If $A = < 7, - 5, 6 >$ $A = <7 ,-5 ,6 >$ , $B = < 4, 4, 9 >$ $B = <4 ,4 ,9 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

Question

1348 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-13T07:17:28

The angle is $= 62.6$ $=62.6$ º

We start by calculating $\to C$ $vecC$

$\to C = \to A - \to B$ $vecC=vecA-vecB$

$= ⟨ 7, - 5, 6 ⟩ - ⟨ 4, 4, 9 ⟩ = ⟨ 3, - 9, - 3 ⟩$ $=〈7,-5,6〉-〈4,4,9〉=〈3,-9,-3〉$

The angle between 2 vectors is given by the dot product.

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

The dot product is $⟨ 7, - 5, 6 ⟩ . ⟨ 3, - 9, - 3 ⟩ = 21 + 45 - 18 = 48$ $〈7,-5,6〉.〈3,-9,-3〉=21+45-18=48$

The modulus of $\to A = ∥ ⟨ 7, - 5, 6 ⟩ ∥ = \sqrt{49 + 25 + 36} = \sqrt{110}$ $vecA=∥〈7,-5,6〉∥=sqrt(49+25+36)=sqrt110$

The modulus of $\to C = ∥ ⟨ 3, - 9, - 3 ⟩ ∥ = \sqrt{9 + 81 + 9} = \sqrt{99}$ $vecC=∥〈3,-9,-3〉∥=sqrt(9+81+9)=sqrt99$

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

$= \frac{48}{\sqrt{110} \sqrt{99}} = 0.46$ $=48/(sqrt110sqrt99)=0.46$

$θ = 62.6$ $theta=62.6$ º

If A = <7 ,-5 ,6 >A=<7,−5,6>A = <7 ,-5 ,6 >, B = <4 ,4 ,9 >B=<4,4,9>B = <4 ,4 ,9 > and C=A-BC=A−BC=A-B, what is the angle between A and C?