Ortogonal basis for a subspace of $R^{4}$ and find a point in a plane closest to the origin.

Question

Let $V \subset R^{4}$ a subspace defined for the equation $x_{1}+3x_{2}-5x_{3}-x_{4}=0$.

a) Find a ortogonal basis for $V$.

b) Wich point in the plane $x_{1}+3x_{2}-5x_{3}-x_{4}=36$ is closest to $(0,0,0,0)$?

For a), any vector $(x_{1},x_{2},x_{3},x_{4})$ in $V$ satisfy $x_{1}=-3x_{2}+5x_{3}+x_{4}$ then

$$x = (x_{1},x_{2},x_{3},x_{4})= x_{2} (-3,1,0,0) + x_{3}(5,0,1,0) + x_{4}(1,0,0,1)$$

so $(-3,1,0,0), (5,0,1,0), (1,0,0,1)$ is a basis for $V$, then I used the Gram-Schmidt process.

For b) I tried to find a basis like in a) and find the Projection Matrix but I failed.

Answer 1

$V$ is the subspace of vectors orthogonal to the vector $(1,3,-5,-1)$.

For a), I would start with the following two orthogonal vectors in $V$: $v_1 = (3,-1,0,0)$, $v_2 = (0,0,-1,5)$ and complete that to an orthogonal basis by adding the vector $v_3 = (26, 78, 50, 10)$ [See the pattern?]

For b), use that the point $A=(1, 3, -5, -1)$ is in the plane. Let $B$ be the projection of $A$ onto the plane $x_1 +3x_2 - 5x_3 -x_4 = 0$. Then $A$, $B$, $O= (0,0,0,0)$, and the point $X$ you need to find form a rectangle; in particular, $X = A-B$.