What is the angle between $< 4, 7, 1 >$ $<4,7,1 >$ and $< 9, 5, 4 >$ $<9,5,4>$ ?

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Answer 1 · 2016-02-19T15:35:31

${33.30}^{o}$ $33.30^o$ approx

The formula for angle between two vectors a and b is $cos θ = \frac{a . b}{| a | | b |}$ $cos theta=(a.b)/(|a| |b|)$

Here $a . b$ $a.b$ is (4.9)+(7.5)+(1.4)= 75

$| a | = \sqrt{4^{2} + 7^{2} + 1^{2}} = \sqrt{66}$ $|a| = sqrt(4^2 +7^2 +1^2)= sqrt 66$

$| b | = \sqrt{9^{2} + 5^{2} + 4^{2}} = \sqrt{122}$ $|b|= sqrt(9^2 +5^2 +4^2)= sqrt122$

$cos θ = \frac{75}{(\sqrt{66}) (\sqrt{122})}$ $cos theta = (75)/((sqrt66)(sqrt122))$
= 0.83581

$θ = {33.30}^{o}$ $theta = 33.30^o$ approx

What is the angle between <4,7,1><4,7,1 > and <9,5,4><9,5,4> ?