Orthogonal Basis and orthogonal projection

Question

I have this problem:

Let $V = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 + 3x_2 -5x_3 - x_4 = 0\}$

• Find an orthogonal basis for $V$.

• What's the closest point to the origin over the plane $x_1 + 3x_2 - 5x_3 - x_4 = 36$?

I found a basis for $V$ given by the vectors:

• $v_1 = (-3, 1, 0, 0) $
• $v_2 = (5, 0, 1, 0) $
• $v_3 = (1, 0, 0, 1) $

Then I used the Gram-Schmidt process to find an orthogonal basis for $V$ given by:

• $w_1 = (-3,1,0,0)$
• $w_2 = (\frac{1}{2},\frac{3}{2},1,0)$
• $w_3 = (\frac{1}{35},\frac{3}{35},\frac{-1}{7},1)$

To find the closest point to the origin over the plane $x_1 + 3x_2 - 5x_3 - x_4 = 36$. I know that I need to find the orthogonal projection of the plane. To find the point I use the fact that the point $(1,3,-5,-1)$ is perpendicular to the plane, then the point $(x_1,x_2,x_3,x_4)$ can be express like this: $$ (x_1,x_2,x_3,x_4) = (0,0,0,0) + t(1,3,-5,-1)$$

Then:

$$t + 3(3t) -5 (-5t) - (-t) = 36t = 36 \leftrightarrow t = 1$$

And so, the closest point to origin is $(1,3,-5,-1)$.

But I know there are plenty of ways to find the closest point, I want to obtain the vector $(1,3,-5,-1)$ by using a theorem that says that the orthogonal projection can be found using an orthonormal basis for $V$, by obtaining a vector $x$ such that

$$x = \sum_{i=1}^{3} <y,u_i> u_i$$

where $y\in V$ and $\{u_1, u_2, u_3\}$ is an orthonormal basis for V.

I'm having problems to apply this method, I just found and orthonormal basis for V, that is:

• $u_1 = \frac{1}{\sqrt{10}} (-3, 1, 0, 0) $
• $u_2 = \frac{2}{\sqrt{14}} (\frac{1}{2},\frac{3}{2},1,0) $
• $u_3 = \frac{\sqrt{35}}{6} (\frac{1}{35},\frac{3}{35},\frac{-1}{7},1) $

but I don't know how to proceed with the method, which vector $y\in V$ should I use to find vector $x = (1,3,-5,-1)$? I hope you can help me by telling me how to proceed.

Answer 1

If $(a,b,c,d)$ is the point of the plane defined by $x_1+3x_2-5x_3-x_4=36$ closest to the origin, then $(a,b,c,d)$ is orthogonal to that plane. Therefore, $(a,b,c,d)=\lambda(1,3,-5,-1)$ for some $\lambda\in\mathbb R$. So, solve the equation$$\lambda+3\times(3\lambda)-5\times(-5\lambda)-(-\lambda)=36.$$In other words, take $\lambda=1$.