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If ##A = <5 ,2 ,8 >##, ##B = <6 ,5 ,3 >## and ##C=A-B##, what is the angle between A and C?
I got ##59.45^@## (using Cosine formula ...see A.S.Adikesavan's answer using Sine formula)
My understanding is that:
##color(blue)"-------------------------------------------------------"## ##color(blue)("| ")"Given vectors: "## ##color(blue)("| ")color(white)("XXX")color(blue)( vecu=< u_1,u_2,u_3 >" and"color(white)("XXXXX")color(blue)("|"## ##color(blue)("| ")color(white)("XXX")color(blue)(vecv = < v_1,v_2,v_3>color(white)("XXXXXxXXX")color(blue)("|")##
##color(blue)("| ")##For an angle ##color(blue)(theta)## between ##color(blue)(vecu)## and ##color(blue)(vecv)####color(white)("XXXXxX")color(blue)("|")## ##color(blue)("| ") color(white)("XXX")color(blue)(sin(theta)=(vecu * vecv)/(abs(vecu)*abs(vecv))color(white)("XXXXXXXxXX")color(blue)("|")##
##color(blue)("| ")##where##color(white)("XXXXXXXXXXXX")## ##color(blue)("| ")color(white)("XXX")color(blue)(vecu * vecv = u_1 * v_1+u_2 * v_2 + u_3 * v_3)color(white)("X")color(blue)("|")## ##color(blue)("| ")color(white)("XXX")color(blue)(abs(vecu)=sqrt(u_1^2+u_2^2+u_3^2))color(white)("XXXXXXxX")color(blue)("|")## ##color(blue)("| ")color(white)("XXX")color(blue)(abs(vecv)=sqrt(v_1^2+v_2^2+v_3^2))color(white)("XXXXXXXX")color(blue)("|")## ##color(blue)"-------------------------------------------------------"##
In this case, we are given ##color(white)("XXX")color(red)(vecA= < 5,2,8 >), color(red)(vecB= < 6,5,3>), and color(red)(vecC=vecA-vecB)## ##color(white)("XXX")rarr color(red)(vecC=< -1,-3,5 >)##
##color(red)(vecA) * color(red)(vecC)=(5)(-1)+(2)(-3)+(8)(5) color(red)(= 29)##
##color(red)(abs(vecA))=sqrt(5^2+2^2+8^2)color(red)(=sqrt(93))## ##color(red)(abs(vecC))=sqrt(1^1+3^2+5^2)color(red)(=sqrt(35))##
Therefore ##color(white)("XXX")color(green)(cos(theta))=color(red)(29)/(color(red)(sqrt(93) * sqrt(35)))##
Using a calculator: ##color(white)("XXX")color(green)(cos(theta))=0.508303## and ##color(white)("XXX")color(green)(theta) = "arccos"(0.508303) = color(green)(59.45^@)##
