Dot product and projection

Question

Given two vectors $\vec{a}=(5,7,-3)$ and $\vec{b}=(1,0,-3)$

Find the projection of $\vec{a}$ onto $\vec{b}$

So we first find $|\vec{a}|=\sqrt{83}$ and then $|\vec{b}|=\sqrt{10}$

We find $\cos \theta$ using $$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|cos\theta $$ Which is: $$14=\sqrt{83}*\sqrt{10}\cos\theta\iff \cos \theta =\frac{14}{\sqrt{830}}$$

Now we want to find $\vec{a}\hat{b}$ where $\hat{b}=\frac{\vec{b}}{|\vec{b}|}$ so we get:

$$\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{14}{\sqrt{10}}$$ but it should have a direction, so the answer is $\frac{14}{\sqrt{10}}\hat{b}$? but we have used $\hat{b}=\frac{\vec{b}}{|\vec{b}|}$ in the calculation

Answer 1

Yes, the projection of $\vec a$ onto $\vec b$ is equal to

$$(\vec a\cdot \hat b)\cdot \hat b$$ where $\hat b$ is the normalized version of vector $\vec b$. Why do you consider this wrong? Sure, you use $\hat b$ both to calculate the length and the direction of the projection, but why exactly do you consider that wrong?

If you want a different formula, the projection of $\vec a$ onto $\vec b$ can also be written as

$$\frac{\vec a\cdot \vec b}{|\vec b|^2}\cdot \vec b$$

or as

$$\frac{\vec a\cdot \vec b}{\vec b\cdot \vec b}\cdot \vec b$$

since all three expressions return the same vector.

Answer 2

The idea of using dot products to find the projection is to take shadow of the required vector onto the unit vector of the required direction on which the projection is to be found. Your are absolutely correct in your statement that the projection vector must have a direction parallel to unit vector of b vector but observe say the projection is termed as P then P=xB and B is the unit vector Along b vector then |P| still remains x as |B|=1