How Do You Determine the Intersection Point of a Line Orthogonal to a Plane?

  • Thread starter Nikitin
  • Start date
In summary, the conversation is about finding the coordinates of the crossing point P between a plane and a line that are orthogonal. The equation of the plane is given as aX + bY + cZ + d = 0, and the line is given by x = ut, y = vt, z = wt. The solution involves finding the values of u, v, and w that are parallel to the directional vector <a, b, c>, and then substituting them into the equation of the plane to solve for t. The final answer is supposed to be in negative form, not absolute value form. There is a discussion about the logic behind this answer.
  • #1
Nikitin
735
27

Homework Statement




A plane has the equation aX + bY + cZ + d = 0. A line L goes from (0,0,0) and crosses the plane at some point. L and plane are orthogonal. express the coordinates of the crossing point P by: a, b, c and d.


Homework Equations



I know that [a,b,c] is the directional vector for the line and that

|d|/sqrt(a^2 + b^2 + c^2) is the distance between point P and origo. I know and understand this at least.

The Attempt at a Solution



I'm lost lol
 
Physics news on Phys.org
  • #2
ah damn, I haven't learned about directional vectors this year. but i was reminded about that stuff by a friend who helped me with this assignment.

So, if I divide [a,b,c] by the length of the directional vector, sqrt(a^2 + b^2 + c^2) and then multiply it by the length of the OP vector, |d|/sqrt(a^2 + b^2 + c^2), I get:

a|d|/a^2 + b^2 + c^2, b|d|/a^2 + b^2 + c^2, c|d|/a^2 + b^2 + c^2 = P.

The answer is supposed to be (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P

*sigh* does this have something to do with all the fools sometimes saying a plane's equation is ax+by+cz -d = 0 instead of ax+by+cz +d = 0 ? or that d is negative if the directional vector is posetive?

has been a really long day at school + after 4 me and I'm sleepy as hell.. hope somebody can fill me in tomorrow
 
  • #3
Nikitin said:

Homework Statement




A plane has the equation aX + bY + cZ + d = 0. A line L goes from (0,0,0) and crosses the plane at some point. L and plane are orthogonal. express the coordinates of the crossing point P by: a, b, c and d.
Let the line be given by x= ut, y= vt, z= wt (since (0,0,0) satifies that). If the plane and line are orthogonal, then <u, v, w> must be orthogonal to the plane and so must be parallel to <a, b, c>. That is u= ka, v=kb, w= kc. x= kat, y= kbv, z= kct. Putting that into the equation of the plane, [itex]k(a^2+ b^2+ c^2)t+ d= 0[/itex]. [itex]t= -d/(a^2+ b^2+ c^2)[/itex]. Put that into the equation of the line and solve for x, y, and z.


Homework Equations



I know that [a,b,c] is the directional vector for the line and that

|d|/sqrt(a^2 + b^2 + c^2) is the distance between point P and origo. I know and understand this at least.

The Attempt at a Solution



I'm lost lol
 
  • #4
hmm thanks but could you, or somebody else, answer the question in post 2?
 
Last edited:
  • #5
Nikitin said:
ah damn, I haven't learned about directional vectors this year. but i was reminded about that stuff by a friend who helped me with this assignment.

So, if I divide [a,b,c] by the length of the directional vector, sqrt(a^2 + b^2 + c^2) and then multiply it by the length of the OP vector, |d|/sqrt(a^2 + b^2 + c^2), I get:

a|d|/a^2 + b^2 + c^2, b|d|/a^2 + b^2 + c^2, c|d|/a^2 + b^2 + c^2 = P.

The answer is supposed to be (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P

*sigh* does this have something to do with all the fools sometimes saying a plane's equation is ax+by+cz -d = 0 instead of ax+by+cz +d = 0 ? or that d is negative if the directional vector is positive?

has been a really long day at school + after 4 me and I'm sleepy as hell.. hope somebody can fill me in tomorrow

Can somebody explain to me the logic of why the answer should be negative [like this: (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P], and not in absolutes [like this (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P]?

I believe the absolutes answer is plain wrong.
 

1. What is an extremely tough vectors-planes problem?

An extremely tough vectors-planes problem is a mathematical problem that involves finding the intersection point between two or more planes in three-dimensional space using vector equations. It often requires advanced mathematical skills and techniques to solve.

2. What makes solving vectors-planes problems difficult?

Vectors-planes problems can be difficult because they require a strong understanding of vector operations, as well as knowledge of plane geometry and algebraic techniques. They also often involve multiple steps and require a lot of calculation and manipulation.

3. How can I approach solving an extremely tough vectors-planes problem?

First, make sure you have a solid understanding of vector operations and plane geometry. Then, carefully analyze the given information and determine which equations and techniques can be used to solve the problem. Work through the problem step by step, and double check your calculations and solutions.

4. Are there any common mistakes to watch out for when solving vectors-planes problems?

Yes, some common mistakes include not fully understanding the given information, using incorrect equations or operations, and making calculation errors. It is important to carefully check your work and make sure all steps are correct.

5. Are there any tips or tricks for solving extremely tough vectors-planes problems?

One helpful tip is to draw a diagram or visualize the problem in three-dimensional space. This can make it easier to understand the given information and determine the correct approach for solving the problem. It is also important to practice and become familiar with different types of vectors-planes problems to improve problem-solving skills.

Similar threads

  • Calculus and Beyond Homework Help
Replies
11
Views
970
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
273
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
997
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
3K
Back
Top