How do you find the angle between the vectors $u=2i-3j$ and $v=8i+3j$ ?

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Answer 1 · 2016-10-18T17:13:41

$76.9^o$

Starting from the saclar (dot) product

the definition:

$vec a .vec b=|a||b|costheta$ -------(1)
where $theta$ is the angle between $veca$ & $vecb$

Also to evaluate the scalar product using components we have the result:

$veca=x_"1"veci+y_"1"vecj$

$vecb=x_"2"veci+y_"2"vecj$

$vec a .vec b=x_"1"x_"2"+y_1"$ $y_"2"$ ------(2)

so for

$vecu=2veci-3vecj$ & $vecv=8veci+3vecj$

using result (2)

$vecu.vecv=(2xx8)+((-3)xx3)$

$vecu.vecv=16-9=7$

Now we use the definition (1)

$vecu.vecv=|u||v|costheta$

so, $7=sqrt(2^2+3^2)xxsqrt(8^2+3^2)xxcostheta$

#

$7=sqrt13xxsqrt73xxcostheta$

$theta=cos^-1(7/(sqrt13sqrt73))$

$theta=76.9^o$

How do you find the angle between the vectors u=2i-3ju=2i-3j and v=8i+3jv=8i+3j?