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 points be A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3).

Let us find the position vectors of these points.

Assume that O is the origin.

Position vector of A is given by,

Know that, two vectors are said to be collinear, if they lie on the same line or parallel lines.

Since, A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3) are collinear, we can say that:

Sum of modulus of any two vectors will be equal to the modulus of third vector.

Now,

Now,

Now,

[∵ by algebraic identity, (a – b)² = a² + b² – 2ab]

[∵ by algebraic identity, (a – b)² = a² + b² – 2ab]

Again, squaring on both sides, we get

⇒ (48)² + k² – 2(48)(k) = (k² – 6k + 234)(10) [∵ by algebraic identity, (a – b)² = a² + b² – 2ab]

⇒ 2304 + k² – 96k = 10k² – 60k + 2340

⇒ 10k² – k² – 60k + 96k + 2340 – 2304 = 0

⇒ 9k² + 36k + 36 = 0

⇒ 9 (k² + 4k + 4) = 0

⇒ k² + 4k + 4 = 0

⇒ k² + 2k + 2k + 4 = 0

⇒ k (k + 2) + 2 (k + 2) = 0

⇒ (k + 2)(k + 2) = 0

⇒ k = -2 or k = -2

Thus, value of k is -2.