If #hat(a_1),hat(a_2)and hat(a_3)# are unit vectors and #2hat(a_1)+2hat(a_2)+hat(a_3) =0# then angle between #hat(a_1) and hat(a_2)# is ?

2 Answers
P dilip_k
Jun 10, 2018

Given #hat(a_1),hat(a_2)and hat(a_3)# are unit vectors

So #abshat(a_1)=abshat(a_2)=abs hat(a_3)=1#

Let the angle between #hat(a_1) and hat(a_2)# is #theta#

Again given

#2hat(a_1)+2hat(a_2)+hat(a_3) =0.....[1]#

So #2abs(hat(a_1)+hat(a_2))=abshat(a_3) #

#=>2sqrt(abshat(a_1)^2+abshat(a_2)^2+2abshat(a_1)abshat(a_2)costheta)=abshat(a_3) #

#=>2sqrt(1^2+1^2+2*1*1*costheta)=1 #

#=>2sqrt(2+2costheta)=1 #

#=>8(1+costheta)=1 #

#=>costheta=1/8-1=-7/8 #

#=>cos^-1(-7/8)~~151^@#

Alternative way

We have

#2hat(a_1)+2hat(a_2)+hat(a_3) =0#

#=>2hat(a_1)*hat(a_1)+2hat(a_2)*hat(a_1)+hat(a_3)hat(a_1) =0#

#=>2+2abshat(a_2)abshat(a_1)costheta+hat(a_3)hat(a_1) =0#

#color(red)(=>2+2costheta+hat(a_3)hat(a_1) =0.......[2])#

similarly

#2hat(a_1)*hat(a_2)+2hat(a_2)*hat(a_2)+hat(a_3) *hat(a_2)=0#

#color(blue)(=>2costheta+2+hat(a_3) *hat(a_2)=0......[3])#

And also

#2hat(a_1)*hat(a_3)+2hat(a_2)*hat(a_3)+hat(a_3) *hat(a_3)=0#

#=>color(green)(2hat(a_1)*hat(a_3)+2hat(a_2)*hat(a_3)+1)=0.......[4])#

Combining [2],[3]and [4] we get

#8(1+costheta)=1 #

#=>costheta=1/8-1=-7/8 #

#=>cos^-1(-7/8)~~151^@#

Shortest method

#2hat(a_1)+2hat(a_2)=-hat(a_3) #

#=>(2hat(a_1)+2hat(a_2))*(2hat(a_1)+2hat(a_2))=hat(a_3) *hat(a_3) #

#=>4(1+2hat(a_2)*hat(a_1)+1)=1 #

#=>8(1+costheta)=1 #

#=>costheta=1/8-1=-7/8 #

#=>cos^-1(-7/8)~~151^@#

Ultrilliam
Jun 10, 2018

#bb( phi = cos^(-1) (-7/8))#

Explanation:

#hat(a_1)* hat(a_2) = cos phi#

#2hat(a_1)+2hat(a_2)+hat(a_3) =0 implies hat(a_1) = - hat(a_2) - 1/2hat(a_3)#

Using: #qquad bb( hat(a_1) = - hat(a_2) - 1/2hat(a_3) )#

  • #bb( hat(a_1)* hat(a_1) )= 1 = - hat(a_1)* hat(a_2) - 1/2hat(a_1)* hat(a_3) = - cos phi - 1/2hat(a_1)* hat(a_3) #

#implies cos phi = -1 - 1/2hat(a_1)* hat(a_3) qquad square#

  • #bb( hat(a_1)* hat(a_2) ) = - a_2^2 - 1/2hat(a_2)* hat(a_3) = - 1 - 1/2hat(a_2)* hat(a_3) #

#implies cos phi = - 1 - 1/2hat(a_2)* hat(a_3) qquad triangle#

  • #bb( hat(a_1)* hat(a_3) )= - hat(a_2)* hat(a_3) - 1/2 qquad circ#

#square + circ implies cos phi = -1 - 1/2(- hat(a_2)* hat(a_3) - 1/2)#

#= -3/4 + 1/2 hat(a_2)* hat(a_3) qquad star#

#triangle + star implies 2 cos phi = -7/4#

#bb( phi = cos^(-1) (-7/8))#

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