Projection of vectors over along their axis

Question

I have difficulties to understand first how does the strong blue vectors appear

Second, how does the light blue vector $w=u\times v$ appears? I thought it was going to be $\vec 0$

Answer 1

We have that $u$ is a unit vector (I assume this from context). The vector $\color{blue}{(p\cdot u)u}$, to the right, is the projection of $p$ in the direction of $u$. It points in the same direction as $u$, since these vectors are parallel (the factor is precisely $(p\cdot u)$).

Now, the vector $\color{blue}{p - (p\cdot u)u}$, to the left, is the difference between $p$ and $\color{blue}{(p\cdot u)u}$, which can be visualized as follows: the vectors $p$ and $\color{blue}{(p\cdot u)u}$ start at the same point, say, $O$. Then draw the vector starting at the end of $\color{blue}{(p\cdot u) u}$ and ending at the tip of $p$. Parallel translate that vector so its start is now at $O$. That is $\color{blue}{p-(p\cdot u)u}$.

Now, to visualize the cross product $\color{teal}{w = u \times v}$, use the right-hand rule: point your index finger in the direction of the first vector, $u$, and your middle finger in the direction of the second vector, $v$. Then your thumb will point in the direction of the cross product $\color{teal}{u \times v}$.

The general formula for a projection of a vector $x$ in the direction of $u$ is $${\rm proj}_ux = \frac{x \cdot u}{u \cdot u}\,u.$$and the fact that $u$ is a unit vector tells us that ${\rm proj}_up = \color{blue}{(p\cdot u)u}$, in the beginning.