Math vectors qn... i forgot how to do??

The points A,B and C, relative to an origin O, have position vectors p + q, 3p - 2q and 6p + mq respectively. Find by a vector method, the value of m for which A, B and C are collinear.

Math vectors qn... i forgot how to do??
If any three distinct points A, B %26amp; C are collinear, then any one of the vectors, say AB, must be equal to "some scalar k times" any other vector, say BC. [Otherwise A, B %26amp; C form a triangle %26amp; shall no more be collinear].





Hence vector AB = k*(vector BC) ( k ≠ 0)





or OB - OA = k(OC - OB)





or (3p - 2q) - (p + q) = k*[(6p + mq) - (3p - 2q)]





or p - q = k*[3p + (m + 2)q]. This implies that the coefficients





of p and q must be proportional:





i.e. 3k/1 = k*(m + 2)/ -1.





Cancelling non-zero k:





3 = - m - 2 or m = - 5.
Reply:Let a = position vector of A


Let b = position vector of B


Let c = position vector of C


a = p + q


b = 3p - 2q


c = 6p + mq


Vector AB = 2p - 3q


Vector AC = 5p + (m - 1) q


If A,B and C are collinear:-


gradient AC = gradient AB


(m - 1) / 5 = (- 3) / 2


m - 1 = -15/2


m = - 13 / 2
Reply:From p+q and 3p-2q, we see that for each p that is added, 3q/2 must be subtracted. So, (p+q)+5p-15q/2, or 6p-13q/2, is on the line.
Reply:First of all we work out the vector of the line, knowing that this takes the form of a + x(b-a) where a,b are position vectors of points A and B and x is a scalar.





work out the vector for the line i.e p + q + x(3p-p + -2q-q)


i.e. p + q + x(2p - 3q)


Then work out the value for x that results in 6p in total


i.e x = 2.5


This means that the total q value is 1 -7.5 = -6.5 i.e. m = -6.5