Question

Find an equation of the plane that passes through the points P(1, 3, 2), Q(3, −1, 6) and
R(5, 2, 0)

Answer 1

To find the equation of a plane that passes through three points \( P(1, 3, 2) \), \( Q(3, -1, 6) \), and \( R(5, 2, 0) \), we can follow these steps:

1. **Find two vectors that lie in the plane**: We can create two vectors using the given points. Let's compute vectors \( \vec{PQ} \) and \( \vec{PR} \):

\[
\vec{PQ} = Q - P = (3 - 1, -1 - 3, 6 - 2) = (2, -4, 4)
\]

\[
\vec{PR} = R - P = (5 - 1, 2 - 3, 0 - 2) = (4, -1, -2)
\]

2. **Find the normal vector to the plane**: The normal vector \( \vec{n} \) to the plane can be found by taking the cross product of the two vectors \( \vec{PQ} \) and \( \vec{PR} \).

Using the formula for the cross product, we can write:

\[
\vec{n} = \vec{PQ} \times \vec{PR} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -4 & 4 \\
4 & -1 & -2
\end{vmatrix}
\]

Calculating this determinant,

\[
\vec{n} = \hat{i} \left((-4)(-2) - (4)(-1)\right) - \hat{j} \left((2)(-2) - (4)(4)\right) + \hat{k} \left((2)(-1) - (-4)(4)\right)
\]

This simplifies to:

\[
\vec{n} = \hat{i} \left(8 + 4\right) - \hat{j} \left(-4 - 16\right) + \hat{k} \left(-2 + 16\right)
\]

\[
\vec{n} = \hat{i}(12) - \hat{j}(-20) + \hat{k}(14)
\]

So,

\[
\vec{n} = (12, 20, 14)
\]

3. **Equation of the plane**: The equation of the plane can be written in the form:

\[
n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0
\]

where \( (x_0, y_0, z_0) \) is a point on the plane (we can use point P) and \( (n_x, n_y, n_z) \) is the normal vector.

Using point \( P(1, 3, 2) \):

\[
12(x - 1) + 20(y - 3) + 14(z - 2) = 0
\]

Expanding this gives:

\[
12x - 12 + 20y - 60 + 14z - 28 = 0
\]

Combining like terms:

\[
12x + 20y + 14z - 100 = 0
\]

Rearranging it gives the equation of the plane:

\[
12x + 20y + 14z = 100
\]

This is the final equation of the plane.