Find a vector parametric equation $\vec r(t)$ for the line through the points $P\equiv(1,0,−4)$ and$ Q\equiv(3,−3,1)$

Question

Find a vector parametric equation $\vec r(t)$ for the line through the points $P\equiv(1,0,−4)$ and $Q\equiv(3,−3,1)$ for each of the given conditions on the parameter $t$

If $\vec r(3)=P$ and $\vec r(7)=Q$

I get how to do it then I have $r(0)$ but im not sure what to do with the $r(3)$

Answer 1

$$r(t) = \frac{(7-t)}{(7-3)}(1,0,-4) + \frac{(t-3)}{(7-3)}(3,-3,1)$$ will do.

Answer 2

Hint: If $$\vec g:[0,1]\to \Bbb R^3$$ and $$s:[a,b]\to [0,1]: t \mapsto \frac{a-t}{a-b},$$ then for $\vec h(t):=\vec g(s(t))$, we have $$\vec h: [a,b]\to \Bbb R^3, \quad \vec h(a)=\vec g(0),\quad\vec h(b)=\vec g(1),\quad \text{ and } \quad\vec h([a,b])=\vec g([0,1])$$

Answer 3

Hint: If your line is the points $(x_0+tx_1, y_0+ty_1, z_0+tz_1)$ then you have the results

• $x_0+3x_1=1$
• $x_0+7x_1=3$
• $y_0+3y_1=0$
• $y_0+7y_1=-3$
• $z_0+3z_1=-4$
• $z_0+7z_1=1$

which are six equations in six unknowns, easuily solved in pairs