• Geometrically, $\scal _\vec {v}(\vec {u})$ gives the signed magnitude of the component of $\vec {u}$ that is parallel to $\vec {v}$; when the angle between $\vec {u}$ and $\vec {v}$ is acute, $\scal _\vec {v}(\vec {u})$ is positive, and $\scal _\vec {v}(\vec {u}) = |\proj _\vec {v}(\vec {u})|$, and when the angle is obtuse, $\scal _\vec {v}(\vec {u})$ is negative and $\scal _\vec {v}(\vec {u}) = -|\proj _\vec {v}(\vec {u})|$

  (think in terms of what this represents; don’t get lost in the notation!).

  Since $\vec {u} \neq \vec {0}$ but $\scal _\vec {v}(\vec {u}) =0$, this means no part of $\vec {u}$ is parallel to $\vec {v}$.

• Algebraically, if $\scal _\vec {v}(\vec {u}) =0$ we have

  $$0 = \scal _\vec {v}(\vec {u}) = \frac {\vec {u} \dotp \vec {v}}{|\vec {v}|}.$$

  Hence $\vec {u} \dotp \vec {v} = 0$ and $\vec {u}$ and $\vec {v}$ are orthogonalparallelneither orthogonal nor parallel , not orthogonalparallelorthogonal nor parallel .

Geometrically, this is difficult to think about, but note that $\vec {u} \dotp \vec {v} = |\vec {u}| \cdot |\vec {v}| \cos (\theta )$, where $\theta$ is the angle between $\vec {u}$ and $\vec {v}$. Since $\cos (\theta ) \leq 1$, $\vec {u} \dotp \vec {v} = |\vec {u}| \cdot |\vec {v}| \cos (\theta )$ $\leq$$\geq$ $|\vec {u}| \cdot |\vec {v}|$

• Geometrically, $\proj _\vec {v}(\vec {u})$ gives the vector component of $\vec {u}$ that is parallel to $\vec {v}$. Since $\vec {u}$ and $\vec {v}$ are parallel by assumption, $\proj _\vec {v}(\vec {u}) =\vec {u}$. Similarly, $\proj _\vec {u}(\vec {v}) =\vec {v}$, so $\proj _\vec {v}(\vec {u}) = \proj _\vec {u}(\vec {v})$ only if $\vec {u}=\vec {v}$whenever $\vec {u}$ and $\vec {v}$ are parallelwhenever $\vec {u}$ and $\vec {v}$ are orthogonalcannot happen .
• Algebraically, we can write out the expressions for both orthogonal projections. $$\begin{align*} \proj_\vec{v}(\vec{u}) &= \left[\frac{\vec{u} \dotp \vec{v}}{\vec{v} \dotp \vec{v}}\right] \vec{v} \\ \proj_\vec{u}(\vec{v}) &= \left[\frac{\vec{v} \dotp \vec{u}}{\vec{u} \dotp \vec{u}}\right] \vec{u} \\ \end{align*}$$

  Since $\proj _\vec {v}(\vec {u})$ is a vector in a direction parallel to $\vec {v}$ and $\proj _\vec {u}(\vec {v})$ is a vector in a direction parallel to $\vec {u}$, the only way $\proj _\vec {v}(\vec {u})$ and $\proj _\vec {u}(\vec {v})$ could be equal is when $\vec {u}$ and $\vec {v}$ are parallel.

  In this case, note that there is a nonzero constant $c$ so $\vec {u} = c \vec {v}$, and thus

  $$\begin{align*} \proj_\vec{v}(\vec{u}) = \proj_\vec{v}(c\vec{v}) &=\left[\frac{c\vec{v} \dotp \vec{v}}{\vec{v} \dotp \vec{v}}\right] \vec{v} = c \vec{v} = \vec{u}\\ \proj_\vec{u}(\vec{v}) = \proj_{c\vec{v}}(\vec{v}) &= \left[\frac{\vec{v} \dotp c\vec{v}}{c\vec{v} \dotp c\vec{v}}\right] c \vec{v} = \vec{v}\\ \end{align*}$$

  Thus, $\proj _\vec {v}(\vec {u}) = \proj _\vec {u}(\vec {v})$ only if $\vec {u}=\vec {v}$whenever $\vec {u}$ and $\vec {v}$ are parallelwhenever $\vec {u}$ and $\vec {v}$ are orthogonalcannot happen .

Take a step back and look at both the algebraic and geometric reasoning used here. Both use the same observation, but it is very easy to lose sight of the geometric intuition in the context of the algebraic argument.

What is true:

The logic that can be used to show this can be extracted from the description given in the previous part of the problem.

• Geometrically, $\scal _\vec {u}(\vec {u})$ gives the signed magnitude of the component of $\vec {u}$ that is parallel to $\vec {u}$, so we expect $\scal _\vec {u}(\vec {u})=$ $\vec {u}$$\vec {0}$ .
• Algebraically, we can write out $\scal _\vec {u}(\vec {u})$.

  $$\scal _\vec {u}(\vec {u}) = \frac {\vec {u} \dotp \vec {u}}{|\vec {u}|} = \frac {|\vec {u}|^2}{|\vec {u}|} = |\vec {u}|.$$

Once again, note that both the algebraic and geometric reasoning lead to the same result, but it’s easy to lose sight of the geometry in the algebra unless you force yourself to consider both perspectives.