Component of a vector along another vector.

[Hijaz Aslam]

Homework Statement

Given ##\vec{A}=2\hat{i}+3\hat{j}## and ##\vec{B}=\hat{i}+\hat{j}##.Find the component of ##\vec{A}## along ##\vec{B}##.

Homework Equations

##\vec{A}.\vec{B}=ABcosθ## where θ is the angle between both the vectors.

The Attempt at a Solution

I attempted the question as follows:
Let the angle between ##\vec{A}## and ##\vec{B}## be 'θ'. So the component of ##\vec{A}## along ##\vec{B}## is given by ##Acosθ\hat{B}## => ##Acosθ(\frac{\vec{B}}{B})##

As ##\vec{A}.\vec{B}=ABcosθ## => ##[( 2\hat{i}+3\hat{j})(\hat{i}+\hat{j})]/B=Acosθ## => ##\frac{5}{\sqrt{2}}=Acosθ##

Therefore the component is : ##\frac{5}{\sqrt{2}}(\frac{\hat{i}+\hat{j}}{\sqrt{2}})## => ##\frac{5}{2}({\hat{i}+\hat{j}})##

But my text produces the solution as follows:
##A_B=(\vec{A}.\vec{B})\hat{B}=\frac{5}{\sqrt{2}}(\hat{i}+\hat{j})##.

 

[RUber]

Therefore the component is : ##\frac{5}{\sqrt{2}}(\frac{\hat{i}+\hat{j}}{\sqrt{2}})## => ##\frac{5}{2}({\hat{i}+\hat{j}})##

But my text produces the solution as follows:
##A_B=(\vec{A}.\vec{B})\hat{B}=\frac{5}{\sqrt{2}}(\hat{i}+\hat{j})##.

I usually see this process broken down into basis components.
That is ##\hat B =\sqrt{2}/2 \hat i + \sqrt{2}/2 \hat j. ##
Then the component is ##A\cdot \hat B_i \hat i + A\cdot \hat B_j \hat j ##.
Somewhere in your process, you divided by the magnitude of B twice.

 

[willem2]

But my text produces the solution as follows:
AB=(A⃗ .B⃗ )B^=52√(i^+j^)A_B=(\vec{A}.\vec{B})\hat{B}=\frac{5}{\sqrt{2}}(\hat{i}+\hat{j}).

You're right and the book is wrong. The book answer as well as the formula for A_B they use.
The length of your answer is smaller than the length of A as it should be. The book answer is larger.
The projection of A on B should only depend on the direction of B, not the magnitude. The formula used for A_B in the book does depend on the magnitude of B.

 

[Delta2]

Book is wrong . We can verify this by standard euclidean geometry easily because by the definition of cosine, it will be cos(θ)=(component of A along B)/A hence Acos(θ)=(component of A along B). And we have to multiply this by the unit vector of B to get the required result.