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Question
Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.
Solution
Let the given points are A(k, 10, 3), B(1, 1, 3) and C(3, 5, 3)
`vec"AB" = (1 - "k")hat"i" + (-1 + 10)hat"j" + (3 - 3)hat"k"`
`vec"AB" = (1 - "k")hat"i" + 9hat"j" + 0hat"k"`
∴ `|vec"AB"| = sqrt((1 - "k")^2 + (9)^2)`
= `sqrt((1 - "k")^2 + 81)`
`vec"BC" = (3 - 1)hat"i" + (5 + 1)hat"j" + (3 - 3)hat"k"`
= `2hat"i" + 6hat"j" + 0hat"k"`
∴ `|vec"BC"| = sqrt((2)^2 + (6)^2)`
= `sqrt(4 + 36)`
= `sqrt(40)`
= `2sqrt(10)`
`vec"AC" = (3 - "k")hat"i" + (5 + 10)hat"j" + (3 - 3)hat"k"`
= `(3 - "k")hat"i" + 15hat"j" + 0hat"k"`
∴ `|vec"AC"| = sqrt((3 - "k")^2 + (15)^2)`
= `sqrt((3 - "k")^2 + 225)`
If A, B and C are collinear, then
`|vec"AB"| + |vec"BC"| = |vec"AC"|`
`sqrt((1 - "k")^2 + 81) + sqrt(40) = sqrt((3 - "k")^2 + 225)`
Squaring both sides, we have
`[sqrt((1 - "k")^2 + 81) + sqrt(40)]^2 = [sqrt((3 - "k")^2 + 225)]^2`
⇒ `(1 - "k")^2 + 81 + 40 + 2sqrt(40) sqrt((1 - "k")^2 + 81) = (3 - "k")^2 + 225`
⇒ `1 + "k"^2 - 2"k" + 121 + 2sqrt(40) sqrt(1 + "k"^2 - 2"k" + 81) =9 + "k"^2 - 6"k" + 225`
⇒ `122 - 2"k" + 2sqrt(40) sqrt("k"^2 - 2"k" + 82) = 234 - 6"k"`
Dividing by 2, we get
⇒ `61 - "k" + sqrt(40) sqrt("k"^2 - 2"k" + 82) = 117 - 3"k"`
⇒ `sqrt(40) sqrt("k"^2 - 2"k" + 82) = 117 - 61 - 3"k" + "k"`
⇒ `sqrt(40) sqrt("k"^2 - 2"k" + 82) = 56 - 2"k"`
⇒ `sqrt(10) sqrt("k"^2 - 2"k" + 82) = 28 - "k"` ...(Dividing by 2)
Squaring both sides, we get
⇒ 10(k2 – 2k + 82) = 784 + k2 – 56k
⇒ 10k2 – 20k + 820 = 784 + k2 – 56k
⇒ 10k2 – k2 – 20k + 56k + 820 – 784 = 0
⇒ 9k2 + 36k + 36 = 0
⇒ k2 + 4k + 4 = 0
⇒ (k + 2)2 = 0
⇒ k = – 2
⇒ k = – 2
Hence, the required value is k = – 2
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