Advanced problems in mathematics Q14

By Veronica Fontama on Saturday, December 07, 2002 - 09:26 pm:

Can anyone help with an explanation for this:

This is the last part of Question 14 from "Advanced Problems in Mathematics" by Dr. Siklos

Here it goes:

Given the following three linear equations:
a² + ay +z = a² (1)
ax + y + bz = 1 (2)
a² bx + y + bz = b (3)

It then states:
Substituting a = 0, in equations (1), (2) & (3), they become:
z = 0, y + bz =1, y + bz = b

Comparing the second and third shows that the equations are inconsistent (and have no solutions) unless b =1, and (since z = 0) y = 1. Geometrically, the solution is respresented by any point on the line

X=(0 1 0)+l×(1 0 0)
I know this is the vector equation of a line that passes through (0 1 0) and parallel to (1,0,0). H

I will be grateful if someone can explain the geormetric interpretation of the above to me.

Thank You
Veronica

By Jeremy Reizenstein on Saturday, December 07, 2002 - 10:46 pm:

A "line that passes through (0 1 0) and parallel to (1,0,0)" looks like the following on an x y z coordinate system:

Line

Solutions to the equation (with a=0 and b=1) are z=0, y=1, and x is any number. That is the same as the coordinates of a point on the line.

By John Grindall on Sunday, December 08, 2002 - 03:59 am:

This says that the three planes
z = 0; y + bz = 1; y + bz = b

i) don't have any points of common intersection unless b = 1.

ii) when b = 1, they become z = 0 and y = 1

Here is a picture of these planes:

Planes

They intersect along Jeremy's line.