Q. 5

# Using vectors, fi

Answer :

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, Position vector of B is given by, Position vector of C 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.

This means, we need to find .

To find : Position vector of B-Position vector of A    Now,  …(i)

To find : Position vector of C-Position vector of B    Now,   …(ii)

To find : Position vector of C-Position vector of A    Now,  …(iii)

Take, Substitute values of from (i), (ii) and (iii) respectively. We get, Or  [ by algebraic identity, (a – b)2 = a2 + b2 – 2ab]  Squaring on both sides,  [ 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.

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