Finding line where two planes intersect

  • Thread starter ChiralSuperfields
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In summary: In vector form, the above is ##L = \begin{bmatrix} \frac 3 7 \\ \frac 1 7 \\ 1 \end{bmatrix}t + \begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}##. To find the points on this line, you would need to go out 3 units on the x-axis and then go in the direction of (3t/7, t/7, t).
  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1682916672512.png

The solution is,
1682916718595.png

However, I could not get that. By getting the system in REF, I got ##x = 3 + \frac{3}{7}z ## and ##y = \frac{1}{7}z##. Therefore z is a free variable so ##x = 3 + \frac{3}{7}t = 0## and ##y = \frac{1}{7}t##.

Thus equation of line is ##x\hat i + y\hat j + z\hat k = 3\hat i + (\frac{3}{7}\hat i + \frac{1}{7}\hat j)t##

Does anybody please know what I did wrong here?

Many thanks!
 
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  • #2
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

By getting the system in REF, I got x=3+37z=0 and y=17z.
You are right to get
[tex]x=3+\frac{3}{7}z[/tex]
but why do you make it zero ?
 
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  • #3
anuttarasammyak said:
You are right to get
[tex]x=3+\frac{3}{7}z[/tex]
but why do you make it zero ?
Thank you for your reply @anuttarasammyak !

Sorry my mistake. That was not meant to be set to zero.

Many thanks!
 
  • #4
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 325740
The solution is,
View attachment 325741
However, I could not get that. By getting the system in REF, I got ##x = 3 + \frac{3}{7}z ## and ##y = \frac{1}{7}z##. Therefore z is a free variable so ##x = 3 + \frac{3}{7}t = 0## and ##y = \frac{1}{7}t##.
I got the same thing but took it one step further, with z = t. So the equation of the line of intersection is ##L = (\frac 3 7 t + 3)\hat i + \frac 1 7 t \hat j + t \hat k##.

In vector form, the above is ##L = \begin{bmatrix} \frac 3 7 \\ \frac 1 7 \\ 1 \end{bmatrix}t + \begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}##

Geometrically, to get points on this line, go out 3 units on the x-axis, and then go off in the direction of <3t/7, t/7, t>.
ChiralSuperfields said:
Thus equation of line is ##x\hat i + y\hat j + z\hat k = 3\hat i + (\frac{3}{7}\hat i + \frac{1}{7}\hat j)t##
Your equation doesn't follow from your work that preceded it.
 
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