Help Needed! Vectors & Planes: Find Unit Vector & Vector v

Click For Summary

Homework Help Overview

The discussion revolves around vector operations involving three vectors and the intersection of two planes. Participants are addressing how to find a unit vector perpendicular to a combination of vectors and a vector parallel to the line of intersection of two planes defined by their Cartesian equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find a unit vector perpendicular to vector a and the sum of vectors b and c, as well as evaluating the dot product of a with the cross product of b and c. Questions arise regarding the order of operations and the nature of the cross product in relation to the vectors involved.

Discussion Status

Some participants have offered guidance on the relationships between the vectors, particularly regarding the use of cross products to find perpendicular vectors. There is an ongoing exploration of how to derive the necessary vectors and the implications of the operations involved.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the amount of direct assistance provided. The discussion includes clarifications on the definitions of vector operations and the geometric interpretations of the results.

~angel~
Messages
150
Reaction score
0
Please help.

Given the 3 vectors a = -21 + 3j - k, b = 41 - j + 2k and c = -3i + 2j - 3k:

1. Find the unit vector perpendicular to a and b + c.

2. Evaluate a . b x c

I'm completely clueless on how to approach the first question. Any help would be great.

I'm not sure which product I'm meant to perform first in the second question.

Also,

3. p1 and p2 are planes with cartesian equations 2x - y + 3z = 5 and
x - 3y + z = -2, respectively, and l is the line of intersection of p1 and p2.

Find a vector v parallel to l.

I've already determined the normals of both planes:

for p1 : 2i - j + 3k and for p2 : i - 3j + k,

but I'm not sure where to go from here. Clearly v will be perpendicular to both normals, but I don't know how to find that vector.

Any help for these questions would be greatly appreciated.
 
Physics news on Phys.org
Several questions by several posters, so let me just get you started on the first one. I assume you can add b + c. A vector perpendicular to a and b + c is the cross product of a with the sum of b + c. The unit vector is found by dividing that vector by its length.

In the second question, it only makes sense if you do the cross product first. If you did a . b there would be no vector to cross with c
 
Last edited:
Thanks. :smile:
 
~angel~ said:
Thanks. :smile:

I think you can get 3 now. You are right that v is perpendicular to both normals, and you know the normals. So what vector do you know for sure is perpendicular to both of them?
 
If you have time, could you tell me how you know that the cross product of b and c is perpendicular to both a and b + c? Thank you.
 
OlderDan said:
I think you can get 3 now. You are right that v is perpendicular to both normals, and you know the normals. So what vector do you know for sure is perpendicular to both of them?

I bet his question is really easy, but I can't seem to get it. :redface:
 
~angel~ said:
If you have time, could you tell me how you know that the cross product of b and c is perpendicular to both a and b + c? Thank you.

I stated it incorrectly. I will go back and edit it. What I meant to say was

The vector that is perpendicular to a and b + c is the cross product of the vector a with the vector that is the sum of the two vectors b + c.
 
Last edited:
So a X (b + c)?
 
~angel~ said:
So a X (b + c)?

Yes. That is it. Then you have to normalize it to get the unit vector. The cross product is by definition perpendicular to the two vectors in the product.
 
  • #10
Yep. Thanks for that.
 

Similar threads

Replies
26
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
13
Views
2K
Replies
3
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K