## richyw Group Title 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. one year ago one year ago

1. richyw Group Title

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

2. helder_edwin Group Title

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

3. TuringTest Group Title

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 Group Title

they are parallel lines

5. TuringTest Group Title

what are parallel line?

6. TuringTest Group Title

lines*

7. richyw Group Title

I need the plane defined by the two lines.

8. TuringTest Group Title

you are given the planes

9. TuringTest Group Title

and you are only given one line

10. richyw Group Title

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 Group Title

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 Group Title

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

13. richyw Group Title

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

14. helder_edwin Group Title

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

15. TuringTest Group Title

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

16. TuringTest Group Title

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

17. richyw Group Title

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 Group Title

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

19. TuringTest Group Title

indeed

20. richyw Group Title

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

21. helder_edwin Group Title

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

22. TuringTest Group Title

you need a point and a vector to define a plane

23. helder_edwin Group Title

but the point cannot belong to the line.

24. TuringTest Group Title

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

25. richyw Group Title

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

26. TuringTest Group Title

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

27. richyw Group Title

but these are not vectors. These are lines!

28. richyw Group Title

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 Group Title

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

30. richyw Group Title

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 Group Title

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

32. TuringTest Group Title

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 Group Title

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

34. TuringTest Group Title

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 Group Title

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

36. TuringTest Group Title

looks reasonable enough

37. TuringTest Group Title

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

38. richyw Group Title

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 Group Title

so two lines. need this plane!

40. TuringTest Group Title

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 Group Title

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 Group Title

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

43. TuringTest Group Title

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 Group Title

... and you're welcome :)