# If A = <5 ,2 ,-5 >, B = <6 ,5 ,3 > and C=A-B, what is the angle between A and C?

Aug 7, 2018

$\therefore \theta = {\cos}^{-} 1 \left(\frac{29}{\sqrt{3996}}\right) \approx$${\left(62.69\right)}^{\circ}$

#### Explanation:

Here,

$A = < 5 , 2 , - 5 > \mathmr{and} B = < 6 , 5 , 3 >$

$\therefore C = A - B = < 5 , 2 , - 5 > - < 6 , 5 , 3 >$

$C = < - 1 , - 3 - 8 >$

Let , $\vec{a} = < 5 , 2 , - 5 > \mathmr{and} \vec{c} = < - 1 , - 3 , - 8 >$ be

the two vectors.

$\therefore | \vec{a} | = \sqrt{{\left(5\right)}^{2} + {\left(2\right)}^{2} + {\left(- 5\right)}^{2}} = \sqrt{25 + 4 + 25} = \sqrt{54}$

$| \vec{c} | = \sqrt{{\left(- 1\right)}^{2} + {\left(- 3\right)}^{2} + {\left(- 8\right)}^{2}} = \sqrt{1 + 9 + 64}$=$\sqrt{74}$

Dot product : $a \cdot c = < 5 , 2 , - 5 > \cdot < - 1 , - 3 , - 8 >$

$\therefore a \cdot c = \left(5\right) \times \left(- 1\right) + \left(2\right) \times \left(- 3\right) + \left(- 5\right) \times \left(- 8\right)$

$\therefore a \cdot c = - 5 - 6 + 40 = 29$

Now the angle between $\vec{a} \mathmr{and} \vec{c}$ is:

$\theta = {\cos}^{-} 1 \left(\frac{a \cdot c}{| \vec{a} | | \vec{c} |}\right) = {\cos}^{-} 1 \left(\frac{29}{\sqrt{54} \sqrt{74}}\right)$

$\therefore \theta = {\cos}^{-} 1 \left(\frac{29}{\sqrt{3996}}\right) \approx$${\left(62.69\right)}^{\circ}$

Aug 7, 2018

Angle between $\vec{A} \mathmr{and} \vec{C}$ is ${62.69}^{0}$

#### Explanation:

$\vec{A} = < 5 , 2 , - 5 > \mathmr{and} \vec{B} = < 6 , 5 , 3 >$

$\vec{C} = \vec{A} - \vec{B} = \left(< 5 , 2 , - 5 >\right) - \left(< 6 , 5 , 3 >\right)$

$\therefore \vec{C} = < 5 - 6 , 2 - 5 , - 5 - 3 > = < - 1 , - 3 , - 8 >$

Let $\theta$ be the angle between $\vec{A} \mathmr{and} \vec{C}$ ; then we

know $\cos \theta = \frac{\vec{A} \cdot \vec{C}}{| | \vec{A} | | \cdot | | \vec{C} | |}$ or

$\cos \theta =$

$\frac{\left(5 \cdot - 1\right) + \left(2 \cdot - 3\right) + \left(- 5 \cdot - 8\right)}{\sqrt{{5}^{2} + {2}^{2} + {\left(- 5\right)}^{2}} \cdot \sqrt{{\left(- 1\right)}^{2} + {\left(- 3\right)}^{2} + {\left(- 8\right)}^{2}}}$

or $\cos \theta = \frac{29}{\sqrt{54} \cdot \sqrt{74}} = \frac{29}{63.21} \approx 0.4588$

$\therefore \theta = {\cos}^{-} 1 \left(0.4588\right) \approx {62.69}^{0}$

The angle between $\vec{A} \mathmr{and} \vec{C}$ is ${62.69}^{0}$ [Ans]