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

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Answer 1 · 2017-10-22T13:02:28

$36.22^o$ **(2,d.p.)

A= $((3),(-1),(8))$

B= $((4),(-3),(-1))$

C= A - B: C= $((3),(-1),(8))-((4),(-3),(-1))=((-1),(2),(7))$

Angle between A and C:

$((3),(-1),(8))*((-1),(2),(7))$

This is called the Dot product, Scaler product or Inner product and is defined as:

$a*b= |a|*|b|cos(theta)$

$|A|= sqrt(3^2+(-1)^2+8^2)= sqrt(74)$

$|C|= sqrt((-1)^2+2^2+7^2)= sqrt(54)=3sqrt(6)$

$A*C= ((3),(-1),(8))*((-1),(2),(7))=((-3),(-2),(56))$

We need to sum the components in the product vector, so:

$-2-3+56=51$

So now we have:

$51 = sqrt(74)*3sqrt(6)cos(theta)=51/(sqrt(74)*3sqrt(6))= cos(theta)$

$-> 51/(3sqrt(444))=cos(theta)$

$theta= arccos(51/(3sqrt(444))) =36.22^o$ (2,d.p.)

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The angle is formed where the vectors are moving in the same relative direction. i.e where both arrow heads are pointing.

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