How do we prove the projection law using vectors?

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Answer 1 · 2018-04-23T04:52:38

Let #vecaandvecb# are two vectors inclined at an angle #theta#.

So

#veca*vecb=absvecaabsvecbcostheta#

Now vector projection of #veca# on to #vecb# will be
#=absvecacostheta*"unit vector of "vecb#

#=absvecacostheta*vecb/(absvecb)#
#=absvecaabsvecbcostheta*vecb/(absvecb^2)#

#=(veca*vecb)vecb/(absvecb^2)#

Answer 2 · 2018-04-23T10:00:56

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To prove for #ABC#
#a=bcos(C)+c cos(B)# by using vector law.

Proof

Let us consider that three vectors #veca,vecb and vecc# are represented respectively in order by three sides #BC,CA andAB# of a #DeltaABC#.

This means

#vec(BC)=veca#

#vec(CA)=vecb#

#vec(AB)=vecc#

So we can write

#vec(BC)+vec(CA)+vec(AB)=0#

#=>veca+vecb+vec c=0#

#=>veca*(veca+vecb+vec c)=0#

#=>veca*veca+veca*vecb+veca*vec c=0#

#=>absvecaabsvecacos0^@+absvecaabsvecbcos(pi-C)+absvecaabsvecc cos(pi-B)=0#

#=>absvecaabsveca*1-absvecaabsvecbcos(C)-absvecaabsvecc cos(B)=0#

#=>absvecaabsveca=absvecaabsvecbcos(C)+absvecaabsvecc cos(B)#

#=>absveca=absvecbcos(C)+absvecc cos(B)#