## richyw 3 years ago Show that the line of intersection of the planes $x+2y-z=2$and$3x+2y+2z=7$ is parallel to the line $\langle x,y,x\rangle =\langle 1+6t,3-5t,2-4t\rangle$ Find the equation of the plane determined by these two lines.

1. richyw

I can do the first part, the trouble I am having is finding the plane. How can I do this?

2. helder_edwin

first, solve the system $\large \begin{cases} x+2y-z=2\\ 3x+2y+2z=7 \end{cases}$

3. TuringTest

I would do it a bit differently I would cross the normal vectors of the two planes. The resulting vector should point in the same direction as the line.

4. richyw

they are parallel lines

5. TuringTest

what are parallel line?

6. TuringTest

lines*

7. richyw

I need the plane defined by the two lines.

8. TuringTest

you are given the planes

9. TuringTest

and you are only given one line

10. richyw

right and I have already found the direction of the line defined by the intersection of the two planes, and shown that it is perpendicular to the given line.

11. richyw

now what I need to do is find the plane defined by the given line, and the line of intersection of the two planes.

12. TuringTest

it's supposed to be parallel to the line, I am getting confused

13. richyw

it is. but two parallel lines can still define a plane.

14. helder_edwin

doing what @TuringTest sugested u get the vector $\large (6,-5,-4)$ which the direction of the given line.

15. TuringTest

I'll let @helder_edwin take it then, I am a bit confused as to what you're asking

16. TuringTest

^right, that's what the Q seems to be asking

17. richyw

so what I did was find a point on the line of intersection which I did by setting z=0 and got the point (5/2,-1/4, 0)

18. helder_edwin

for the last part. two parallel lines CANNOT define a plane. u need two independent directions (vectors)

19. TuringTest

indeed

20. richyw

why can they not? I see no reason why they can't.

21. helder_edwin

a plane is two-dimensional whareas a line is one-dimensional

22. TuringTest

you need a point and a vector to define a plane

23. helder_edwin

but the point cannot belong to the line.

24. TuringTest

right, I mean a point in the plane and a normal vector|dw:1349383884636:dw|

25. richyw

but I have two lines. so I could get a displacment vector from one line and a point on the other line?

26. TuringTest

vectors can be moved, so having two does not define any point

27. richyw

but these are not vectors. These are lines!

28. richyw

I have one line, and If I could find the other, then I would have TWO LINES. I cannot visualize how there is more than one unique plane determined by these two lines

29. TuringTest

but you can't get another line because these vectors are linearly dependent

30. richyw

ok. what I need to do is find the equation of the line formed by the intersection of the two planes. This line has the direction (6,-5,-4), now I just need a point along that line. So I can just solve for where those planes equal eachother right?

31. richyw

doing that I find that the point (5/2,-1/4,0) is on the intersection of the two planes.

32. TuringTest

you just need one point of intersection, then use the vector you get from crossing the two normal vectors of the plane to write the equation of the line

33. richyw

now I have a point and a direction, meaning I have a line. How do I find this line?

34. TuringTest

use r(t)=P+tv where P is a point of intersection of the planes and v is the vector that points in that direction

35. richyw

ok I have done this and get $\vec{r}(t)=\langle 5/2+6t, -1/4-5t,-4t \rangle$

36. TuringTest

looks reasonable enough

37. TuringTest

oh I see, so now you want the plane between this line and the given one, eh?

38. richyw

ok now I have another line givin in the initial question.$\langle x,y,x\rangle =\langle 1+6t,3-5t,2-4t\rangle$

39. richyw

so two lines. need this plane!

40. TuringTest

take the given point P1=(1,3,2) and the point you just found P2=(5/2,-1/4,0) and make a vector u=P1-P2 then cross that with the vector for the line of intersection uXv=n (n is the normal vector of the plane you are looking for) then you can use n(dot)P2=n(dot)P where P=(x,y,z) to find the equation of that plane

41. richyw

alright cool thanks a lot! I got super confused when helder_edwin said "two parallel lines CANNOT define a plane. u need two independent directions (vectors)". So just to confirm I can get a second direction vector by subtracting a point on line 1 from a point on line two, correct?

42. helder_edwin

yes. and by doing that u got the second direction u needed.

43. TuringTest

yes @richyw that is right. Sorry for the laps in communication earlier, I was thinking we were just talking about two parallel vectors

44. TuringTest

... and you're welcome :)