If $A = <7 ,-8 ,-2 >$ , $B = <4 ,-8 ,9 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-06-13T15:34:23

$69.6^@$

Given
$vecA=<7,-8,-2>$

$vecB=<4,-8,9>$
$=>vecA-vecB=<(7-4),(-8+8),(-2-9)>$
$=<3,0,-11>$

$|vecA|=sqrt(7^2+(-8)^2+(-2)^2)$
$=sqrt117$

$|vecC|=sqrt(3^2+(0)^2+(-11)^2)$
$=sqrt130$

Now if angle beween $vecA and vecC$ be $theta$ then

$costheta=(vecA*vecC)/(|vecA||vecC|)$

$=(7*3+(-8)*0+(-2)*(-11))/sqrt(117*130)$

$=43/(39sqrt10)$

$:.theta= 69.6^@$

If A = <7 ,-8 ,-2 >A = <7 ,-8 ,-2 >, B = <4 ,-8 ,9 >B = <4 ,-8 ,9 > and C=A-BC=A-B, what is the angle between A and C?