# Dot Product and Cross Product

Describe the surface defined by the equation: (a) $$\vec{r}\cdot \hat{x}= 2,$$ where $$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$$; (b) $$\left \| \vec{r} \times \hat{x}\right \|=2$$

For the first one, I know that is interpreted as the projection of the r vector onto the x axis is equal to two. However, I am not sure what this means graphically. I have no ideal what the cross product looks like. Any help would be greatly appreciated.

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jbunniii
Homework Helper
Gold Member
For (a), write out what $\vec{r}\cdot \hat{x}$ reduces to:

$$\vec{r}\cdot \hat{x} = (x\hat{x}+y\hat{y}+z\hat{z}) \cdot \hat{x} = ?$$

Proceed similarly with (b). You don't need to know what the cross product "looks like". You just need to keep in mind these facts:

(1) the cross product obeys the distributive law
(2) $$\hat{x} \times \hat{y} = \hat{z}, \quad \hat{y} \times \hat{x} = -\hat{z}$$
(3) $$\hat{y} \times \hat{z} = \hat{x}, \quad \hat{z} \times \hat{y} = -\hat{x}$$
(4) $$\hat{z} \times \hat{x} = \hat{y}, \quad \hat{x} \times \hat{z} = -\hat{y}$$
(3) $$\hat{x} \times \hat{x} = \hat{y} \times \hat{y} = \hat{z} \times \hat{z} = 0$$

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Yeah, so basically any any vector with an x component of "2" satisfies part (a). However, even knowing that, I still cannot wrap my head around the surface created by such an equation.

jbunniii
Homework Helper
Gold Member
Yeah, so basically any any vector with an x component of "2" satisfies part (a). However, even knowing that, I still cannot wrap my head around the surface created by such an equation.
Try not to think about vectors. The equation is simply

$$x = 2$$

What sort of surface in (x,y,z) space satisfies this equation?

Hint: In 2 dimensions, $x = 2$ is a line. In 3 dimensions, $x = 2$ is a ...?

So actually doing the cross product gives the equation: $$\left \| -z\hat{y} - y\hat{z}\right \|=2$$ So this is the equation is for a circle in in the y-z plane. Did I do that right?

jbunniii
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So actually doing the cross product gives the equation: $$\left \| -z\hat{y} - y\hat{z}\right \|=2$$ So this is the equation is for a circle in in the y-z plane. Did I do that right?
I get

$$\vec{r} \times \hat{x} = y (\hat{y} \times \hat{x}) + z(\hat{z} \times \hat{x}) = -y \hat{z} + z \hat{y}$$

so the equation would be

$$||-y \hat{z} + z \hat{y}|| = 2$$

You can evaluate the left-hand side to make it look more like the equation for a circle. What does

$$||-y \hat{z} + z \hat{y}||$$

simplify to?