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anonymous
 one year ago
SIMPLE MATRIX QUESTION PLEASE HELP!! are the given vectors normal? a = (5,2) and b = (6,15)
anonymous
 one year ago
SIMPLE MATRIX QUESTION PLEASE HELP!! are the given vectors normal? a = (5,2) and b = (6,15)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the question asking if they are perpendicular to each other?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@greg_d all it is asking is if it is "normal"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't determine what or why or how it would be considered normal.. the rest of the homework was working on matrices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the question is, are they "normal" (as perpendicular) to each other, you can use the dot product...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is honestly the last problem and i don't know how i could do that..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, IF that is the question, we need to multiply them using the dot product. \[a.b=a_xb_x+a_yb_y\] since\[a.b=abcos(\theta) \] with theta beign the angle between the vectors. if that product is 0, said angle is 90º and they are indeed "normal" to each other

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could you possibly solve that for me so i can use it next time i see something like this?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Similar Example: Let's say you had the two vectors u = <1,2> v = <5,7> The dot product of u and v is u dot v = 1*5 + 2*7 = 5 + 14 = 19 Since the dot product is not 0, this means that u and v are not perpendicular

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3In general u = <a,b> v = <c,d> u dot v = a*c + b*d u,v are vectors a,b,c,d are scalars

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 @greg_d Thank you!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:D that examples will lead you to the answer !
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