# Vectors, find implied dot product?

The angle between two vectors $\vec{w}$ and $\vec{r}$ is less than 90 degrees. Vector $\vec{w}$ is given by $\vec{w} = \vec{u} + \vec{v}$ where $\vec{u} \parallel \vec{r}$ and $\vec{v}$ $\perp$ $\vec{r}$. Find $\vec{w} \cdot \vec{r}$ when $\vert\vec{u}\vert = 4$ and $\vert\vec{r}\vert = 3$

I've been looking at this on and off all day. If it's really easy break it to me easy. I already feel really dumb. yey.

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Yes, this is simple. Here's how to go about it:

\begin{align*}\vec{w} \cdot \vec{r}&=(\vec{u}+\vec{v}) \cdot \vec{r} \\&= \vec{u} \cdot\vec{r}+\vec{v} \cdot \vec{r} \\&=\vec{u}\cdot\vec{r}+0 ~~~\text{since } \vec{v} \perp\vec{r} \\&=|\vec{u}|\cdot|\vec{r}| \cdot \cos \theta\\&=3\cdot 4\cdot \cos 0\\&=12 ~~(\text{as}~~ \cos 0=1)\end{align*}

In this computation, I have used the properties that

• scalar product or the dot product distributes over vector addition

• two vectors are perpendicular if and only if their dot product is zero.

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Third line from bottom, shouldn't it be length of vector u in the beginning? Thanks! – Algific Feb 1 '12 at 23:02
That was a typo @Algific. Thank You, Fixed – user21436 Feb 1 '12 at 23:05

The dot product satisfies $$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos(\theta)$$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, and it distributes over addition: $$\vec{a}\cdot(\vec{b}+\vec{c})=\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}.$$ Thus $$\vec{w}\cdot\vec{r}=(\vec{u}+\vec{v})\cdot\vec{r}=\vec{u}\cdot\vec{r}+\vec{v}\cdot\vec{r}=|\vec{u}||\vec{r}|\cos(0)+|\vec{v}||\vec{r}|\cos(\tfrac{\pi}{2})=(4\cdot 3\cdot1)+ (|\vec{v}|\cdot 3\cdot 0)=12,$$ since the statements that $\vec{u} \parallel \vec{r}$ and $\vec{v}\perp\vec{r}$ just mean that the angle between $\vec{u}$ and $\vec{r}$ is $0$, and the angle between $\vec{v}$ and $\vec{r}$ is $\frac{\pi}{2}$.

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You can use $$\Vert r\Vert \Vert w\Vert \cos\theta=r\cdot w$$ to conclude (draw a triangle): $$\underbrace{\Vert w\Vert\cos\theta}_{\color{darkgreen}{\Vert u\Vert}}= {r\cdot w\over \Vert r\Vert}.$$

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