Let $A = (1; 2; 1)$ and $B = (2;-1;-1)$. Find the vector parametric equation of the line through $A$ and $B$. If $C = (3;-4;-3)$, show using your equation that $A, B$ and $C$ are collinear.
I missed the class on all this and now I am lost. Can someone help me.
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$\begingroup$A parametric equation of a given line requires one point $P_0=(x_0,y_0,z_0)$ on the line, and a vector giving the direction of the line $\vec{u}=(\alpha,\beta,\gamma)$. Then a parametric equation of the line is $P_0+\mathbb{R}\vec{u}$, that is $$ x=x_0+t\alpha\qquad y=y_0+t\beta\qquad z=z_0+t\gamma\qquad(t\in\mathbb{R}). $$ Note that a line has infinitely many parametric equations.
If you are given two points $A,B$ on the line, you can take, for instance $P_0=A$ and $\vec{u}=\vec{AB}$. In your case, this yields: $$ x=1+t\qquad y=2-3t\qquad z=1-2t\qquad(t\in\mathbb{R}). $$
As $A\neq B$, the three points $A,B,C$ are collinear, if and only if $C$ belongs to the line $(A,B)$ above. This amounts to solving for $t$ in the system of three equations $$ 3=1+t\qquad -4=2-3t\qquad -3=1-2t. $$ All three equations yield $t=2$. So $C=A+2\;\vec{AB}$ belongs to the line $(A,B)$, hence $A,B,C$ are collinear.
Note: if you know determinants, there is a convenient symmetric characterization of collinearity for three points $A_j=(x_j,y_j,z_j)$ ($j=1,2,3$). It is $$ \left| \matrix{x_1&x_2&x_3\\y_1&y_2&y_3\\z_1&z_2&z_3}\right|=0. $$See here for details and more.
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