What is the Geometric Interpretation of Dot and Cross Products?

[Xkaliber]

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.

 

[jbunniii]

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

 

[Xkaliber]

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]

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 ...?

 

[Xkaliber]

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]

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?