C4 vectors

Cannot do this question part b;



My working for part a)


My working for part b, using two failed attempts:


Method 2


I don't understand how to get the other values, despite looking at the various methods on the markscheme.

Markecheme:


Thanks!

-Jay




Posted from TSR Mobile

I haven't worked this question through, but it looks like they're using the fact that PA is perpendicular to l to observe that (since A and B are both on l) you have an isosceles right-angled triangle and to find the coordinates you can either use the fact that the 2 non-hypotenuse sides have equal length (magnitude) or the fact that the length of the hypotenuse is root(2) times the length of one of the other sides.

Bump, anyone??


Posted from TSR Mobile

Where are you getting stuck now?

Are you happy that PAB is a right-angled triangle with angle BPA = 45 degrees and angle PBA = 45 degrees, so length of PA = length of AB.

Can you work out length of PA from the info given?

There are a couple of ways of finding possible positions of B - I chose the longer, "stupider" method which involves working out the length of AB in terms of lambda and solving a quadratic equation, but looking at the mark scheme it's much simpler to note that B lies either side of A and a certain distance away from it using a multiple of the direction vector of l.

I have no idea where I went wrong, if you see my working above for method 2...

If possible, can you post your working or take a picture, I need to see where that λ comes from to form a quadratic. I prefer it that way. (method you did)