anonymous
  • anonymous
Let u = <-5, 1>, v = <7, -4>. Find 9u - 6v.
Mathematics
jamiebookeater
  • jamiebookeater
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jdoe0001
  • jdoe0001
9u <-- scalar multiplication, same for 6v 9u - 6v <---- vector addition do the scalar mutiplication first, to get a resultant vector then "add" them up, or subtract in this case
anonymous
  • anonymous
You might get annoyed, but I'm new to this topic, so I don't know what to do. Maybe baby steps? :)
jdoe0001
  • jdoe0001
one sec

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anonymous
  • anonymous
Sure thing.
jdoe0001
  • jdoe0001
\(\bf { \begin{array}{llll} u=<-5,1>\\ v=<7,-4> \end{array}\qquad \begin{array}{llll} 9u\implies <9\cdot -5,9\cdot 1>\implies <-45,9>\\ -6v\implies <-6\cdot 7,-6\cdot 4>\implies <-42,24> \end{array} \\ \quad \\ 9u-6v\implies <-45,9> + <-42,24>\\ \quad \\ <-45+(-42)\quad ,\quad 9+24> }\)
jdoe0001
  • jdoe0001
well... could have been done using the - I gather
anonymous
  • anonymous
Ohhh, so does it look like I'm distributing 9u to the first one (u), and -6v to v?
anonymous
  • anonymous
And the answer is <-87, 33>?
jdoe0001
  • jdoe0001
yeap
anonymous
  • anonymous
Thank you so much! It became clear now. :)
jdoe0001
  • jdoe0001
\(\begin{array}{llll} u=<-5,1>\\ v=<7,-4> \end{array}\qquad \begin{array}{llll} 9u\implies <9\cdot -5,9\cdot 1>\implies <-45,9>\\ 6v\implies <6\cdot 7,6\cdot 4>\implies <42,-24> \end{array} \\ \quad \\ 9u-6v\implies <-45,9> - <42,-24>\implies <-45-42\quad ,\quad 9-(-24)>\) would have done the same thing anyhow
jdoe0001
  • jdoe0001
yw

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