In a 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

Given ,`vec(AB) =(3hati+hatj+hatk) and vec(AC)=(hati+2hatj+hatk)`
From the triangle law of verctor , `vec(AB)+vec(BC)=vec(AC)`
or, `vec(BC)=vec(AC)-vec(AB)=(hati+2hatj+hatk)-(3hati+hatj+hatk)=-2hati+hatj`
`angle ABC` is the between `vec(BA) and vec(BC)`.

`therefore vec(BA)*vec(BC)=|vec(BA)||vec(BC)| cos theta`
`or, (6-1) = sqrt(3^2+1^1+1^1)xxsqrt((-2)^2+1^2) xx cos theta `
or, `5/sqrt(55)= cos theta or, theta = cos^(-1) (sqrt(5/(11)))`