How do you find the angle between the vectors #u=2i-3j# and #v=8i+3j#?

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Answer 1 · 2016-10-18T17:13:41

#76.9^o#

Starting from the saclar (dot) product

the definition:

#vec a .vec b=|a||b|costheta#-------(1)
where #theta# is the angle between #veca# &#vecb#

Also to evaluate the scalar product using components we have the result:

#veca=x_"1"veci+y_"1"vecj#

#vecb=x_"2"veci+y_"2"vecj#

#vec a .vec b=x_"1"x_"2"+y_1"##y_"2"#------(2)

so for

#vecu=2veci-3vecj# &# vecv=8veci+3vecj#

using result (2)

#vecu.vecv=(2xx8)+((-3)xx3)#

#vecu.vecv=16-9=7#

Now we use the definition (1)

#vecu.vecv=|u||v|costheta#

so, # 7=sqrt(2^2+3^2)xxsqrt(8^2+3^2)xxcostheta#

#

# 7=sqrt13xxsqrt73xxcostheta#

#theta=cos^-1(7/(sqrt13sqrt73))#

#theta=76.9^o#