With respect to an origin O, the pts. A and B have the following position vectors
The line has vector equation
I. Prove that the line l does not intersect the line through A and B
II. Find the equation of the plane containing l and the point A giving your answer in the form ax+by+cz=d
Hello I-Think...(1)where is a scalar parameter. Do you understand this bit? It simply says:
So, from (1), the equation of the line through is:
Next, we need to prove that this line doesn't intersect line . We do this by showing that there are no values of and that will bring us to the same point on each of the lines. So, suppose that the lines meet. Then, for some value of and of :
Comparing the coefficients of and , we get:
If we now compare the coefficients of :
On the LHS:Since these are not equal, there are no values of and that satisfy the equation, and hence the lines do not intersect.
On the RHS:
We have a choice of two methods here. The first uses simple algebra; the second, rather more sophisticated, uses a vector and scalar product.
For the first method, we find a couple of points on the line by taking two values of , and then solve a set of simultaneous equations.
So, with and , we get the points and lying on , and therefore also in the plane. And the point also lies in the plane.
So, if the plane is (which is simply the equation re-written when divided throughout by , using new values of ), we get:This gives:
and, when we get rid of fractions, gives the plane:
If you want to see an explanation of the vector method, look at this page.