Question

已知a的模=4,b的模=5,a 与b 的夹角为60'.求3a -b 的模

Answer 1

设a=(xa,ya),b=(xb.yb)

|a|=√(xa^2+ya^2)=4

xa^2+ya^2=16

|b|=√(xb^2+yb^2)=5

xb^2+yb^2=25

cos(a,b)=a*b/|a|*|b|=1/2

a*b=1/2*|a|*|b|=1/2*5*4=10

a*b=xaxb+yayb=10

|3a-b|=|(3xa-xb,3ya-yb)|=√[(3xa-xb)^2+(3ya-yb)^2]

=√(9xa^2-6xaxb+xb^2+9ya^2-6yayb+yb^2)

=√[9(xa^2+ya^2)+(xb^2+yb^2)-6(xaxb+yayb)]

=√(9*16+25-6*10)

=√109