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 2 years ago
Given \(\vec{g} = 3\vec{i} + 6\vec{j}  12\vec{k}\) and \(\vec{h} = 5\vec{i}  10\vec{j} + 20\vec{k}\), how could I prove that \(\vec{g}\) and \(\vec{h}\) are parallel? Please explain the reasoning.
 2 years ago
Given \(\vec{g} = 3\vec{i} + 6\vec{j}  12\vec{k}\) and \(\vec{h} = 5\vec{i}  10\vec{j} + 20\vec{k}\), how could I prove that \(\vec{g}\) and \(\vec{h}\) are parallel? Please explain the reasoning.

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myko
 2 years ago
Best ResponseYou've already chosen the best response.1take their vector product, or prove they are multiple of each other

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0Well does a vector product of 0 allow me to conclude that they are parallel?

myko
 2 years ago
Best ResponseYou've already chosen the best response.1g x h=ghsin(theta) where theta is angle between them. It will be = 0 only if angle is 0º or 180º

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, just clarifying, is it because using the formal definition of the magnitude of a vector product, if \(\theta\) = 0º or 180º, it would cause the entire thing to become 0, and therefore parallel since an angle of 0º or 180º would be the same line or parallel?

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0Oh, never mind, thank you!

myko
 2 years ago
Best ResponseYou've already chosen the best response.1more easy: 5/3(3i+6j12h)=5i10j+20k

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0Ok. that is the reason the book gave me, so I'm curious as to why that works...I don't see where 5/3 comes from.

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0Oh. so you're multiplying \(\vec{g}\) by 5/3 to make it equal \(\vec{h}\)?

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0ok, once again, thank you :)
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