Finding the Orthonormal Basis for a Given Subspace W

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Homework Statement


Find the orthogonal projection of the given vector on the given subspace W of the inner product space V?
V=R3, u = (2,1,3), and W = {(x,y,z): x + 3y - 2z = 0}
I don't understand how to find the orthonormal basis for W?


Homework Equations


I don't understand how to find the orthonormal basis for W?
I know once you have an orthonormal basis then you know that the projection is just
proj. = <u,v1>*v1 + <u,v2>*v2
where v1,v2 are the orthonormal vectors of the basis for W

The Attempt at a Solution


Since x + 3y -2z = 0 I took three vectors that are a solution to that system and then used the gram-schmidth to make them orthogonal, then normalized them, then used it to calculate the porjection but it came out wrong. It is supposed to be 1/17*(26 104) where 26 and 104 are in a column vector.
So basically, how to do you find the right basis for W?
 
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How do you end up with a two-dimensional vector when you're working in R3?

It would help if you showed us your actual work rather than just describing what you did.
 
  1. Find a basis for W (does it has to be orthonormal? do not think so), note dim(W) = 2
  2. find a non-zero vector z orthogonal to the basis vectors of W
  3. Solve cz + x = du, where x is a unknown vector in W and c and d are unknown scalars.
  4. Now x is your projection.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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