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Question
If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ABC. [∠ABC is the angle between the vectors `bar(BA)` and `bar(BC)`].
Solution
Let O be the origin, then the position vector of A, `vec(OA) = hati + 2hatj + 3hatk`, the position vector of B, `vec(OB) = -hati` and the position vector of C, `vec(OC) = hatj + 2hatk`
`vec(BA) = vec(OA) - vec(OB) = (hati + 2hatj + 3hatk) + hati`
= `2hati + 2hatj + 3hatk`
`|vec(BA)| = sqrt(2^2 + 2^2 + 3^2) `
`= sqrt(4 + 4 + 9) `
`= sqrt17`
`vec(BC) = vec(OC) - vec(OB) = (hatj + 2hatk) - (hati)`
= `hati + hatj + 2hatk`
`|vec(BC)| = sqrt(1^2 + 1^2 + 2^2) `
`= sqrt(1 + 1 + 4)`
` = sqrt6`
Angle between vectors `vec(BC), vec(BA)` ∠ABC
cos ABC = `(vec(BC) xx vec(BA))/(|vec(BC)| |vec(BA)|)`
`= ((hati + hatj + 2hatk) xx (2hati + 2hatj + 3hatk))/(sqrt(1^2 + 1^2 + 2^2 ) sqrt (2^2 + 2^2 + 3^2))`
`= ((1) (2) + (1) (2) + (2) (3))/(sqrt (1 + 1 + 4) sqrt(4 + 4 + 9))`
`= (2 + 2 + 6)/(sqrt6 sqrt7)`
`= 10/sqrt102`
`∠ABC = cos^-1(10/sqrt102)`
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