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.