# If A = <7 ,-3 ,6 >, B = <4 ,-7 ,-5 > and C=A-B, what is the angle between A and C?

Mar 24, 2016

$\alpha \cong 27 , {04}^{o}$

#### Explanation:

$\text{there are five steps:}$
$1. \text{find A-B=C}$
$2. \text{find dot product A.B}$
$3. \text{find ||A|| (magnitude of A)}$
$4. \text{find ||B|| (magnitude of C)}$
$5. \text{use " A.C=||A||*||C||*cos alpha" formula}$

$S t e p - 1 :$
$A = < 7 , - 3 , 6 > \text{ } B < 4 , - 7 , - 5 >$
$C = A - B$
${C}_{x} = {A}_{x} - {B}_{x} = 7 - 4 \text{ ; } {C}_{x} = 3$
$C - y = {A}_{y} - {B}_{y} = - 7 + 3 = - 4 \text{ ; } {C}_{y} = - 4$
${C}_{z} = {A}_{z} - {B}_{z} = 6 + 5 = 11 \text{ ; } {C}_{z} = 11$
$C = < 3 , - 4 , 11 >$
$\text{Step-2:}$
$\text{now let's find the dot product of A.C}$
$A . C = {A}_{x} \cdot {C}_{x} + {A}_{y} \cdot {C}_{y} + {A}_{z} \cdot {C}_{z}$
$A . C = 7 \cdot 3 + \left(- 3\right) \cdot \left(- 4\right) + 6.11$
$A . C = 21 + 12 + 66$
$A . C = 99$
$\text{let's find the magnitude of A and C}$
$\text{Step-3:}$
$| | A | | = \sqrt{{A}_{x}^{2} + {A}_{y}^{2} + {A}_{z}^{2}} \text{ } | | A | | = \sqrt{{7}^{2} + {\left(- 3\right)}^{2} + {6}^{2}}$

$| | A | | = \sqrt{49 + 9 + 36} \text{ } | | A | | = \sqrt{94}$
$\text{Step-4:}$
$| | C | | = \sqrt{{C}_{x}^{2} + {C}_{y}^{2} + {C}_{z}^{2}} \text{ } | | C | | = \sqrt{{3}^{2} + {\left(- 4\right)}^{2} + {11}^{2}}$
$| | C | | = \sqrt{9 + 16 + 121} \text{ } | | C | | = \sqrt{146}$

$\text{now,let's use dot product formula}$
$\text{Step-5:}$
$A . C = | | A | | \cdot | | C | | \cdot \cos \alpha$
$A . C = 99$
$| | A | | = \sqrt{94}$
$| | C | | = \sqrt{146}$

$99 = \sqrt{94} \cdot \sqrt{146} \cdot \cos \alpha$
$99 = \sqrt{94 \cdot 146} \cdot \cos \alpha$
$99 = 111 , 15 \cdot \cos \alpha$
$\cos \alpha = \frac{99}{111 , 15} = 0 , 8906882591$
$\alpha \cong 27 , {04}^{o}$