[SOLVED] Equation of a plane - Given z intercept and parametric equation

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May 23rd 2010, 10:05 AM
Kakariki

[SOLVED] Equation of a plane - Given z intercept and parametric equation

Hey,

I have been working on this question for the past couple days (searching the internet, searching for examples in textbook etc.) but I cannot figure out how to answer this question. I will likely have follow up questions to your solution, so please keep an eye on this thread following helping me!

Question
Determine a scalar equation of a plane that contains the line r = (3, -1, 1) + t(2, 4, -5) and whose z-intercept is -2.

Solution
The textbook gives a hint in the back of the book that I need 2 direction vectors in order to find the solution. The question is giving me one direction vector "(2, 4, -5)" but I am unsure of how to find the second one. After I find the second direction vector (which I am unsure how to do) I find the cross product of the two direction vectors to find the normal. After having the normal I think I need a point on the plane The z intercept is listed, so I know a point is (0, 0, -2), right?
Knowing this I can simply sub into the D = -Ax - By - Cz. (which A B C and are from the normal and the x y z are from the point on the plane)
This is how I am understanding what I need to do. PLEASE correct me if I am mistaken here! I am unsure how to do certain steps in my explanation, which is what I am needing help with.

Thank you for helping!!!
May 23rd 2010, 10:44 AM
earboth

Quote:

Originally Posted by Kakariki

Hey,

I have been working on this question for the past couple days (searching the internet, searching for examples in textbook etc.) but I cannot figure out how to answer this question. I will likely have follow up questions to your solution, so please keep an eye on this thread following helping me!

Question
Determine a scalar equation of a plane that contains the line r = (3, -1, 1) + t(2, 4, -5) and whose z-intercept is -2.

Solution
The textbook gives a hint in the back of the book that I need 2 direction vectors in order to find the solution. The question is giving me one direction vector "(2, 4, -5)" but I am unsure of how to find the second one. After I find the second direction vector (which I am unsure how to do) I find the cross product of the two direction vectors to find the normal. After having the normal I think I need a point on the plane The z intercept is listed, so I know a point is (0, 0, -2), right?
Knowing this I can simply sub into the D = -Ax - By - Cz. (which A B C and are from the normal and the x y z are from the point on the plane)
This is how I am understanding what I need to do. PLEASE correct me if I am mistaken here! I am unsure how to do certain steps in my explanation, which is what I am needing help with.

Thank you for helping!!!

1. The line passes through P(3, -1, 1) and the plane passes through Q(0,0,-2). Then the vector $\overrightarrow{PQ}$ is the second direction vector which span he plane.

2. $p: \vec r = (3,-1,1) + t(2,4,-5)+s(-3,1,-3)$

3. Now calculate the cross product of the two direction vectors:

$(2,4,-5) \times (-3,1,-3) = (-7,21,14) = 7(-1,3,2)$

4. Use the "reduced" vector to determine the equation of the plane.
May 23rd 2010, 10:46 AM
Kakariki

Quote:

Originally Posted by earboth

1. The line passes through P(3, -1, 1) and the plane passes through Q(0,0,-2). Then the vector $\overrightarrow{PQ}$ is the second direction vector which span he plane.

2. $p: \vec r = (3,-1,1) + t(2,4,-5)+s(-3,1,-3)$

3. Now calculate the cross product of the two direction vectors:

$(2,4,-5) \times (-3,1,-3) = (-7,21,14) = 7(-1,3,2)$

4. Use the "reduced" vector to determine the equation of the plane.

Thank you! I just figured this out on my own, but thank you for answering so quickly!