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Question
Show that the points A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.
Solution
The given points are A (1, –2, –8), B (5, 0, –2), and C (11, 3, 7).
∴ `vec(AB) = (5 - 1)hati + (0 + 2)hatj + (-2 + 8)hatk `
`= 4hati + 2hatj + 6hatk`
`vec(BC) = (11 - 5)hati + (3 - 0)hatj + (7 + 2)hatk`
`= 6hati + 3hatj + 9hatk`
`vec(AC) = (11 - 1)hati + (3 + 2)hatj + (7 + 8)hatk`
`= 10hati + 5hatj + 15hhatk`
`|vec(AB)| = sqrt(4^2 + 2^2 + 6^2) = sqrt(16 + 4 + 36) `
`= sqrt56 `
`= 2sqrt14`
`|vec(BC)| = sqrt(6^2 + 3^2 + 9^2) `
`= sqrt(36 + 9 + 81)`
` = sqrt126 `
`= 3sqrt14`
`|vec(AC)| = sqrt(10^2 + 5^2 + 15^2)`
` = sqrt(100 + 25 + 225) `
`= sqrt350 `
`= 5sqrt14`
∴ `|vec(AC)| = |vec(AB)| + |vec(BC)|`
Thus, the given points A, B, and C are collinear.
Now, let point B divide AC in the ratio `lambda: 1`. Then, we have:
`vec(OB) = (lambdavec(OC) + vec(OA))/((lambda + 1))`
`⇒ 5hati - 2hatk = (lambda(11hati + 3hatj + 7hatk) + (hati - 2hatj - 8hatk))/(lambda + 1)`
⇒ `(lambda + 1)(5hati - 2hatk) = 11lambdahati + 3lambdahatj + 7lambdahatk + hati - 2hatj - 8hatk`
`⇒ 5(lambda + 1)hati - 2(lambda + 1)hatk = (11lambda + 1)hati + (3lambda - 2)hatj + (7lambda - 8)hatk`
On equating the corresponding components, we get:
5(λ + 1) = 11λ + 1
⇒ 5λ + 5 = 11λ + 1
⇒ 6λ = 4
⇒ λ = `4/6 + 2/3`
Hence, point B divides AC in the ratio of 2: 3
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