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Find the angle between two vector `A=2hati+hatj-hatk andB=hati-hatk.` |
Answer» `a=|A|=sqrt((2)^(2)+(1)^(2)+(-1)^(2))=sqrt(6)` `B=|b|=sqrt((1)^(2)+(-1)^(2))=sqrt(2)` `A.B(2hati+hatj-hatk).(hati-hatk)` `Now , cos theta=(A.B)/(AB)=(3)/(sqrt(6).sqrt(2))=(3)/(sqrt(12))=(sqrt(3))/(2)` `thereforetheta=30^(@)` |
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