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Dot Product and Cross Product

  • Thread starter Xkaliber
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Describe the surface defined by the equation: (a) [tex]\vec{r}\cdot \hat{x}= 2,[/tex] where [tex]\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}[/tex]; (b) [tex]\left \| \vec{r} \times \hat{x}\right \|=2[/tex]

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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  • #2
jbunniii
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For (a), write out what [itex]\vec{r}\cdot \hat{x}[/itex] reduces to:

[tex]\vec{r}\cdot \hat{x} = (x\hat{x}+y\hat{y}+z\hat{z}) \cdot \hat{x} = ?[/tex]

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) [tex]\hat{x} \times \hat{y} = \hat{z}, \quad \hat{y} \times \hat{x} = -\hat{z}[/tex]
(3) [tex]\hat{y} \times \hat{z} = \hat{x}, \quad \hat{z} \times \hat{y} = -\hat{x}[/tex]
(4) [tex] \hat{z} \times \hat{x} = \hat{y}, \quad \hat{x} \times \hat{z} = -\hat{y}[/tex]
(3) [tex]\hat{x} \times \hat{x} = \hat{y} \times \hat{y} = \hat{z} \times \hat{z} = 0[/tex]
 
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  • #3
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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.
 
  • #4
jbunniii
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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.
Try not to think about vectors. The equation is simply

[tex]x = 2[/tex]

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

Hint: In 2 dimensions, [itex]x = 2[/itex] is a line. In 3 dimensions, [itex]x = 2[/itex] is a ...?
 
  • #5
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So actually doing the cross product gives the equation: [tex]\left \| -z\hat{y} - y\hat{z}\right \|=2[/tex] So this is the equation is for a circle in in the y-z plane. Did I do that right?
 
  • #6
jbunniii
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So actually doing the cross product gives the equation: [tex]\left \| -z\hat{y} - y\hat{z}\right \|=2[/tex] So this is the equation is for a circle in in the y-z plane. Did I do that right?
I get

[tex]\vec{r} \times \hat{x} = y (\hat{y} \times \hat{x}) + z(\hat{z} \times \hat{x}) = -y \hat{z} + z \hat{y}[/tex]

so the equation would be

[tex]||-y \hat{z} + z \hat{y}|| = 2[/tex]

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

[tex]||-y \hat{z} + z \hat{y}||[/tex]

simplify to?
 

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