Find a number a so that the vectors (a,2+a,6) and (2,1,-3) are parallell. How do I do that?
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Originally Posted by a4swe Find a number a so that the vectors (a,2+a,6) and (2,1,-3) are parallel. Two vectors are parallel if they are non-zero multiples of each other. (a,2+a,6)=t(2,1,-3)=(2t,t,-3t) Thus t=-2. a=?
Originally Posted by a4swe Find a number a so that the vectors (a,2+a,6) and (2,1,-3) are parallell. How do I do that? Also if and only if their dot product is zero. The dot product in this case is, 2a+(2+a)1+6(-3)=0
Originally Posted by ThePerfectHacker Also if and only if their dot product is zero. No, if their dot product is zero then they are perpendicular not parallel.
Originally Posted by Plato No, if their dot product is zero then they are perpendicular not parallel. Thank you Plato. --- I meant if their cross product is a zero vector. But that seems more complicated than finding scalar multiples.
Thanks. But can I say anything about parallelity considering the soultion set of some particullar linear equation system? And Plato, what you said is equavilent with: "Two vectors are parallell if they are linear dependent" right?
I am almost afraid to answer not really understanding the question. But yes two parallel vectors are linearly dependent. I do not understand what you mean by ‘system’.
By system I mean a collection of two or more equations. But that was another question.
And just now I realized that my first question in my second post was a stupid one, problem completly solved, thank you Plato.
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