Given two vectors find a unit vector that is perpendicular

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Homework Help Overview

The discussion revolves around finding a unit vector that is perpendicular to two given vectors, \(\vec{A}\) and \(\vec{B}\). The subject area pertains to vector mathematics, specifically the use of the cross product to determine orthogonal vectors.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to utilize the cross product to find a perpendicular vector but expresses uncertainty about the subsequent steps. Other participants suggest calculating the cross product and finding its magnitude, with one participant questioning the need to adjust the signs of the components.

Discussion Status

Participants are actively engaging in the problem-solving process, with some providing guidance on the necessary calculations and confirming the correctness of the vector product and magnitude. There is no explicit consensus on the final form of the unit vector, as discussions about the signs of the components are ongoing.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the extent of assistance provided. The original poster's confusion about the steps following the cross product calculation indicates a need for clarification on vector normalization.

1man
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Given two vectors \vec{A} = - 2.00 \hat{ i } + 3.00 \hat{ j } + 4.00 \hat{k} and \vec{B} = 3.00 \hat{ i } + 1.00 \hat{ j } - 3.00 \hat{k}. Obtain a unit vector perpendicular to these two vectors.

Express your answer as a unit vector N_unit in the form N_x, N_y, N_z where the x, y, and z components are separated by commas.

i know this involves the cross product but I am stuck on what to do, can someone help me please?
 
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Fist calculate the cross product of the two vectors

[tex]\vec{A} = - 2.00 \hat{ i } + 3.00 \hat{ j } + 4.00 \hat{k}[/tex]

and

[tex]\vec{B} = 3.00 \hat{ i } + 1.00 \hat{ j } - 3.00 \hat{k}[/tex]


ehild
 
i get -13, 6, -11, if my calculations are correct. do i need to find the magnitude of this? sqrt of 326... hen divide by this amount for each. Do I also need to change the +/- signs to make it perpendicular
 
Last edited:
The vector product is correct. The magnitude is correct. Yes, divide each component with the magnitude. You can use the signs as they are or change all to opposite, the vector stays perpendicular.

ehild
 
ty so much for your help
 

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