9709/May Jun/2008/3/Q10

The points A and B have position vectors, relative to the origin O, given by

OA = i + 2j + 3k  and  OB = 2i + j + 3k.

The line l has vector equation

r = (1 − 2t)i + (5 + t)j + (2 − t)k.

(i) Show that l does not intersect the line passing through A and B.

(ii) The point P lies on l and is such that angle PAB is equal to 60^o. Given that the position vector

of P is (1 − 2t)i + (5 + t)j + (2 − t)k, show that 3t² + 7t + 2 = 0. Hence find the only possible

position vector of P.

Solution:

Reference: PYQ - May/Jun 2008 Paper 3 Q10