Solving Line-Plane Intersection: Tips and Guidance

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To find the intersection of a line and a plane, substitute the parametric equations of the line into the plane's equation. The line is defined as r = (2, 7, -5) + t(1, 2, -1), which translates to the parametric equations x = 2 + t, y = 7 + 2t, and z = -5 - t. By replacing x, y, and z in the plane equation 2x + 3y - z = 3, you can derive an equation in terms of t. Solving this equation will yield the value of t at which the intersection occurs. The discussion emphasizes the importance of showing some effort in problem-solving to receive help.
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Help!

Hi everyone,
This is my very first post on physics forum! I was wondering if anyone could help explain to me how to find the coordinates of a point where a line intersects a plane.
The question I'm trying to do has given the line as r= (2, 7, -5) + t(1, 2, -1) and the plane as 2x + 3y - z=3.
Any advice or help would be much appreciated, thank you.
 
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The x,y and z coordinates of the points on the line are (2+t, 7+2t, -5-t) right?
Now those coordinates have to satisfy 2x+3y-z=3 to be in the plane. Just plug it in and see for what value it t they are.
 
E=m(C)^2 said:
Hi everyone,
This is my very first post on physics forum! I was wondering if anyone could help explain to me how to find the coordinates of a point where a line intersects a plane.
The question I'm trying to do has given the line as r= (2, 7, -5) + t(1, 2, -1) and the plane as 2x + 3y - z=3.
Any advice or help would be much appreciated, thank you.

Saying the line is given by the vector equation r= (2,7,-5)+ t(1, 2, -1) is the same as saying it is given by the parametric equations x= 2+ t, y= 7+ 2t, z= -5- t. Replace x, y, z in the equation of the plane to get an equation for t and solve.
 
Hi there E=mc2 and welcome to PF,

According to the rules of this forum, you must show some of your own efforts in order to gain assistance. However, if I may offer you the hint that any point of intersection must satisfy the equation of both the line and plane.

Edit: Halls beat me to it; I must learn to type faster. :smile:
 
Please don't double post E=mc2.
 
Thanks a lot guys, i really appreciate it. Sorry about the double post Hootenanny and believe me i made some effort but yeah didn't really show it, sorry about that. Won't happen again.
Thank you again.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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