Multivariable Calculus - Scalar projection

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Homework Help Overview

The discussion revolves around finding the scalar and vector projection of the vector b=(3,5,3) onto the vector a=(0,1,-5) within the context of multivariable calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the scalar projection using the dot product and the magnitude of vector a. Questions are raised regarding the correctness of the initial calculations and the reasoning behind dividing by the magnitude of vector a.

Discussion Status

Some participants have provided guidance on the formula for scalar projection and confirmed that the approach taken is generally correct. However, there remains a request for clarification on the underlying concepts and calculations.

Contextual Notes

There is an emphasis on understanding the relationship between the dot product and the angle between the vectors, as well as the implications of using the magnitude in the calculation.

Larrytsai
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Homework Statement


Find the scalar and vector projection of the vector b=(3,5,3) onto the vector a=(0,1,-5) .


Homework Equations





The Attempt at a Solution



What I've tried is multiplying all the i's and j's and k's together and adding up everything because you get a scalar answer in the end so basically...

3*0 + 1*5 + 3*(-5)
= -10
then i tried to divide by the magnitude of vector |a|

|a|=sqrt(26)

so i get scalar projection as -10/(sqrt(26)) as my answer
 
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Ok, so the scalar projection is b.a/|a|. That seems ok. What's your question?
 
Can you please explain why we divide by the magnitude of vector "a" and if my work is correct or not please?
 
Larrytsai said:
Can you please explain why we divide by the magnitude of vector "a" and if my work is correct or not please?

The scalar projection should be |b|*cos(theta) where theta is the angle between a and b. Since the dot product is given by (a.b)=|a||b|cos(theta) the scalar projection should be (a.b)/|a|. Yes, you are doing fine so far.
 

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