If $A = < 1, 6, - 8 >$ $A = <1 ,6 ,-8 >$ , $B = < - 9, 4, 1 >$ $B = <-9 ,4 ,1 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2018-04-28T13:37:11

$θ = arccos (\frac{94}{\sqrt{18685}}) \approx {46.55}^{\circ}$ $theta = arccos(94/sqrt{18685})approx 46.55^circ$

$A \cdot C = | A | | C | cos θ$ $A cdot C = |A||C| cos theta$

The dot product divided by the magnitudes gives the cosine of the angle.

$cos θ = \frac{A \cdot C}{| A | | C |}$ $cos theta = { A cdot C}/{|A||C|}$

$C = A - B = (1, 6, - 8) - (- 9, 4, 1) = (10, 2, - 9)$ $C = A - B = (1,6,-8) -(-9,4,1) = (10, 2, -9)$

$cos theta = {(1,6,-8) cdot (10, 2, -9 ) } / {|((1,6,-8))||((10, 2, -9 )) |}$

$cos theta = {1(10) + 6(2) + -8(-9) }/{ sqrt{ (1^2+6^2+8^2 ) (10^2+2^2+9^2 ) } }$

$cos theta = 94/sqrt(18685)$

$theta = arccos(94/sqrt{18685})$

$theta approx 46.55^circ$

If A = <1 ,6 ,-8 >A=<1,6,−8>A = <1 ,6 ,-8 >, B = <-9 ,4 ,1 >B=<−9,4,1>B = <-9 ,4 ,1 > and C=A-BC=A−BC=A-B, what is the angle between A and C?