## Distance of a point from a plane

### Task number: 2713

Determine the distance of the point \(A=(5, 5, 3, 3)^T\) from the plane passing through the origin and the points \(B=(8, -1, 1, -2)^T\) and \(C=(4, -2, 2, -1)^T\).

#### Hint

Use Gram-Schmidt ortonormalization. The distance is the norm of the vector \(\mathbf y_3\), where vectors \(\mathbf x_1\), \(\mathbf x_2\) and \(\mathbf x_3\) corresponds to the points \(B\), \(C\) and \(A\).

#### Resolution

With a minor pre-computation: vectors \(\mathbf x_1=(8, -1, 1, -2)^T\), \(\mathbf x_2=(4, -2, 2, -1)^T\) are linearly independent and indeed \( \begin{pmatrix} 8 & -1 & 1 & -2 \\ 4 & -2 & 2 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 8 & -4 & 4 & -2 \\ 8 & -1 & 1 & -2 \\ \end{pmatrix} \sim \begin{pmatrix} 8 & -4 & 4 & -2 \\ 0 & 3 & -3 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 4 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ \end{pmatrix}\).

The last two vectors \(\mathbf x_1'\), \(\mathbf x_2'\) are perpendicular and generate the same plane as \(\mathbf x_1\) and \(\mathbf x_2\).

We now project \(\mathbf x_3=(5, 5, 3, 3)^T\):

\(p(\mathbf x_3)=\frac{\langle \mathbf x_3|\mathbf x_1' \rangle}{||\mathbf x_1'||^2}\mathbf x_1'+\frac{\langle \mathbf x_3|\mathbf x_2' \rangle}{||\mathbf x_2'||^2}\mathbf x_2' =\frac{17}{17}(4, 0, 0, -1)^T+\frac{2}{2}(0, 1, -1, 0)^T=(4, 1, -1, -1)^T \)

(We have used the non-normalized basis, only orthogonal hence the denominator contains the norm \(||\mathbf x_i'||\) twice: once due to the normalization of \(\mathbf x_i'\) and in the second time due to the product with the normalized \(\mathbf x_i'\).)

We get \(||A-p(A)||=||\mathbf x_3-p(\mathbf x_3)||=||\mathbf y_3||=\sqrt{1+16+16+16}=7\).