Question

Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.

Answer 1

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