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Let y=(3,5) and u=(6,2). Write y as the sum of a vector in Span{u} and a vector orthogonal to u.

If someone could do this problem as an example, it would be great.

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$y=u_\perp + u_\parallel$

$u_\perp = (y - u_\parallel)$ and $u_\parallel = proj_u y = \frac{y\cdot u}{y \cdot y} y$.

The rest is just plugging into these equations.

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Since these are vectors in the $xy$-plane, you can also approach it this way. A vector $\vec{v}$ orthogonal to $\vec{u} = < 6, 2 >$ must produce the dot product $\vec{u} \cdot \vec{v} = 0$. The easiest way to accomplish this is to choose something like $\vec{v} = < 2, -6 >$ . (Note that this is related to the fact that the product of the slopes of perpendicular lines is $-1$ .)

We want $\vec{y} = < 3 , 5 >$ then to be some linear combination of $\vec{u}$ and $\vec{v}$, that is,

$$< 3 , 5 > = a < 6 , 2 > + b < 2 , -6 > .$$

You would now solve for $a$ and $b$ . However, an infinite number of choices are permissible, since we could have multiplied $\vec{v}$ by any non-zero number to make it any length we wished (or even reversed its direction). So you could pick either $a$ or $b$ to be 1 (or really any non-zero value you like), and solve for the remaining coefficient. The vector in Span(u) is $a\vec{u}$ and your orthogonal vector will be $\vec{w} = b \vec{v}$.

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