Linear Algebra

Biochemistry

To determine the equation of a plane in three-dimensional space, you need to have either: 1. A point on the plane and a normal vector to the plane, or 2. Three non-collinear points on the plane. For the purpose of this explanation, let's assume we have three non-collinear points $ A(x_1, y_1, z_1) $, $ B(x_2, y_2, z_2) $, and $ C(x_3, y_3, z_3) $ on the plane. Here are the steps to find the equation of the plane: ### Step 1: Find two vectors on the plane First, we need to find two vectors that lie on the plane. We can do this by subtracting the coordinates of point A from points B and C to get vectors $ \vec{AB} $ and $ \vec{AC} $: $$ \vec{AB} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle $$ $$ \vec{AC} = \langle x_3 - x_1, y_3 - y_1, z_3 - z_1 \rangle $$ ### Step 2: Find the normal vector to the plane The normal vector to the plane is perpendicular to any vector lying on the plane. We can find the normal vector by taking the cross product of $ \vec{AB} $ and $ \vec{AC} $: $$ \vec{n} = \vec{AB} \times \vec{AC} $$ The cross product is given by: $$ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \\ \end{vmatrix} $$ $$ \vec{n} = \left( (y_2 - y_1)(z_3 - z_1) - (z_2 - z_1)(y_3 - y_1) \right) \mathbf{i} - \left( (x_2 - x_1)(z_3 - z_1) - (z_2 - z_1)(x_3 - x_1) \right) \mathbf{j} + \left( (x_2 - x_1)(y_3 - y_1) - (y_2 - y_1)(x_3 - x_1) \right) \mathbf{k} $$ ### Step 3: Write the equation of the plane The equation of a plane with a normal vector $ \vec{n} = \langle a, b, c \rangle $ and passing through a point $ P_0(x_0, y_0, z_0) $ is given by: $$ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 $$ Using the normal vector $ \vec{n} $ we found and point A (which lies on the plane), we can write the equation of the plane as: $$ n_x(x - x_1) + n_y(y - y_1) + n_z(z - z_1) = 0 $$ Where $ n_x, n_y, $ and $ n_z $ are the components of the normal vector $ \vec{n} $. ### Step 4: Simplify the equation Finally, distribute the components of the normal vector across the equation and simplify to get the standard form of the plane's equation: $$ ax + by + cz + d = 0 $$ Where $ d $ is a constant that can be found by substituting the coordinates of point A into the equation. By following these steps, you can determine the equation of a plane in three-dimensional space using either a point and a normal vector or three non-collinear points.

Question

Utilize a calculator to determine the equation of a plane in three-dimensional space.

Solution

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