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Question
If the points A(3, 0, p), B(–1, q, 3) and C(–3, 3, 0) are collinear, then find
- the ratio in which the point C divides the line segment AB
- the values of p and q.
Solution
Let `veca, vecb, vecc` be the position vectors of A, B and C respectively.
Then `veca = 3hati + 0.hatj + phatk`,
`vecb = -hati + qhatj + 3hatk` and
`vecc = - 3hati + 3hatj + 0.hatk`
(i) As the points A, B, C are collinear, suppose the point C divides line segment AB in the ratio λ:1.
∴ By the section formula,
`vecc = (λ.vecb + 1.veca)/(λ + 1)`
∴ `-3hati + 3hatj + 0.hatk = (λ(-hati + qhatj + 3hatk) + (3hati + 0.hatj + phatk))/(λ + 1)`
∴ `(λ + 1)(- 3hati + 3hatj + 0.hatk) = (- λhati + λqhatj + 3λhatk) + (3hati + 0.hatj + phatk)`
∴ `-3(λ + 1)hati + 3(λ + 1)hatj + 0.hatk = (- λ + 3)hati + λqhatj + (3λ + p)hatk`
By equality of vectors, we have,
– 3 (λ + 1) = – λ + 3 ...(1)
3(λ + 1) = λq ...(2)
0 = 3λ + p ...(3)
From equation (1),
– 3λ – 3 = – λ + 3
∴ – 2λ = 6
∴ λ = – 3
∴ C divides segment AB externally in the ratio 3:1.
(ii) Putting λ = – 3 in equation (2), we get
3(– 3 + 1) = – 3q
∴ – 6 = – 3q
∴ q = 2
Also, putting λ = – 3 in equation (3), we get
0 = – 9 + p
∴ p = 9
Hence p = 9 and q = 2.
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