If #A = <7 ,-5 ,6 >#, #B = <4 ,-8 ,9 ># and #C=A-B#, what is the angle between A and C?

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Oct 13, 2016

Answer:

#theta ~~ 102.7°#

Explanation:

Compute vector C:

#barC=barA -barB = (7-4)hati + (-5 - -8)hatj + (6 - 9)hatk#

#barC = 3hati + 3hatj - 3hatk#

Compute the dot-product of vectors A and C:

#barA*barC = (7)(3) + (-5)(3) + (6)(-3)#

#barA*barC = (7)(3) + (-5)(3) + (6)(-3)#

#barA*barC = -12#

There is another form for the equation of the dot-product that contains the angle between the two vectors:

#barA*barC = |barA||barC|cos(theta) = -12#

Compute the magnitude of vector A:

#|barA| = sqrt(7^2 + (-5)^2 + 6^2)#

#|barA| = sqrt(110)#

Compute the magnitude of vector C:

#|barC| = sqrt(3^2 + 3^2 + (-3)^2)#

#|barC| = 3sqrt(3)#

#sqrt(110)(3sqrt(3))cos(theta) = -12#

#cos(theta) = -12/(sqrt(110)(3sqrt(3))#

#theta = cos^-1(-4/(sqrt(330)))#

#theta ~~ 102.7°#

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