If A=3i+2j+3k and B=i-j+2k then what is angel between them?

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Mar 9, 2018

Answer:

The angle is #=52.5^@#

Explanation:

The angle between #2#vectors, #vecA# and #vecB# is given by the dot product definition.

#vecA.vecB=∥vecA∥*∥vecB∥costheta#

Where #theta# is the angle between #vecA# and #vecB#

The dot product is

#vecA.vecB=〈3,2,3〉.〈1,-1,2〉=(3)*(1)+(2)*(-1)+(3)*(2)=7#

The modulus of #vecA#= #∥〈3,2,3〉∥=sqrt(9+4+9)=sqrt22#

The modulus of #vecB#= #∥〈1,-1,2〉∥=sqrt(1+1+4)=sqrt6#

So,

#costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=7/(sqrt22*sqrt6)=0.609#

#theta=arccos(0.609)=52.5^@#

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