A-level math vector problem

[p***@gmail.com]

Hi

I am having difficulty with this question. I hope you can help,
please? I use ( )-> for directional vector notation.

The points A, B and C have position vectors a, b and c respectively
relative to the origin O. P is the point on BC such that (PC)-> =
1/10(BC)->

a) show that the position vector of P is 1/10(9c + b)

1/10(BC)->=1/10(c-b)
(PC)->=c-1/10(c-b)
=c-1/10c+1/10b
=1/10(9c+b)

OK, so far!
---------------

b)Given that the line AP is perpendicular to the line BC, show that
(9c + b).(c-b)=10a.(c-b).

(1/10(9c+b)-a).(c-b)=0
therefore
(9c + b).(c-b)=10a.(c-b)

still OK!
------------

c)Given also that OA, OB and OC are mutually perpendicular, prove that
OC = 1/3OB

Difficulty!
++++++

[Dr A. N. Walker]

In article <***@n2g2000hse.googlegroups.com>,
<***@gmail.com> wrote:
[...]

Post by p***@gmail.com
(9c + b).(c-b)=10a.(c-b)
c)Given also that OA, OB and OC are mutually perpendicular, prove that
OC = 1/3OB
Difficulty!

You're almost there! What do you now know about a.c, for
example? If you expand out the last result and substitute in, what
do you get?

Reminds me a bit of an interesting and little-known extension
to Pythagoras: if OA, OB and OC are mutually perpendicular, show that
the square of the area of ABC is the sum of the squares of the areas
of OBC, OCA and OAB. Well, it's not that hard to do algebraically,
but it's very easy using vectors. The vector area of ABC is minus the
sum of the vector areas of the other three sides [vector area of a
closed surface is zero], square both sides, and use the perpendicular
result for dot products. [Different from the usual 3D Pythagoras,
which is about the squares of various *lengths*.]