Vector help: OA = a, OB = b, and APB is a straight line....

[darksabbath2]

I know its kind of rude to ask for help of you guys but I am really unsure of vectors and I have a exam on monday afternoon.

Could I please ask for some help in solving this vector question. I have no clue on how to solve it as I was away in Hospital for 2 months with a broken neck when we went through this topic.

The questions for it are:

a) Write AB -> in terms of a & b _______________________

P divides AB in the ratio 1:2

b) Find OP -> in terms of a and b, writing your answer in its simplest form ____________________

OPQ is also a straight line and OQ -> = 3/2 OP ->

c) Find AQ -> in terms of a and b, writing your answer in its simplest form
________________

d) what conclusions can you make about the line segments AQ and OB?


THank you in advance I am worried about this topic as I dont understand it!

Cheers

**Cris**

 

[morson]

Hey; thanks for providing a picture, and I hope your neck (broken? :? ) is okay.

a) AB is the vector sum of -a and b. Hence, it is b - a. If you're unsure of how this result was achieved, try getting from "A" to "B," and you'll see that this works by travelling the opposite direction of the vector a and travelling in the direction of b.

b) OP is the vector sum of a and AP. But you're told AP = \(\displaystyle \frac{1}{3}\ AB\), hence OP is a + \(\displaystyle \frac{1}{3}\(b - a)\), from above.

c) AQ is the vector sum of AP and PQ. We're given \(\displaystyle OP = \frac{2}{3}\ OQ\), but OP is the vector sum of b and BP = \(\displaystyle b + (-PB) = b - \frac{2}{3}\ AB = b - \frac{2}{3}\(b - a)\).

Hence, \(\displaystyle b - \frac{2}{3}\ (b - a) = \frac{2}{3}\ OQ\)

So \(\displaystyle \frac{2}{3}\ OQ = b - \frac{2}{3}\ (b - a)\)

And therefore:

\(\displaystyle OQ = \frac{1}{2}\ b - (b - a)\)

But \(\displaystyle PQ = \frac{1}{3}\ OQ = \frac{1}{3}\[\frac{1}{2}\ b - (b - a)]\)

So, finally, \(\displaystyle AQ = \frac{1}{3}\ (b - a) + \frac{1}{3}\[\frac{1}{2}\ b - (b - a)]\)

d) Your simplification of c should have yielded:

\(\displaystyle AQ = \frac{1}{6}\ b\)

Hence, AQ is parallel to b, as its direction has been adjusted by a factor of a sixth. This shows how useful vectors can be in geometrical proofs.