Question

Find the equation of the plane containing the points (2,0,1), (−1,2,3) and (0,2, −2).

Answer 1

The equation of the plane containing the points (2, 0, 1), (−1, 2, 3), and (0, 2, −2) is 4x - 3y - 2z = 7.

Step-by-step explanation:

To find the equation of the plane passing through three non-collinear points, we can use the following steps. Let the given points be
`\( P_1(x_1, y_1, z_1) \), \( P_2(x_2, y_2, z_2) \)`, and
`\( P_3(x_3, y_3, z_3) \)`. The normal vector
`\( \mathbf{N} \)` to the plane can be found by taking the cross product of the vectors
`\( \overrightarrow{P_1P_2} \)` and
`\( \overrightarrow{P_1P_3} \)`.

`\[ \mathbf{N} = \overrightarrow{P_1P_2} * \overrightarrow{P_1P_3} \]`

Once we have the normal vector, we can use it to write the equation of the plane in the form Ax + By + Cz = D, where (A, B, C) is the normal vector. The coordinates of any of the given points can then be substituted into the equation to find the value of D.

In this case, the normal vector
`\( \mathbf{N} \)` is
`\( \langle 4, -3, -2 \rangle \)`, and choosing a point
`\( P_1(2, 0, 1) \)` to substitute into the equation, we get 4(2) - 3(0) - 2(1) = 7. Therefore, the equation of the plane is 4x - 3y - 2z = 7.