Question

Hi, please I need help with: if I have 2 lines with each one having x, y, z known coordinates, 1)how to find a third line orthogonal to both? 2)how to prove that there is a plane containing both lines, 3)if so what's the equation? thanks.

Answer 1

a cross product

Answer 2

there may not be a plane containing both lines unless the lines actually meet.

Answer 3

Yes, I need to 1st prove that they intersect (the 2 lines) but...I don't know how to solve after that! how to find the 3rd line orthogonal to both...and the rest, please help!

Answer 4

do you have any specific lines? or do youjust keep it general?

Answer 5

yep...1st line: x=5-3t, y=2, z=2-t (parameter t) the 2nd line: x=s, y=3-4s, z=1+s (parameter s)

Answer 6

x = 5 - 3t
y = 2 + 0t
z = 2 - 1t
^ ^ vector t = (-3,0,-1)
point
-----------------------------

x = 0 + 1s
y = 3 - 4s
z = 1 + 1s
^ ^vector s = (1, -4, 1)
point

does this make sense so far?

Answer 7

yep, so far so good

Answer 8

we can go a long route and try to compare point values, or we can go straight to defining a plane with at least one of the lines it it

Answer 9

how to do that? could you please choose the simplest way to solve it?

Answer 10

a cross product is a determinant of a matrix; in this case, a 3x3 matrix

x y z x y
-3, 0, -1 -3 0
1, -4, 1 1 -4

take the first 2 columns and repeat them; then run the diagonals

- (0z 4x -3y)
x y z x y
-3, 0, -1 -3 0
1, -4, 1 1 -4
+ (0x -1y 12z)

add like terms

-4x, 2y, 12z; our othogonal vectors is: (2,-1,-3) if i did that right

Answer 11

yay i did it right lol

http://www.wolframalpha.com/input/?i=determinant+%7B%7Bx%2Cy%2Cz%7D%2C%7B-3%2C0%2C-1%7D%2C%7B1%2C-4%2C1%7D%7D

Answer 12

(2,-1,-6) is better since i reduced it all by -2

Answer 13

no we use that orthogonal vector and one of the point values to define a plane with one of the lines in it: lets use the top point (5,2,2)

2 (x-5) -1(y-2) -6(z-2) = 0

Answer 14

oh! I never thought that algebra applies to this problem...

Answer 15

it does :)

my strategy was to define a plane that has the greatest chance of containing both lines, i know that one line is definantly in the plane, its only left to determine of the other xyz defnition fit into it, or if i only get a single solution for s

Answer 16

using the xyz values of:

x = 0 + 1s
y = 3 - 4s
z = 1 + 1s

lets input them into the plane equation, and solve for s

2 (x-5) -1(y-2) -6(z-2) = 0

2 (0 + s - 5) -1(3 - 4s -2) -6(1+ s - 2) = 0

if s can only be one value, then this passes thru the plane and never is contained by it

Answer 17

could you then find the equation???

Answer 18

if the lines cross, then they will be defined for all values of t or s since they exist in one plane.


if they are parallel lines, the determinant we get would have been 0, or a normal vector of (0,0,0) which produces a true result regardless