( r - a ) x b = 0

By Andrew Hodges (P4403) on Sunday, May 20, 2001 - 03:58 pm :

I have my P6 AS level maths exam in a couple of weeks (on matrices and vector geometry). My maths teacher has covered most of the topics before, and so has taught it well, however, there was one topic he had not come across before; expressing a line in the form (r - a ) x b = 0 . Our teacher could not really see any use for it - is there?

On a practice exam paper involving it, it was connected with describing a locus, the question was:

Two points A and P have position vectors a and r , relative to an origin O.

A vector b is such that (r -a ) x b = O.

Given that a and b are fixed vectors and r varies in such a way that the above equation is satisfied, describe in words, the locus of P.

Please could you help me with this, as I hopefully would like to understand it in time for my exam!

By William Astle (Wja24) on Sunday, May 20, 2001 - 04:43 pm :

As you say a line is defined by the set of all points R such that

(r - a) \times b = 0

There are two things of note.

Firstly, A is on the line since if $r = a$ then $0 \times b = 0$ .

Secondly, if R is not A then the vector $r - a$ is parallel to the line (since R and A are both points on the line.) We know $0 = | (r - a) \times b | = | r - a | | b | \sin φ$ (where $φ$ is the angle between $r - a$ and $b$ ) so we must have $φ = n π$ for some integer $n$ (these are the only zeros of sine).

In other words $(r - a) \times b = 0$ defines a line parallel to $b$ which passes through the point A.

Good luck with the exam.

By Andrew Pontzen (P4476) on Sunday, May 20, 2001 - 09:48 pm :

Response to "Our teacher could not really see any use for it - is there?"

Cross products are incredibly useful. Quick example, if you have a plane defined as

r =a +tb

then a vector perpendicular to the plane is

a x b

By Andrew Hodges (P4403) on Monday, May 21, 2001 - 05:18 pm :

Thank you very much for your help - your answer makes sense and agrees with the answer booklet!!