Questions: 2. [12 points] Find an equation for the plane that contains the points (1,2,4), (3,2,3) and (6,4,3).

Solution Steps

To find the equation of a plane given three points, we can use the following steps:

1. Find two vectors that lie on the plane by subtracting the coordinates of the given points.
2. Compute the cross product of these two vectors to get a normal vector to the plane.
3. Use the normal vector and one of the points to write the equation of the plane in the form \(Ax + By + Cz = D\).

We are given three points in 3D space:

• \( p_1 = (1, 2, 4) \)
• \( p_2 = (3, 2, 3) \)
• \( p_3 = (6, 4, 3) \)

We can find two vectors that lie on the plane by subtracting the coordinates of the points:

• \( \mathbf{v_1} = p_2 - p_1 = (3 - 1, 2 - 2, 3 - 4) = (2, 0, -1) \)
• \( \mathbf{v_2} = p_3 - p_1 = (6 - 1, 4 - 2, 3 - 4) = (5, 2, -1) \)

The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of \( \mathbf{v_1} \) and \( \mathbf{v_2} \): \[ \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & -1 \\ 5 & 2 & -1 \end{vmatrix} = (2, -3, 4) \]

The equation of the plane can be expressed in the form: \[ Ax + By + Cz = D \] where \( A, B, C \) are the components of the normal vector, and \( D \) can be calculated using one of the points, say \( p_1 \): \[ D = \mathbf{n} \cdot p_1 = 2 \cdot 1 + (-3) \cdot 2 + 4 \cdot 4 = 2 - 6 + 16 = 12 \]

Thus, the equation of the plane is: \[ 2x - 3y + 4z = 12 \]

Final Answer