{l} \left| \begin{array}{ccc} x - 1 & y - 2 & z - 3 \\ 1 & 2 & 2 \\ 3 & 3 & 4 \end{array} \right| = (x - 1) \left| \begin{array}{cc} 2 & 2 \\ 3 & 4 \end{array} \right| - (y - 2) \left| \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right| + (z - 3) \left| \begin{array}{cc} 1 & 2 \\ 3 & 3 \end{array} \right| = \\ = (x - 1)(2 \cdot 4 - 3 \cdot 2) - (y - 2)(1 \cdot 4 - 3 \cdot 2) + (z - 3)(1 \cdot 3 - 3 \cdot 2) \\ = 2(x - 1) + 2(y - 2) - 3(z - 3) = 2x + 2y - 3z + (-2 - 4 + 9) \\ = 2x + 2y - 3z + 3 = 0 \\ \end{array}$$

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