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

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

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Answer 1 · 2016-03-11T18:31:14

${20.438}^{\circ}$ $20.438^@$

Explanation:

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

$= (10, 12, 17)$ $=(10,12,17)$ .

Hence the angle $θ$ $theta$ between $A and C$ $A and C$ may be given in terms of the Euclidean inner product as follows

$θ = {cos}^{- 1} (\frac{A \cdot C}{| | A | | | | C | |})$ $theta=cos^(-1)((A*C)/(||A||||C||))$

$= {cos}^{- 1} ⎛ ⎜ ⎝ \frac{1 \times 10 + 6 \times 12 + 9 \times 17}{\sqrt{1^{2} + 6^{2} + 9^{2}} (\sqrt{10^{2} + 12^{2} + 17^{2}})} ⎞ ⎟ ⎠$ $=cos^(-1)((1xx10+6xx12+9xx17)/(sqrt(1^2+6^2+9^2)(sqrt(10^2+12^2+17^2))))$

$= {cos}^{- 1} (\frac{235}{\sqrt{118} \sqrt{533}})$ $=cos^(-1)((235)/(sqrt118sqrt533))$

$= {20.438}^{\circ}$ $=20.438^@$

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

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Explanation:

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

1 Answer

Explanation:

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