# Express the vector $\vec{v}$ as the sum of a vectors parallel to $\vec{b}$ and a vector component of $\vec{v}$ orthogonal to $\vec{b}$.

$$\vec{v}=2i-4j,\qquad \vec{b}=i+j$$

I have no idea where to start on this.

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Where to start: Find two vectors perpendicular to $\vec{b}$ :) – John Stalfos Sep 9 '12 at 22:19
How would I do that? – questionguy Sep 9 '12 at 22:19

First step is to draw the vectors so you can visualize possible solutions to the problem. The second step is to find a vector that is orthogonal to $\vec{b}$, such as $\vec{u} = i - j$. The thirst step is to write $\vec{v}$ in terms of $\vec{b}$ and $\vec{u}$, i.e.

$\vec{v} = \alpha \,\vec{b} + \beta \,\vec{u}$

and find scalars $\alpha$ and $\beta$. You will have to solve two equations in two unknowns.

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How do you have two equations? Isn't it 2i-4j= a(i+j) + b(i-j)? – questionguy Sep 9 '12 at 22:29
@ questioguy: note that I used $\alpha$ and $\beta$, not $a$ and $b$. Having said that, rearrange the right-hand side (RHS), and you obtain two equations in $\alpha$ and $\beta$ when you match the coefficients of $i$ and $j$ in the RHS with those of $i$ and $j$ in the LHS. – Rod Carvalho Sep 9 '12 at 22:36
Could you work out what you are saying? I don't know what you mean – questionguy Sep 9 '12 at 22:37
@ questionguy: I despise the notation you're using. Write $b = (1,1)$ and $u = (1,-1)$, without the stupid $i$ and $j$, and then you naturally obtain two equation in two unknowns from $(2,-4) = (\alpha + \beta, \alpha - \beta)$. – Rod Carvalho Sep 9 '12 at 22:48

There's a more direct route to this when you only need to express one vector relative to $b$ in this way. Take $\frac{(v.b)}{(b.b)}b$, noticing that's parallel to $b$, and add $v-\frac{(v.b)}{(b.b)}b$. You should be able to check directly that the second vector I gave is perpendicular to $b$; and obviously they add up to $v$.

The first vector is called the projection of $v$ onto $b$, and it's quite an important notion to have in hand as you're learning linear algebra.

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What is (v.b)/(b.b)b? – questionguy Sep 9 '12 at 22:24
@questionguy $$\frac{v\cdot b}{b\cdot b}$$ will be a scalar. You then multiply this scalar by the vector $b$. Note $b\cdot b=||b||^2$, so you're normalizing and multiplying by a suitable scalar, namely $\langle a,b\rangle /||b||$ – Pedro Tamaroff Sep 9 '12 at 22:31