# If A = <8 ,3 ,-7 >, B = <6 ,-9 ,5 >, and C=A-B, what is the angle between A and C?

Jan 27, 2018

The angle is $= {43.9}^{\circ}$

#### Explanation:

Let's start by calculating

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

vecC=〈8,3,-7〉-〈6,-9,5〉=〈2,12,-12〉

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=〈8,3,-7〉.〈2,12,-12〉=16+36+84=136

The modulus of $\vec{A}$= ∥〈8,3,-7〉∥=sqrt(64+9+49)=sqrt122

The modulus of $\vec{C}$= ∥〈2,12,-12〉∥=sqrt(4+144+144)=sqrt292

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=136/(sqrt122*sqrt292)=0.72

$\theta = \arccos \left(0.72\right) = {43.9}^{\circ}$

Jan 27, 2018

${43.9}^{\circ}$

#### Explanation:

$A = \left[\begin{matrix}8 \\ 3 \\ - 7\end{matrix}\right]$

$B = \left[\begin{matrix}6 \\ - 9 \\ 5\end{matrix}\right]$

$C = A - B = \left[\begin{matrix}8 \\ 3 \\ - 7\end{matrix}\right] - \left[\begin{matrix}6 \\ - 9 \\ 5\end{matrix}\right] = \left[\begin{matrix}8 - 6 \\ 3 - \left(- 9\right) \\ - 7 - 5\end{matrix}\right] = \left[\begin{matrix}2 \\ 12 \\ - 12\end{matrix}\right]$

We can find the angle between vectors using the Dot Product

The dot product states that for vectors a and b:

$\textcolor{b l u e}{a \cdot b = | | a | | \cdot | | b | | \cdot \cos \left(\theta\right)}$

The dot product is sometimes called the inner product, because of the way the vectors a and b are multiplied and summed.

We are used to multiplying brackets in the following way.

$\left(a + b\right) \left(c + d\right) = a c + a d + b c + b d$

In the dot product we multiply the vectors in the following way.

$\left(a + b + c\right) \cdot \left(d + e + f\right) = a d + b e + c f$

So we are multiplying corresponding components and then adding them together.

Let $a = \left[\begin{matrix}x \\ y \\ z\end{matrix}\right]$

Magnitude of $a = | | a | |$

$\textcolor{b l u e}{| | a | | = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}$

From our example:

First find the product of:

$A \cdot C$

$\left[\begin{matrix}8 \\ 3 \\ - 7\end{matrix}\right] \cdot \left[\begin{matrix}2 \\ 12 \\ - 12\end{matrix}\right] = \left[\begin{matrix}8 \times 2 \\ 3 \times 12 \\ - 7 \times - 12\end{matrix}\right]$

$= \left[\begin{matrix}16 \\ 36 \\ 84\end{matrix}\right] = 16 + 36 + 84 = 136$

We now find the magnitudes of A and C:

$| | A | | = \sqrt{{\left(8\right)}^{2} + {\left(3\right)}^{2} + {\left(- 7\right)}^{2}} = \sqrt{122}$

$| | C | | = \sqrt{{\left(2\right)}^{2} + {\left(12\right)}^{2} + {\left(- 12\right)}^{2}} = \sqrt{292} = 2 \sqrt{73}$

So we have for:

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

$136 = \sqrt{122} \cdot 2 \sqrt{73} \cdot \cos \left(\theta\right)$

$\cos \left(\theta\right) = \frac{136}{\sqrt{122} \cdot 2 \sqrt{73}}$

$\theta = \arccos \left(\cos \left(\theta\right)\right) = \arccos \left(\frac{136}{\sqrt{122} \cdot 2 \sqrt{73}}\right) = {43.9}^{\circ}$
( 2 .d.p.)

The angle between vectors A and C is ${43.9}^{\circ}$

From the diagram we can see that the angle found by the dot product, is the angle between the vectors where they are heading in the same direction.