# How can I visualize $c\mathbf{v} + (1 - c)\mathbf{w}$?

Draw the line of all combinations that has $$c\mathbf{v} + d\mathbf{w}$$ and $$c + d = 1$$.

Solution: All combinations with $$c + d = 1$$ are on the line that passes through $$\mathbf{v}$$ and $$\mathbf{w}$$.

1. $$c\mathbf{v} + d\mathbf{w} \quad \& \quad c + d = 1 \implies c\mathbf{v} + (1 - c)\mathbf{w}$$.
But how do I draw $$(1 - c)\mathbf{w}$$?

2. My $$c = 1.5$$ sketch fits the solution. But why doesn't my sketch for $$c = -0.5$$?

I additionally tried $$c\mathbf{v} + (1 - c)\mathbf{w} = \color{#318CE7}{c(\mathbf{v} - \mathbf{w}) + \mathbf{w}}$$ in picture 2 but it doesn't agree with solution?

1. How can I get the solution algebraically, without pictures?
• I think in picture 2 if you shifted the initial point of c(v-w) to origin then your observation will agree with the result.. – Ritu Feb 20 '15 at 7:08
• And for $c=-0.5$, start $-0.5v$ from origin... – Ritu Feb 20 '15 at 7:16

## 2 Answers

Maybe this little illustration will help too (sorry for the quality). The problem in issue#2 is the wrong addition of vectors.

Suppose $AB$ and $AC$ are your vectors $v$ and $w$, respectively. Then $AD$ is your convex combination $uv+(1-u)w$, where $uv$ is $AF$, and $AE$ is $(1-u)w$. The similarities between the triangles $BFD$, $EDC$ and $ABC$ explains it.

The fact that $BD+DC=BC$ is related to $(1-u)+u=1$, because you get $BD/BC+DC/BC=1$. Now, the triangles similarities gives you that $BD/BC=FD/AC=AE/AC=1-u$ and $DC/BC=ED/AB=AF/AB=u$.