If $A = < 5, 6, - 3 >$ $A = <5 ,6 ,-3 >$ , $B = < - 9, 6, - 5 >$ $B = <-9 ,6 ,-5 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

Question

1433 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-14T10:59:13

The angle is $= 57.3$ $=57.3$ º

Let's start calculating $\to C$ $vecC$

$\to C = \to A - V e c B = ⟨ 5, 6, - 3 ⟩ - ⟨ - 9, 6, - 5 ⟩$ $vecC=vecA-VecB=〈5,6,-3〉-〈-9,6,-5〉$
$= ⟨ 14, 0, 2 ⟩$ $=〈14,0,2〉$

To calculate the angle $θ$ $theta$ between $\to A$ $vecA$ and $\to C$ $vecC$ ,
we use the dot product definition:

$\to A . \to C = ∥ \to A ∥ \cdot ∥ \to C ∥ cos θ$ $vecA.vecC=∥vecA∥*∥vecC∥costheta$

The dot product $= ⟨ 5, 6, - 3 ⟩ . ⟨ 14, 0, 2 ⟩ = 70 + 0 - 6 = 64$ $=〈5,6,-3〉.〈14,0,2〉=70+0-6=64$

The modulus of $\to A = ∥ ⟨ 5, 6, - 3 ⟩ ∥ = \sqrt{25 + 36 + 9} = \sqrt{70}$ $vecA=∥〈5,6,-3〉∥=sqrt(25+36+9)=sqrt70$

The modulus of $\to C = ∥ ⟨ 14, 0, 2 ⟩ ∥ = \sqrt{196 + 0 + 4} = \sqrt{200}$ $vecC=∥〈14,0,2〉∥=sqrt(196+0+4)=sqrt200$

$:. cos theta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=64/(sqrt70sqrt200)$

$costheta=0.54$

$theta=57.3$ º

If A = <5 ,6 ,-3 >A=<5,6,−3>A = <5 ,6 ,-3 >, B = <-9 ,6 ,-5 >B=<−9,6,−5>B = <-9 ,6 ,-5 > and C=A-BC=A−BC=A-B, what is the angle between A and C?