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

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Answer 1 · 2016-11-28T16:00:24

The angle is $= 19.7$ $=19.7$ º

To calculate the angle between 2 vectors, we use the dot product.

$\to A . \to C = ∥ \to A ∥ \cdot \to C ∥ \cdot cos (\to A, \to C)$ $vecA.vecC=∥vecA∥*vecC∥*cos(vecA,vecC)$

$\to C = \to A - \to B = ⟨ 3, 8, 8 ⟩ - ⟨ 4, - 3, - 1 ⟩ = ⟨ - 1, 11, 9 ⟩$ $vecC=vecA-vecB=〈3,8,8〉-〈4,-3,-1〉=〈-1,11,9〉$

The dot product is

$\to A . \to C = ⟨ 3, 8, 8 ⟩ . ⟨ - 1, 11, 9 ⟩ = - 3 + 88 + 72 = 157$ $vecA.vecC=〈3,8,8〉.〈-1,11,9〉=-3+88+72=157$

The modulus of $\to A$ $vecA$ is

$∥ ∥ ∥ \to A ∥ = ∥ ⟨ 3, 8, 8 ⟩ ∥ = \sqrt{9 + 64 + 64} = \sqrt{137}$ $∥vecA∥=∥〈3,8,8〉∥=sqrt(9+64+64)=sqrt137$

The modulus of $\to C$ $vecC$ is

$∥ ∥ ∥ \to C ∥ = ∥ ⟨ - 1, 11, 9 ⟩ ∥ = \sqrt{1 + 121 + 81} = \sqrt{203}$ $∥vecC∥=∥〈-1,11,9〉∥=sqrt(1+121+81)=sqrt203$

Therefore,
$cos (\to A, \to C) = \frac{\to A . \to C}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to C ∥ ∥ ∥}$ $cos(vecA,vecC)=(vecA.vecC)/(∥vecA∥*∥vecC∥)$

$= \frac{157}{\sqrt{137} \sqrt{203}} = 0.94$ $=157/(sqrt137sqrt203)=0.94$

$(\to A, \to C) = 19.7$ $(vecA,vecC)=19.7$ º

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