Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

I have no idea where to start on this.

share|cite|improve this question
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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.