If #A = <6 ,6 ,-3 >#, #B = <-9 ,7 ,-2 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2018-03-01T11:42:54

The angle between #vecA and vec C# is #50.09^0#

#vecA =<6,6, -3> , vecB =<-9,7, -2>#

#vec C=vecA- vecB = (< 6,6,-3 >) - (< -9,7,-2 >)#

#=(< 6+9,6-7,-3+2 >) =<15,-1,-1 >#

#vecA =<6,6, -3> and vecC =<15,-1,-1> ; theta# be the

angle between them ; then we know

#cos theta= (vecA*vecC)/(||vecA||*||vecC||)#

#=((6.15)+(6* -1)+(-3*-1))/(sqrt(6^2+6^2+(-3)^2)* (sqrt((15)^2+(-1)^2+(-1)^2))#

#= 87/(sqrt81*sqrt227)~~87/135.6 ~~0.6416#

#:. theta=cos^-1(0.6416)~~50.09^0#

The angle between #vecA and vec C# is #50.09^0# [Ans]

Answer 2 · 2018-03-01T11:44:17

From,the given informations,we can write,

#vec A = 6 hat i+6 hat j -3 hat k# so, #|vec A|=9#

and, #vec B = -9hati+7hatj-2hatk# so, #|vec B|=sqrt(134)#

So,if the angle between, #vec A# and #vec B# be #theta#,

then, # vec A* vec B = |vec A| |vec B| cos theta#

or, #-54+42+6 = 9*sqrt(134) cos theta#

or, #cos theta = (-6/(9*sqrt(134)))#

So,#theta = 93.30^@#

Given, #vec C = vec A - vec B#

So,if the angle between #vec C# and #vec A# be #phi#,

then, #tan phi = (sqrt(134) sin 93.30)/(9-sqrt(134) cos 93.30)#

so, #phi=50^@#