If A = <6 ,6 ,-3 >, B = <-9 ,7 ,-2 > and C=A-B, what is the angle between A and C?

Mar 1, 2018

The angle between $\vec{A} \mathmr{and} \vec{C}$ is ${50.09}^{0}$

Explanation:

$\vec{A} = < 6 , 6 , - 3 > , \vec{B} = < - 9 , 7 , - 2 >$

$\vec{C} = \vec{A} - \vec{B} = \left(< 6 , 6 , - 3 >\right) - \left(< - 9 , 7 , - 2 >\right)$

$= \left(< 6 + 9 , 6 - 7 , - 3 + 2 >\right) = < 15 , - 1 , - 1 >$

vecA =<6,6, -3> and vecC =<15,-1,-1> ; theta be the

angle between them ; then we know

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

=((6.15)+(6* -1)+(-3*-1))/(sqrt(6^2+6^2+(-3)^2)* (sqrt((15)^2+(-1)^2+(-1)^2))

$= \frac{87}{\sqrt{81} \cdot \sqrt{227}} \approx \frac{87}{135.6} \approx 0.6416$

$\therefore \theta = {\cos}^{-} 1 \left(0.6416\right) \approx {50.09}^{0}$

The angle between $\vec{A} \mathmr{and} \vec{C}$ is ${50.09}^{0}$ [Ans]

Mar 1, 2018

From,the given informations,we can write,

$\vec{A} = 6 \hat{i} + 6 \hat{j} - 3 \hat{k}$ so, $| \vec{A} | = 9$

and, $\vec{B} = - 9 \hat{i} + 7 \hat{j} - 2 \hat{k}$ so, $| \vec{B} | = \sqrt{134}$

So,if the angle between, $\vec{A}$ and $\vec{B}$ be $\theta$,

then, $\vec{A} \cdot \vec{B} = | \vec{A} | | \vec{B} | \cos \theta$

or, $- 54 + 42 + 6 = 9 \cdot \sqrt{134} \cos \theta$

or, $\cos \theta = \left(- \frac{6}{9 \cdot \sqrt{134}}\right)$

So,$\theta = {93.30}^{\circ}$

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

So,if the angle between $\vec{C}$ and $\vec{A}$ be $\phi$,

then, $\tan \phi = \frac{\sqrt{134} \sin 93.30}{9 - \sqrt{134} \cos 93.30}$

so, $\phi = {50}^{\circ}$