Why does the formula hold?

[evinda]

Hello! (Wave)
Given two vectors $a$ and $b$ why does it hold that the vector projection of $a$ on $ b$ is $$\frac{\vec{a} \cdot \vec{b}}{||\vec{b}||^2} \cdot \vec{b}$$

?
Could you explain to me why the formula holds?

 

[Greg]

$$\dfrac{\vec{a}\cdot\vec{b}}{||\vec{b}||^2}$$ may be written as $$\dfrac{||\vec{a}||\cos\theta}{||\vec{b}||}$$.

Does that help?

 

[evinda]

$$\dfrac{\vec{a}\cdot\vec{b}}{||\vec{b}||^2}$$ may be written as $$\dfrac{||\vec{a}||\cos\theta}{||\vec{b}||}$$.

Does that help?

How do we use this?

 

[Greg]

$||\vec{a}||\cos\theta$ is the horizontal component of $\vec{a}$. Dividing by $||\vec{b}||$ gives us the ratio by which we want to scale $\vec{b}$.

 

[evinda]

$||\vec{a}||\cos\theta$ is the horizontal component of $\vec{a}$. Dividing by $||\vec{b}||$ gives us the ratio by which we want to scale $\vec{b}$.

Could you explain to me what you mean by horizontal component?

 

[Greg]

Every vector may be expressed as the sum of two perpendicular vectors. The horizontal component is the (signed) magnitude (length) of the vector that is adjacent to the angle $\theta$. It may be negative. For example, the vector <-3, 1> has a horizontal component of -3.

 

[evinda]

Every vector may be expressed as the sum of two perpendicular vectors. The horizontal component is the (signed) magnitude (length) of the vector that is adjacent to the angle $\theta$. It may be negative. For example, the vector <-3, 1> has a horizontal component of -3.

Is this the only way to see it?

 

[Evgeny.Makarov]

The vector projection has the signed length $\|a\|\cos\varphi$ where $\varphi$ is the angle between the two vectors. Since the projection goes in the same direction as $b$ or in the opposite direction, you multiply the length by the unit vector in the direction of $b$, i.e., by $\dfrac{b}{\|b\|}$. The result is
\[
\|a\|\cos\varphi\cdot\frac{b}{\|b\|}=\frac{\|a\|\|b\|\cos\varphi}{\|b\|^2}b=\frac{a\cdot b}{\|b\|^2}b.
\]

 

[evinda]

The vector projection has the signed length $\|a\|\cos\varphi$ where $\varphi$ is the angle between the two vectors. Since the projection goes in the same direction as $b$ or in the opposite direction, you multiply the length by the unit vector in the direction of $b$, i.e., by $\dfrac{b}{\|b\|}$. The result is
\[
\|a\|\cos\varphi\cdot\frac{b}{\|b\|}=\frac{\|a\|\|b\|\cos\varphi}{\|b\|^2}b=\frac{a\cdot b}{\|b\|^2}b.
\]

The vector projection has the signed length $\|a\|\cos\varphi$

What exactly do you mean by "signed length"? (Thinking)

Since the projection goes in the same direction as $b$ or in the opposite direction, you multiply the length by the unit vector in the direction of $b$, i.e., by $\dfrac{b}{\|b\|}$.

Why do we have to multiply the length by the unit vector in the direction of $b$? (Thinking)

 

[Evgeny.Makarov]

What exactly do you mean by "signed length"?

Suppose that a plane or space has a scale that allows measuring the length of every line segment, in particular, the length of every vector. For example, we may stipulate that all lengths are measured in centimeters.

Suppose that a vector $e$ is given on a plane or in space. Given another vector $v$ that is parallel to $e$, by the signed length of $v$ with respect to $e$ I mean the length of $v$ (in the given scale) if $v$ points in the same direction as $e$ and negative length of $v$ if $v$ points in the opposite direction from $e$. In the previous post, I referred to the vector projection of $a$ on $b$ (thus by definition the projection is parallel to $b$) and its signed length with respect to $b$.

Why do we have to multiply the length by the unit vector in the direction of $b$?

If $\|e\|=1$ and $v$ is parallel to $e$, then $v$ is equal to $e$ multiplied by the signed length of $v$ with respect to $e$. This follows from the definition of multiplication of a vector by a number.