# If A = <5 ,2 ,8 >, B = <2 ,7 ,6 > and C=A-B, what is the angle between A and C?

Apr 7, 2018

The angle is $= {69.3}^{\circ}$

#### Explanation:

Start by calculating

$\vec{C} = \vec{A} - \vec{B}$

vecC=〈5,2,8〉-〈2,7,6〉=〈3,-5,2〉

The angle between $\vec{A}$ and $\vec{C}$ is given by the dot product definition.

vecA.vecC=∥vecA∥*∥vecC∥costheta

Where $\theta$ is the angle between $\vec{A}$ and $\vec{C}$

The dot product is

vecA.vecC=〈5,2,8〉.〈3,-5,2〉=15-10+16=21

The modulus of $\vec{A}$= ∥〈5,2,8〉∥=sqrt(25+4+64)=sqrt93

The modulus of $\vec{C}$= ∥〈3,-5,2〉∥=sqrt(9+25+4)=sqrt38

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=21/(sqrt93*sqrt38)=0.35

$\theta = \arccos \left(0.35\right) = {69.3}^{\circ}$

Apr 7, 2018

color(blue)(69.31^@ 2 d.p.

#### Explanation:

$\boldsymbol{A} = \left[\begin{matrix}5 \\ 2 \\ 8\end{matrix}\right] \setminus \setminus \setminus \setminus \setminus \boldsymbol{B} = \left[\begin{matrix}2 \\ 7 \\ 6\end{matrix}\right]$

$\boldsymbol{C} = \boldsymbol{A} - \boldsymbol{B} = \left[\begin{matrix}5 \\ 2 \\ 8\end{matrix}\right] - \left[\begin{matrix}2 \\ 7 \\ 6\end{matrix}\right] = \left[\begin{matrix}5 - 2 \\ 2 - 7 \\ 8 - 6\end{matrix}\right] = \left[\begin{matrix}3 \\ - 5 \\ 2\end{matrix}\right]$

The angle between two vectors can be found using the Dot Product.

$\boldsymbol{a} \cdot \boldsymbol{b} = | | \boldsymbol{a} | | \cdot | | \boldsymbol{b} | | \cdot \cos \left(\theta\right)$

$\cos \left(\theta\right) = \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{| | \boldsymbol{a} | | \cdot | | \boldsymbol{b} | |}$

We first find the magnitudes of $\boldsymbol{A} \mathmr{and} \boldsymbol{B}$

This is given by the Distance Formula:

$| | \boldsymbol{v} | | = \sqrt{{\left({v}_{1}\right)}^{2} + {\left({v}_{2}\right)}^{2} + {\left({v}_{3}\right)}^{2}}$

Where ${v}_{1} , {v}_{2} , {v}_{3}$ are the components of the vector.

$| | \boldsymbol{A} | | = \sqrt{{\left(5\right)}^{2} + {\left(2\right)}^{2} + {\left(8\right)}^{2}} = \sqrt{93}$

$| | \boldsymbol{C} | | = \sqrt{{\left(3\right)}^{2} + {\left(- 5\right)}^{2} + {\left(2\right)}^{2}} = \sqrt{38}$

Now the product of:

$\boldsymbol{A} \cdot \boldsymbol{C}$

This is found by multiplying:

$\left[\begin{matrix}5 \\ 2 \\ 8\end{matrix}\right] \cdot \left[\begin{matrix}3 \\ - 5 \\ 2\end{matrix}\right]$

For this we multiply corresponding components and the n sum these:

$\left[\begin{matrix}5 \\ 2 \\ 8\end{matrix}\right] \cdot \left[\begin{matrix}3 \\ - 5 \\ 2\end{matrix}\right] = \left[\left(5 \cdot 3 + 2 \cdot - 5 + 8 \cdot 2\right)\right] = 21$

Using:

$\cos \left(\theta\right) = \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{| | \boldsymbol{a} | | \cdot | | \boldsymbol{b} | |}$

$\cos \left(\theta\right) = \frac{\boldsymbol{A} \cdot \boldsymbol{C}}{| | \boldsymbol{A} | | \cdot | | \boldsymbol{C} | |}$

$\cos \left(\theta\right) = \frac{21}{\sqrt{93} \cdot \sqrt{38}}$

$\theta = \arccos \left(\frac{21}{\sqrt{93} \cdot \sqrt{38}}\right) \approx 69.313579120997$

color(blue)(69.31^@ 2 d.p.

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