- #1

- 486

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Nice, simple, straighforward problem:

[itex]j \cdot {-j}[/itex]

Difficulty: I solved it two ways and got two answers.

[itex]\vec{a} \cdot \vec{b}=ab\cos\phi[/itex]

So in this case, I should get [tex]j \cdot {-j}=(1)(-1)(cos(180^o))=(1)(-1)(-1)=1[/itex]

However, using the formula [itex]\vec{a} \cdot \vec{b}=a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}[/itex], I got [itex]j \cdot {-j}=a_{y}b_{y}=(1)(-1)=-1[/itex]

Can anyone tell me why?

[itex]j \cdot {-j}[/itex]

Difficulty: I solved it two ways and got two answers.

[itex]\vec{a} \cdot \vec{b}=ab\cos\phi[/itex]

So in this case, I should get [tex]j \cdot {-j}=(1)(-1)(cos(180^o))=(1)(-1)(-1)=1[/itex]

However, using the formula [itex]\vec{a} \cdot \vec{b}=a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}[/itex], I got [itex]j \cdot {-j}=a_{y}b_{y}=(1)(-1)=-1[/itex]

Can anyone tell me why?

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