write an indirect proof to prove that if there is a line and a point not on a line, then exactly one plane contains them
Maybe point (a,b,c)
line like z = m x + n y + k
find vector along line
find vector from point to random spot on line
do cross product, show independent of spot chosen on line. That cross product is perpendicular to the plane containing the two lines
To prove that if there is a line and a point not on that line, then exactly one plane contains them, we can use an indirect proof. Here's how to approach the proof:
Indirect Proof:
1. Assume that there are two planes that contain the line and the point not on the line.
(Let's call the line 'l' and the point 'P'.)
2. This means that we have two planes, let's call them 'Plane A' and 'Plane B', that both contain 'l' and 'P'.
3. Since 'Plane A' contains 'l' and 'P', any line that lies in 'Plane A' and contains 'P' must also lie in 'Plane A'. Therefore, we can conclude that 'Plane A' contains all the lines that pass through 'P'.
4. Similarly, 'Plane B' contains all the lines that pass through 'P' as well, since it also contains 'l' and 'P'.
5. Now, consider a line, 'm', that passes through 'P' but does not lie on 'l'. This is possible because 'P' is not on the line 'l'.
6. Since 'm' passes through 'P', it must lie in either 'Plane A' or 'Plane B', as both planes contain all the lines that pass through 'P'. However, both 'Plane A' and 'Plane B' already contain 'l'.
7. This creates a contradiction, because 'l' and 'm' are two distinct lines that both pass through 'P', but according to our assumption, they cannot both lie in the same plane.
8. Hence, our assumption that there are two planes containing 'l' and 'P' is incorrect.
9. Therefore, by contradiction, there can only be one plane that contains both the line 'l' and the point 'P'.
10. This completes the proof.