Can any one solve this math problem?

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1 Answer
Mar 14, 2018

The length of the diagonals are #=8.39u# and #=7.72u#

Explanation:

If the parallelogram is built on #veca# and #vecb#

The diagonals are #vecd_1=(veca+vecb)# and #vecd_2=(veca-vecb)#

As

#veca=vecp-2vecq#

and

#vecb=2vecp+vecq#

Then,

#vecd_1=vecp-2vecq+2vecp+vecq=3vecp-vecq#

#vecd_2=vecp-2vecq-2vecp-vecq=-vecp-3vecq#

#|vecp|=2#

#|vecq|=3#

#(vecp,vecq)=3/4pi#

Appling the cosine rule

#|vecd_1|=sqrt(36+9-2*6*3*cos(3/4pi))=8.39#

#|vecd_2|=sqrt((4+81-2*2*9*cos(1/4pi)))=7.72#

I hope that this will help!!!