## zordoloom Group Title Find a vector equation and parametric equations for the line segment that joins P(2, 0, 0) Q(6, 2, 2) one year ago one year ago

1. No-data Group Title

Remember that to get vector equation you need any point that lies on to the line and a vector that goes in the same direction.

2. No-data Group Title

You could pick point P or point Q.

3. across Group Title

You could try$\mathbf{r}(t)=\mathbf{P}+t(\mathbf{Q}-\mathbf{P}),\qquad0\leq t\leq1.$Notice that if $$t=0$$, then $$\mathbf{r}(t)=\mathbf{P}$$, and if $$t=1$$, then $$\mathbf{r}(t)=\mathbf{Q}$$.

4. No-data Group Title

How do you make the letters bold @across ?

5. zordoloom Group Title

@across Do I just plug in the values P and Q for that formula?

6. across Group Title

\mathbf{}

7. sami-21 Group Title

just write Large before you type

8. No-data Group Title

$\vec{r}=\large{P}$

9. sami-21 Group Title

@zordoloom did you get it now ?

10. zordoloom Group Title

So the formula that across provided, Can I use that to find the equations?

11. sami-21 Group Title

first find the required parallel vector to the line PQ=<6-2,2-0,2-0)= <4,2,2>

12. sami-21 Group Title

yes you can use that .

13. zordoloom Group Title

so, i end up with r(t)=(2,0,0)+t(4,2,-2). Is this correct?

14. sami-21 Group Title

r(t)=(2,0,0)+t(4,2,2 ) not -2 !!!

15. zordoloom Group Title

ok

16. zordoloom Group Title

Thanks!!

17. sami-21 Group Title

welcome :)