# If A = <6 ,4 ,-3 >, B = <7 ,1 ,-4 > and C=A-B, what is the angle between A and C?

Nov 21, 2016

The angle between A and C is ${\cos}^{-} 1 \left(\frac{3}{\sqrt{671}}\right)$.

#### Explanation:

Let us consider A=(6,4,-3) into a vector $\vec{A} = 6 \hat{i} + 4 \hat{j} - 3 \hat{k}$ and similarly with B as $\vec{B} = 7 \hat{i} + \hat{j} - 4 \hat{k}$.
For C=A-B as given above for this,
The difference of the vectors $\vec{A}$ and $\vec{B}$ is given by
$\vec{A} - \vec{B} = \left({a}_{1} - {b}_{1}\right) \hat{i} + \left({a}_{2} - {b}_{2}\right) \hat{j} + \left({a}_{3} - {b}_{3}\right) \hat{k}$

Substituting the values and calculating we get,
$\therefore C = - \hat{i} + 3 \hat{j} + \hat{k}$
So, now we can find the angle between the A and C and this is given by,
$\theta = {\cos}^{-} 1 \left\{\frac{{a}_{1} {b}_{1} + {a}_{2} {b}_{2} + {a}_{3} {b}_{3}}{\sqrt{{\left({a}_{1}\right)}^{2} + {\left({a}_{2}\right)}^{2} + {\left({a}_{3}\right)}^{2}} \sqrt{{\left({b}_{1}\right)}^{2} + {\left({b}_{2}\right)}^{2} + {\left({b}_{3}\right)}^{2}}}\right\}$

substituting and calculating we get,

the angle between the A and C is $\theta = {\cos}^{-} 1 \left(\frac{3}{\sqrt{671}}\right)$.