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

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Answer 1 · 2016-10-18T17:13:41

${76.9}^{o}$ $76.9^o$

Starting from the saclar (dot) product

the definition:

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

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

$\to a = x_{1} \to i + y_{1} \to j$ $veca=x_"1"veci+y_"1"vecj$

$\to b = x_{2} \to i + y_{2} \to j$ $vecb=x_"2"veci+y_"2"vecj$

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

so for

$\to u = 2 \to i - 3 \to j$ $vecu=2veci-3vecj$ & $\to v = 8 \to i + 3 \to j$ $vecv=8veci+3vecj$

using result (2)

$\to u . \to v = (2 \times 8) + ((- 3) \times 3)$ $vecu.vecv=(2xx8)+((-3)xx3)$

$\to u . \to v = 16 - 9 = 7$ $vecu.vecv=16-9=7$

Now we use the definition (1)

$\to u . \to v = | u | | v | cos θ$ $vecu.vecv=|u||v|costheta$

so, $7 = \sqrt{2^{2} + 3^{2}} \times \sqrt{8^{2} + 3^{2}} \times cos θ$ $7=sqrt(2^2+3^2)xxsqrt(8^2+3^2)xxcostheta$

#

$7 = \sqrt{13} \times \sqrt{73} \times cos θ$ $7=sqrt13xxsqrt73xxcostheta$

$θ = {cos}^{- 1} (\frac{7}{\sqrt{13} \sqrt{73}})$ $theta=cos^-1(7/(sqrt13sqrt73))$

$θ = {76.9}^{o}$ $theta=76.9^o$

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