If $A = {2, - 7, 5}$ $A = {2,-7 ,5}$ , $B = {5, - 7, 4}$ $B = {5,-7,4}$ and $C = A - B$ $C = A - B$ , what is the angle between A and C?

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Answer 1 · 2016-04-05T11:01:31

$α = 90, 02^{o}$ $alpha=90,02 ^o$

$a) find C=A-B$ $"a) find C=A-B"$

$C = (A_{x} - B_{x}), (A_{y} - B_{y}), (A_{z} - B_{z})$ $C=(A_x-B_x),(A_y-B_y),(A_z-B_z)$

$C = (2 - 5), (- 7 + 7), (5 - 4)$ $C=(2-5),(-7+7),(5-4)$

$C = {- 3, 0, 1}$ $C={-3,0,1}$

$b) find the magnitude of A and C:$ $"b) find the magnitude of A and C:"$

$| | A | | = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}$ $||A||=sqrt(A_x^2+A_y^2+A_z^2)$

$| | A | | = \sqrt{2^{2} + {(- 7)}^{2} + 5^{2}} = \sqrt{4 + 49 + 25} = \sqrt{78}$ $||A||=sqrt(2^2+(-7)^2+5^2)=sqrt(4+49+25)=sqrt78$

$| | C | | = \sqrt{C_{x}^{2} + C_{y}^{2} + C_{z}^{2}}$ $||C||=sqrt(C_x^2+C_y^2+C_z^2)$

$| | C | | = \sqrt{{(- 3)}^{2} + 0^{2} + 1^{2}} = \sqrt{9 + 0 + 1} = \sqrt{10}$ $||C||=sqrt((-3)^2+0^2+1^2)=sqrt(9+0+1)=sqrt10$

$c) find dot product of A.C$ $"c) find dot product of A.C"$

$A \cdot C = A_{x} \cdot C_{x} + A_{y} \cdot C_{y} + A_{z} \cdot C_{z}$ $A*C=A_x*C_x+A_y*C_y+A_z*C_z$

$A \cdot C = 2 \cdot (- 3) + (- 7) \cdot 0 + 5 \cdot 1$ $A*C=2*(-3)+(-7)*0+5*1$

$A \cdot C = - 6 + 0 + 5 = - 1$ $A*C=-6+0+5=-1$

$d) use the dot product formula$ $"d) use the dot product formula"$

$A . C = | | A | | \cdot | | C | | \cdot cos α$ $A.C=||A||*||C||*cos alpha$

$A \cdot C = - 1$ $A*C=-1$
$| | A | | = \sqrt{78}$ $||A||=sqrt78$
$| | C | | = \sqrt{10}$ $||C||=sqrt10$

$- 1 = \sqrt{78} \cdot \sqrt{10} \cdot cos α$ $-1=sqrt78*sqrt10*cos alpha$

$cos α = - \frac{1}{\sqrt{78} \cdot \sqrt{10}}$ $cos alpha=-1/(sqrt78*sqrt10)$

$cos α = - \frac{1}{\sqrt{780}}$ $cos alpha=-1/sqrt780$

$cos α = - \frac{1}{27, 93}$ $cos alpha=-1/(27,93)$

$α = 90, 02^{o}$ $alpha=90,02 ^o$

If A={2,−7,5}A = {2,-7 ,5}, B={5,−7,4}B = {5,-7,4} and C=A−BC = A - B, what is the angle between A and C?