Perpendicular vector using dot not cross product.

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Homework Help Overview

The original poster is working with two 3D vectors, A and B, and is attempting to find a third vector, C, that is perpendicular to both A and B using the properties of the dot product.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the condition that the dot product of perpendicular vectors equals zero, leading to two equations involving the components of vector C. The original poster expresses uncertainty about how to proceed with solving for the unknowns given the equations.

Discussion Status

Some participants have provided guidance on using the dot product to establish the necessary conditions for vector C. The original poster acknowledges their understanding but indicates they are still grappling with the implications of the equations and the freedom to choose components of vector C.

Contextual Notes

The original poster notes a potential limitation in the information available to solve for all components of vector C, suggesting a need to choose one component freely.

Alex1976
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Homework Statement


I have 2 (3d)vectors A and B and I want to find a vector C perpendicular to both of them.
A = 3i-2j+4k
B = -2i+5j-2k
C = Cx+Cy+Cz

Homework Equations


So we know A dot C = 3Cx-2Cy+4Cz and B dot C = -2Cx+5Cy-2Cz

The Attempt at a Solution

 
Last edited:
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The dot product of perpendicular vectors is equal to zero. Use this to find vector C.

ehild
 
edit...
 
Last edited:
Alex1976 said:
Yup I know that so we have:
A dot C = 0 = 3Cx-2Cy+4Cz
B dot C = 0 = -2Cx+5Cy-2Cz
But then I'm stuck.
I can isolate any of these obviously but I can't see how there's enough information for me to solve for my unknowns...

If \vec{C} is perpendicular to \vec{A} and to \vec{B}\,, then so is k\,\vec{C}, where k is a scalar, so in general you will be free to choose one of the components of \vec{C}\,.
 
Wait. I see it now, thank's all.
 
Last edited:

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