Question

how to find the equation of the line where two planes meet? (row operations help please!)

Answer 1

P1: 2x –y +4z = 14
P2: 3x +5y + 6z = 21

Answer 2

ive reduced it down to

1 | -0.5 | 2 : 7
0 | 1 | 0 : 0

Answer 3

therefore y=0

Answer 4

but im not sure what to sub next

Answer 5

i have to show the equaiton of the line is x=3-2t y=0 z=2+t

Answer 6

taking the cross product of normal vectors of given planes gives you the direction vector of intersection right ?

Answer 7

yeah i could do that.....

Answer 8

a cross b = [-26, 0, 13]

Answer 9

and i have two points on the line A(3, 0, 2) B(1, 0, 3)

Answer 10

and a cross b = [-2, 0, 1]

Answer 11

voila - i just have to use point a

Answer 12

thanks @ganeshie8

Answer 13

yes :) direction vector and a point is sufficient to fully define a line

Answer 14

row reduction should also work

Answer 15

....it didn't seem to...

Answer 16

just express the pivot variables in terms of free variables

Answer 17

okay..... i got y=0
then x+2z = 7

Answer 18

so let z=t

Answer 19

then x=7-2t ... which isn't right?

Answer 20

2x –y +4z = 14
3x +5y + 6z = 21
--------------------------


say the direction vector of required plane = <p,q,r>
since the required line is perpendicular to both the planes, the dot product of direction vector with the normal vectors of planes is 0 :

2p-q++4r = 0
3p+5q+6r = 0

Answer 21

you need to solve above homogeneous system right ?

Answer 22

ooh i didn't think of it that way....

Answer 23

the matrix equation would be :

\[
\left[\begin{array}\\
2&-1&4 \\
3&5&6 \\
\end{array}\right]
\left[\begin{array}\\
p\\
q\\
r
\end{array}\right]
=
\left[\begin{array}\\
0\\
0\\

\end{array}\right]

\]

Answer 24

you can row reduce and the infinitely many solutions give you the direction vector of intersecting line

Answer 25

thats the idea right ?

Answer 26

the matrix equation would be :

\[
\left[\begin{array}\\
2&-1&4 \\
3&5&6 \\
\end{array}\right] \\~\\

\left[\begin{array}\\
2&-1&4 \\
0&13/2&0 \\
\end{array}\right] \\~\\


\left[\begin{array}\\
2&-1&4 \\
0&1&0 \\
\end{array}\right]~\\~\\


\left[\begin{array}\\
2&0&4 \\
0&1&0 \\
\end{array}\right]~\\~\\

\left[\begin{array}\\
1&0&2 \\
0&1&0 \\
\end{array}\right]~\\~\\


\]

Answer 27

p = -2
q = 0
r = 1

as expected, eh ?

Answer 28

there was a serious typo in my earlier reply, corrected below :


2x –y +4z = 14
3x +5y + 6z = 21
--------------------------


say the direction vector of require \(\color{red}{\text{line}}\) = <p,q,r>
since the required line is perpendicular to both the planes, the dot product of direction vector with the normal vectors of planes is 0 :

2p-q++4r = 0
3p+5q+6r = 0