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

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Answer 1 · 2016-02-01T11:05:02

${cos}^{- 1} (\frac{2 \sqrt{65}}{45})$ $cos^{-1}(frac{2sqrt65}{45})$

The first step is to find $C$ $C$ .

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

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

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

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

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

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

$| C | = \sqrt{{(- 2)}^{2} + {(- 6)}^{2} + 5^{2}} = \sqrt{65}$ $|C| = sqrt( (-2)^2 + (-6)^2 + 5^2 ) = sqrt65$

Therefore,

$cos θ = \frac{26}{9 \sqrt{65}} = \frac{2 \sqrt{65}}{45}$ $cos theta = frac{26}{9sqrt65} = frac{2sqrt65}{45}$

$θ = {cos}^{- 1} (\frac{2 \sqrt{65}}{45})$ $theta = cos^{-1}(frac{2sqrt65}{45})$

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?