# Orthogonal Basis and orthogonal projection

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

$$x = \sum_{i=1}^{3} u_i$$

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.

• $w_3$ isn't orthogonal to $w_2$. – Chris Custer Oct 6 '18 at 22:30
• $\{u_1,u_2,u_3\}$ isn't an orthonormal basis: they're not orthogonal and don't all have unit length... – Chris Custer Oct 6 '18 at 22:39
• @ChrisCuster you're right I checked the math errors and realized I made some mistakes while writing the question. I just edited it. – frl93 Oct 6 '18 at 23:10
• Ok. Looks better. – Chris Custer Oct 6 '18 at 23:26

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 $$\lambda=1$$.
• I think you lost a couple minus signs. I get $t=1$, in agreement with the OP. – Chris Custer Oct 6 '18 at 22:59