Find Projection of (A+C) in B's Direction

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SUMMARY

The discussion focuses on calculating the projection of the vector sum (A + C) in the direction of vector B. The vectors are defined as A = <2, 1, 1>, B = <1, -2, 2>, and C = <3, -4, 2>. The sum (A + C) results in the vector <5, -3, 3>. The dot product of (A + C) with B is computed to be 16, which is a crucial step in finding the projection. The projection is determined using the formula for the component of one vector in the direction of another.

PREREQUISITES
  • Understanding of vector addition and representation
  • Familiarity with the dot product of vectors
  • Knowledge of vector projection concepts
  • Basic trigonometry, specifically cosine relationships
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LocalStudent
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Homework Statement



The following are all vectors:
A = <2, 1, 1>
B = <1, -2, 2>
C = <3, -4, 2>

Find the projection of (A + C) in the direction of B

Homework Equations



Dot product?


The Attempt at a Solution



I was not sure what the meant in this question.

I added A and C and I got (A+C) = <5, -3, 3>

Then I did (A+C)dot(B) and I got that equal to 16



I was also thinking of dotting (A+C) with the unit vector of B?
 
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Dot product is exactly what you want to do. :smile:

\vec{X}\cdot\vec{Y} = |X|\cdot|Y|cos\theta

You have the dot product, |X|, |Y|, and you need cosine of the angle :wink:
 
Infinitum said:
Dot product is exactly what you want to do. :smile:

\vec{X}\cdot\vec{Y} = |X|\cdot|Y|cos\theta

You have the dot product, |X|, |Y|, and you need cosine of the angle :wink:


So is the question basically asking "What is the angle between (A+C) and B?"
 
LocalStudent said:
So is the question basically asking "What is the angle between (A+C) and B?"

No, it is asking you for the projection(component) of A+C on B. What is the component of a vector X on another vector Y when the angle between them is θ??
 
Infinitum said:
No, it is asking you for the projection(component) of A+C on B. What is the component of a vector X on another vector Y when the angle between them is θ??

ok, I see. Thanks for the help.
 

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