If $A = <4 ,1 ,8 >$ , $B = <6 ,7 ,3 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-02-01T11:05:02

$cos^{-1}(frac{2sqrt65}{45})$

The first step is to find $C$ .

$C = <4,1,8> - <6,7,3>$

$= <-2,-6,5>$

To find the acute angle between $A$ and $C$ , you could use either the dot product or the cross product. I prefer to use the dot product.

$A*C = |A| |C| cos theta$

$A*C = (4)(-2) + (1)(-6) + (8)(5) = 26$

$|A| = sqrt( 4^2 + 1^2 + 8^2 ) = 9$

$|C| = sqrt( (-2)^2 + (-6)^2 + 5^2 ) = sqrt65$

Therefore,

$cos theta = frac{26}{9sqrt65} = frac{2sqrt65}{45}$

$theta = cos^{-1}(frac{2sqrt65}{45})$

If A = <4 ,1 ,8 >A = <4 ,1 ,8 >, B = <6 ,7 ,3 >B = <6 ,7 ,3 > and C=A-BC=A-B, what is the angle between A and C?