If $vecu=(3,-1)$ and $vecv=(-6,4)$ , then how do you find the angle between $u$ and $v$ vectors?

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If $vecu=(3,-1)$ and $vecv=(-6,4)$ , then how do you find the angle between $u$ and $v$ vectors?

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Answer 1 · 2018-06-14T13:21:31

See below

Explanation:

We know that dot product of two vectors is defined as

$vecv·vecu=abs(u)·abs(v)·costheta$ (1)

with $theta$ angle between both vectors

By other hand, we know also that $absu=sqrt(u_1^2+u_2^2$

Then $absu=sqrt(9+1)=sqrt10$
$absv=sqrt(36+16)=sqrt52$

And $vecv·vecu=v_1u_1+v_2u_2=abs(u)·abs(v)·costheta$ (1)

Applying this to our case

$vecu·vecv=-18-4=-22$ (2)

$vecu·vecv=sqrt10·sqrt52·costheta$ (3)

Combining (2) and (3) $costheta=-22/(sqrt10·sqrt52)=-0.96476382$

$theta=arccos -0.96476382=164º 44´41.57´´$

Answer 2 · 2018-06-14T13:39:12

Compute the dot product using $vecu*vecv=(u_x)(v_x)+(u_y)(v_y)$
Compute the magnitudes, $|vecu| and |vecv|$
Use the dot product formula, $vecu*vecv= |vecu||vecv|cos(theta)$ , to find the angle.

Explanation:

Compute the dot product using $vecu*vecv=(u_x)(v_x)+(u_y)(v_y)$

$vecu*vecv=(3)(-6)+(-1)(4)$

$vecu*vecv=-22$

Compute the magnitudes, $|vecu| and |vecv|$

$|vecu| = sqrt(u_x^2+u_y^2)$

$|vecu| = sqrt(3^2+(-1)^2)$

$|vecu| = sqrt10$

$|vecv| = sqrt(v_x^2+v_y^2)$

$|vecv| = sqrt((-6)^2+4^2)$

$|vecv| = 2sqrt13$

Use the dot product formula, $vecu*vecv= |vecu||vecv|cos(theta)$ , to find the angle.

$-22= (sqrt10)(2sqrt13)cos(theta)$

$theta = cos^-1(-22/((sqrt10)(2sqrt13)))$

$theta = 164.7^@$

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If $vecu=(3,-1)$ and $vecv=(-6,4)$ , then how do you find the angle between $u$ and $v$ vectors?