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

1 Answer
Mar 24, 2016

#alpha~=27,04 ^o#

Explanation:

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

#Step-1:#
#A= <7,-3,6>" "B<4,-7,-5>#
#C=A-B#
#C_x=A_x-B_x=7-4" ; "C_x=3#
#C-y=A_y-B_y=-7+3=-4" ; "C_y=-4 #
#C_z=A_z-B_z=6+5=11" ; "C_z=11#
#C= <3,-4, 11>#
#"Step-2:"#
#"now let's find the dot product of A.C"#
#A.C=A_x*C_x+A_y*C_y+A_z*C_z#
#A.C=7*3+(-3)*(-4)+6.11#
#A.C=21+12+66#
#A.C=99#
#"let's find the magnitude of A and C"#
#"Step-3:"#
#||A||=sqrt(A_x^2+A_y^2+A_z^2)" "||A||=sqrt(7^2+(-3)^2+6^2)#

#||A||=sqrt(49+9+36)" "||A||=sqrt(94)#
#"Step-4:"#
#||C||=sqrt(C_x^2+C_y^2+C_z^2)" "||C||=sqrt(3^2+(-4)^2+11^2)#
#||C||=sqrt(9+16+121)" "||C||=sqrt(146)#

#"now,let's use dot product formula"#
#"Step-5:"#
#A.C=||A||*||C||*cos alpha#
#A.C=99#
#||A||=sqrt94#
#||C||=sqrt146#

#99=sqrt94 *sqrt 146 *cos alpha#
#99=sqrt(94*146)*cos alpha#
#99=111,15*cos alpha#
#cos alpha=99/(111,15)=0,8906882591#
#alpha~=27,04 ^o#