Orthogonal vector projection and Components in Orthogonal Directions ....

[Math Amateur]

I am reading Miroslav Lovric's book: Vector Calculus ... and am currently focused n Section 1.3: The Dot Product ...

I need help with an apparently simple matter involving Theorem 1.6 and the section on the orthogonal vector projection and the scalar projection ...My question is as follows:

It seems to me that \(\displaystyle a_v\) in Lovric's Theorem 1.6 should be equal to the scalar projection, \(\displaystyle \| \text{ proj}_v a \|\) ... BUT ... they are not given as equal ...'

... indeed ...

\(\displaystyle a_v = a \cdot v / \| v \|^2\)

while

\(\displaystyle \| \text{ proj}_v a \| = a \cdot v / \| v \|\)BUT ... ... geometrically they seem to be referring to the same quantity ... what then accounts for the difference in the formulae ...?Hope someone can help ...

Peter========================================================================================
The above post refers to Theorem 1.6 so I am providing the text of the same ... as follows:
View attachment 8677
View attachment 8678
The above post also refers to the text on the orthogonal vector projection and the scalar projection ... so I am providing the text of the same ... as follows:

View attachment 8679
View attachment 8680
Hope someone can clarify the issue in the above post ...

Peter

 

[GJA]

Hi Peter,

It seems to me that \(\displaystyle a_v\) in Lovric's Theorem 1.6 should be equal to the scalar projection, \(\displaystyle \| \text{ proj}_v a \|\) ... BUT ... they are not given as equal ...'

... indeed ...

\(\displaystyle a_v = a \cdot v / \| v \|^2\)

while

\(\displaystyle \| \text{ proj}_v a \| = a \cdot v / \| v \|\)BUT ... ... geometrically they seem to be referring to the same quantity ... what then accounts for the difference in the formulae ...?

Your intuition is certainly on the right track, nicely done. The reason for the confusion stems from the fact that $v$ is not assumed to be a unit vector. If $v$ were a unit vector, then $a_{v}$ and $\|\text{Proj}_{v} a\|$ would, as you suspect, be equal.

Take a quick example. Suppose $a = (2, 2)$ and $v= (4, 0).$ Then $\|\text{Proj}_{v} a\| = 2.$ Now, by definition, $\text{Proj}_{v} a$ is the vector in the direction of $v$ with length $2$. Since $v$ itself has length $4$, this vector is not $2v$. Instead, we must scale $v$ down to a unit vector (i.e. $v\mapsto v/4$) and then multiply is by 2. This leads to $\text{Proj}_{v} a = 2\cdot\frac{v}{4}.$ Hence $2/4 = a_{v}\neq \|\text{Proj}_{v} a\| = 2.$

In short, the relationship between $a_{v}$ and $\|\text{Proj}_{v} a\|$ is given by $$a_{v}=\frac{\|\text{Proj}_{v} a\|}{\|v\|},$$ because $a_{v}$ accounts for the scaling "fudge factor" we must employ when $v$ is not a unit vector. From this we can see that $a_{v}=\|\text{Proj}_{v} a\|$ in the case when $v$ is a unit vector.

Edit. If it helps, the only information $v$ conveys to the function $\|\text{Proj}_{v} a\|$ is a direction relative to $a$. In other words, the length of $v$ is not an input variable that $\|\text{Proj}_{v} a\|$ utilizes, that job is left for $a_{v}$. This can be seen explicitly in the formula $$\|\text{Proj}_{v} a\| = \|a\|\cos\theta,$$ where $\theta$ is the angle between $a$ and $v$. Indeed, in the above example, $v$ can be replaced with $v(x) = (x,0)$ where $x>0$. Then $\|\text{Proj}_{v(x)} a\| = 2$ for all $x>0.$ However, $a_{v(x)} = 2/x$ for $x>0$.

 

[Math Amateur]

Hi Peter,
Your intuition is certainly on the right track, nicely done. The reason for the confusion stems from the fact that $v$ is not assumed to be a unit vector. If $v$ were a unit vector, then $a_{v}$ and $\|\text{Proj}_{v} a\|$ would, as you suspect, be equal.

Take a quick example. Suppose $a = (2, 2)$ and $v= (4, 0).$ Then $\|\text{Proj}_{v} a\| = 2.$ Now, by definition, $\text{Proj}_{v} a$ is the vector in the direction of $v$ with length $2$. Since $v$ itself has length $4$, this vector is not $2v$. Instead, we must scale $v$ down to a unit vector (i.e. $v\mapsto v/4$) and then multiply is by 2. This leads to $\text{Proj}_{v} a = 2\cdot\frac{v}{4}.$ Hence $2/4 = a_{v}\neq \|\text{Proj}_{v} a\| = 2.$

In short, the relationship between $a_{v}$ and $\|\text{Proj}_{v} a\|$ is given by $$a_{v}=\frac{\|\text{Proj}_{v} a\|}{\|v\|},$$ because $a_{v}$ accounts for the scaling "fudge factor" we must employ when $v$ is not a unit vector. From this we can see that $a_{v}=\|\text{Proj}_{v} a\|$ in the case when $v$ is a unit vector.

Edit. If it helps, the only information $v$ conveys to the function $\|\text{Proj}_{v} a\|$ is a direction relative to $a$. In other words, the length of $v$ is not an input variable that $\|\text{Proj}_{v} a\|$ utilizes, that job is left for $a_{v}$. This can be seen explicitly in the formula $$\|\text{Proj}_{v} a\| = \|a\|\cos\theta,$$ where $\theta$ is the angle between $a$ and $v$. Indeed, in the above example, $v$ can be replaced with $v(x) = (x,0)$ where $x>0$. Then $\|\text{Proj}_{v(x)} a\| = 2$ for all $x>0.$ However, $a_{v(x)} = 2/x$ for $x>0$.

Well ... that's a VERY helpful explanation ...

Thanks so much for such a helpful post ...

Peter