geometrymultivariable-calculusvector analysis
Given $P_1=(1, 1, 1)$, $P_2=(2, -1, 2)$ and $P_3=(3,0,1)$, I need to prove that these three points are not on the same line.
What I tried – I showed that $\vec{P_1P_2}$, $\vec{P_1P_3}$ and $\vec{P_2P_3}$ are different (meaning they have different directions?). Does that conclude the proof?
Thanks.
Hint:- Let $P_2$ be (h, 0).
1) $x_1(h - x_1) = y_1^2$. Therefore, $(h - x_1)^2 = \dfrac {y_1^4}{x_1^2}$.
2) Use Pythagoras theorem. Then, $d^2 = ... = \dfrac {y_1^2}{x_1^2}(x_1^2 + y_1^2) = \dfrac {y_2^2}{x_2^2}a^2$.
The last equality shows the use of "a".
Best Answer
The points $P_1$, $P_2$ and $P_3$ are not on the same line iff $\vec{P_1P_2}$ and $\vec{P_1P_3}$ are not colinear (ie. there exists $\lambda \in \mathbb{R}$ such that $\vec{P_1P_2} = \lambda \vec{P_1P_3}$).
For example, $A=(0,0,0)$, $B=(1,1,1)$ and $C=(2,2,2)$ are on the same line, but $\vec{AB}$ and $\vec{AC}$ are different; however, $\vec{AC}=2 \vec{AB}$.