Vectors ; specifically cross product application

Click For Summary

Homework Help Overview

The discussion revolves around the application of the cross product in vector analysis, specifically focusing on finding the magnitude of the component of vector A that is perpendicular to vector B while remaining in the plane defined by both vectors A and B.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the cross product to find the required component but expresses confusion about their approach. Some participants suggest finding a vector that is perpendicular to B within the plane of A and B, proposing the use of the expression (AxB)xB. Others explore the geometric implications of this approach and the use of the dot product to find the component of A in the direction of B.

Discussion Status

The discussion is active, with participants providing insights and suggestions for approaching the problem. There is a collaborative exploration of concepts, and while one participant claims to have solved the problem, the conversation reflects ongoing inquiry and clarification of the underlying principles.

Contextual Notes

Participants are navigating the complexities of vector magnitudes and directions, emphasizing the need for clarity in the geometric relationships between the vectors involved. The original poster indicates a struggle with the fundamental approach, highlighting the challenges inherent in the problem setup.

kapitanma
Messages
4
Reaction score
0

Homework Statement



vector A = 1.5i + 6.7j - 7.4k
vector B= -8.2i + 6.5j + 2.3k

(f) What is the magnitude of the component of vector A perpendicular to the direction of vector B but in the plane of vector A and B.


The Attempt at a Solution



This part of the problem has me kinda stumped. My attempt at the solution was using the application of the cross product : C = ABsin(theta).

I calculated the angle between the two as 82.43 degrees, and realized that this formula gave me the same thing as simply taking the magnitude of the cross product vector which I calculated to be 107.207.

This shows me that my fundamental approach to this problem is incorrect, but I have no idea where to go with it. Any pointers would be appreciated.
 
Physics news on Phys.org
Magnitudes of vectors isn't all you need, you also need directions. It would be nice to find a vector that is perpendicular to B but in the plane of A and B, right? Then you could just find the magnitude of A along that direction. How about (AxB)xB? Can you see why that works?
 
Looking at a graph of AxB, I think I can see why that (AxB)xB would give me a vector perpendicular to B in the plane, and I'd just have to apply the dot product to get the component of A in the direction of B.
 
kapitanma said:
Looking at a graph of AxB, I think I can see why that (AxB)xB would give me a vector perpendicular to B in the plane, and I'd just have to apply the dot product to get the component of A in the direction of B.

Sure. (AxB)xB is perpendicular to B, and it's also perpendicular to AxB which is the normal to plane containing A and B. And, yes, from here you can use a dot product.
 
I successfully solved this problem, thanks for the nudge in the right direction.
 

Similar threads

Replies
26
Views
2K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
21
Views
4K