Equation of Plane Passing Through Three Points
Equation of a Plane Through Three Points
To derive the equation of a plane passing through three points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3), follow these steps:
- Form two vectors from the points, v1 = (x2 - x1, y2 - y1, z2 - z1) and v2 = (x3 - x1, y3 - y1, z3 - z1).
- Compute the normal vector n using the cross product n = v1 × v2.
- With the normal vector coefficients (a, b, c), the plane's equation can be expressed as ax + by + cz = d.
- Substitute any point into the equation to solve for d, for example using point (x1, y1, z1): d = ax1 + by1 + cz1.
The final equation of the plane is ax + by + cz = d.
Below is the derivation of the plane equation passing through the given points \(x(1, -2, 1)\), \(y(4, -2, -2)\), and \(z(4, 2, 4)\):
- Define vectors from point \(x\) to \(y\) and \(x\) to \(z\): \[ \vec{v}_1 = (4 - 1, -2 + 2, -2 - 1) = (3, 0, -3) \] \[ \vec{v}_2 = (4 - 1, 2 + 2, 4 - 1) = (3, 4, 3) \]
- Compute the cross product of \(\vec{v}_1\) and \(\vec{v}_2\) to find the normal vector \(\vec{n}\): \[ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & -3 \\ 3 & 4 & 3 \end{vmatrix} \] \[ \vec{n} = \mathbf{i} \times (0 \times 3 - (-3) \times 4) - \mathbf{j} \times (3 \times 3 - (-3) \times 3) + \mathbf{k} \times (3 \times 4 - 0 \times 3) \] \[ \vec{n} = (12, -18, 12) \Rightarrow (2, -3, 2) \]
- Plug in any of the three points into the plane equation to solve for \(d\): \[ 2(1) - 3(-2) + 2(1) = d \] \[ 2 + 6 + 2 = 10 \]