How do I express λ in terms of μ?

Question

447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-19T12:46:14

#color(blue)(lamda=(7+2mu)/3)#

I'm reading #(bb(a*)bb(b))# as the dot product.

#s=(bb(a*)bb(b))bbc+bbd#

#(bb(a*)bb(b))=(2*1)+(3*2)+(-1*5)=3#

#(bb(a*)bb(b))bbc=3bbc=3((3),(1),(lambda))=((9),(3),(3lambda))#

#(bb(a*)bb(b))bbc+bbd=((9),(3),(3lambda))+((mu),(-2),(15))=((9+mu),(1),(15+3lamda))#

If this is perpendicular to #bba#, then the dot product is zero:

#((9+mu),(1),(15+3lamda))bb*bba=0#

#((9+mu),(1),(15+3lamda))bb* ((2),(3),(-1))=0#

#(2(9+mu)+(3)+(-(15+3lambda)=0#

#18+2mu+3-15-3lamda=0#

#7+2mu-3lamda=0#

#lamda=(7+2mu)/3#