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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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mykoBest ResponseYou've already chosen the best response.1
take their vector product, or prove they are multiple of each other
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
Well does a vector product of 0 allow me to conclude that they are parallel?
 one year ago

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

CalcmathleteBest ResponseYou've already chosen the best response.0
Ok, 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?
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
Oh, never mind, thank you!
 one year ago

mykoBest ResponseYou've already chosen the best response.1
more easy: 5/3(3i+6j12h)=5i10j+20k
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
Ok. 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.
 one year ago

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

CalcmathleteBest ResponseYou've already chosen the best response.0
ok, once again, thank you :)
 one year ago
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