 Ellen Chang

2022-07-07

Find the coordinates of $\stackrel{\to }{v}$ if $\stackrel{\to }{v}=2\stackrel{\to }{a}-3\stackrel{\to }{b}+4\stackrel{\to }{c}$ and $\stackrel{\to }{a}\left(4;1\right),\stackrel{\to }{b}\left(1;2\right)$ and $\stackrel{\to }{c}\left(2;7\right)$ gutinyalk

Expert

Step 1
From what I understand, $\stackrel{\to }{u}\left(p;q\right)$ means that $\stackrel{\to }{u}$ is a vector in ${\mathbb{R}}^{2}$ , with coordinates (p, q). So,
$\stackrel{\to }{v}=2\stackrel{\to }{a}-3\stackrel{\to }{b}+4\stackrel{\to }{c}=2\left(4,1\right)-3\left(1,2\right)+4\left(2,7\right)=\left(13,24\right)$
In your notation, $\stackrel{\to }{v}\left(13;24\right)$ Mylee Underwood

Expert

Step 1
I am supposing that by "find the coordinates" it means cartesian coordinates, and that $\stackrel{\to }{a}\left(4;1\right)$ is the same as writing:
$\stackrel{\to }{a}=\left(\begin{array}{c}4\\ 1\end{array}\right)$
where 4 is the x component, and 1 is the y component. Then you have a vector determined by three others, but, because you are in the plane ${\mathbb{R}}^{2}$, it is redundant. So you are looking to reduce the number of "coordinates" (numbers by which you label your vector) to two, which is the minimum. As delta-divine has said, the result would be
$\stackrel{\to }{v}=\left(\begin{array}{c}13\\ 24\end{array}\right)$
which is $\stackrel{\to }{v}\left(13;24\right)$ in your notation.

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