Ina triangle ABC, the sides AB and AC are represented by the vectors `3hati+hatj+hatk` and `hati+2hatj+hatk` respectively. Calculate the angle `angleABC. `

Options

A. `cos^(-1)sqrt((5)/(11))`
B. `cos^(-1)sqrt((6)/(11))`
C. `(90^(@)-cos^(-1)sqrt((5)/(11)))`
D. `(180^(@)-cos^(-1)sqrt((5)/(11)))`

Answer

A

Explanation

`vec(AB)= (3hati+hatj+hatk)`
`vec(AC)=(hati+2hatj+hatk)`
`:. Vec(CB) = vec(AB)- vec(AC)`

`= 3hati+hatj+hatj+hatk-hati-2hatj-hatk=2hati-hatj`
`because angleABC=theta` is the angle between `vec(AB) ` and `vec(CB)` ,
`:. bar(AB) . bar(CB) = |bar(AB)|.|bar(CB)|cos theta`
`implies ( 3hati+hatj+hatk).(2hati-hatj)`
`=(sqrt((3)^(2)+(1)^(2)+(1)^(2)))(sqrt((2)^(2)+(-1)^(2)))costheta`
`=6+(-1)+0=sqrt(11) . sqrt(5)cos theta`
`:. cos theta= (5)/(sqrt(11).sqrt(5))=(sqrt(5))/(sqrt(11))or theta= cos^(-1)[sqrt((5)/(11))]`