Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

I've some questions about the proof of $\displaystyle w = u + \{(p-u) \cdot T\}T = p - \{(p-u) \cdot N\}N$

As you can see there are two parts to the proof.

Let $\displaystyle L$ be a line and let $\displaystyle p$ be a point. There is a unique point $\displaystyle w$ on $\displaystyle L$ for which $\displaystyle (w - p)$ is orthogonal to $\displaystyle L$. Moreover, $\displaystyle w$ is given by

the following formulas:

$\displaystyle w = u + \{(p - u) \cdot T\}T = p - \{(p - u) \cdot N\}N ,$

where $\displaystyle u$ is any point on $\displaystyle L$, $\displaystyle T$ is a unit vector in the direction of $\displaystyle L$, and $\displaystyle N$ is a unit vector orthogonal to $\displaystyle T$.

Proof:

$\displaystyle L$ can be parameterized as $\displaystyle z(t) = u + tT$(Vector equation of line). We seek a point $\displaystyle z(t)$ for which $\displaystyle z(t) - p$ is orthogonal to the direction $\displaystyle T$ of the line, i.e. for which

$\displaystyle 0 = (p - z(t)) \cdot T = (p - u - tT) \cdot T$ This relation has a unique solution $\displaystyle t = (p - u) \cdot T$ Substituting for $\displaystyle t$ in $\displaystyle z(t)$ gives the first

formula(first proof) for $\displaystyle w$ in the statement of the result.

My question is:

How did the author find the solution $\displaystyle t = (p - u) \cdot T$ from the equation? What method he used for this?

How did he conclude that $\displaystyle w = u + \{(p-u) \cdot T\}T$?

And back to the proof.

Substituting $\displaystyle v = p - u$ into:

$\displaystyle v = (v \cdot T) T + (v \cdot N) N$

gives the second part of the proof.

How to prove $\displaystyle w = p - \{(p-u) \cdot N\}N$ by substituting $\displaystyle v = p - u$ into $\displaystyle v = (v \cdot T) T + (v \cdot N) N$?

Re: Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

Don't worry I've found the answers.

For the first question i had: we know that $\displaystyle T$ is a unit vector. So $\displaystyle T \cdot T = 1$

$\displaystyle \begin{align*}(p - z(t)) \cdot T =& 0 \\ (p - u - tT) \cdot T =& 0 \\ (p - u) \cdot T - t (T \cdot T) =& 0 \\ t.1 =& (p - u) \cdot T \\ t =& (p - u) \cdot T \end{align*}$

If I plugin $\displaystyle t = (p - u) \cdot T$ into $\displaystyle z(t) = u + tT$ I get:

$\displaystyle z(t) = u + \{(p - u) \cdot T\}T$

Now for the second part of the proof if I plugin $\displaystyle v = p - u$ into $\displaystyle v = (v \cdot T) T + (v \cdot N) N$ I get:

$\displaystyle \begin{align*} p - u = & \{(p - u) \cdot T\} T + \{(p - u) \cdot N\} N \\ p - u = & t T + \{(p - u) \cdot N\} N \text{ because I know that } t = & (p - u) \cdot T \\ u + tT = & p - \{(p - u) \cdot N\} N \text{ because } z(t) = u + tT \\ z(t) = & p - \{(p - u) \cdot N\} N \end{align*}$

And that's it. Q.E.D.

Re: Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

I understand the proof. But what makes one think that $\displaystyle z(t)$ is actually the projection of $\displaystyle p$ which is $\displaystyle w$ and not just any other point on the line $\displaystyle L$?

Because $\displaystyle z(t) = u + tT$ why is this not a projection of point $\displaystyle p$? Why do only $\displaystyle z(t)$ that satisfies this $\displaystyle w = z(t) = u + \{(p - u) \cdot T\}T = p - \{(p - u) \cdot N\}N$ have to be the projection of $\displaystyle p$?

Re: Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

Quote:

Originally Posted by

**x3bnm** I understand the proof. But what makes one think that $\displaystyle z(t)$ is actually the projection of $\displaystyle p$ which is $\displaystyle w$ and not just any other point on the line $\displaystyle L$?

Because $\displaystyle z(t) = u + tT$ why is this not a projection of point $\displaystyle p$? Why do only $\displaystyle z(t)$ that satisfies this $\displaystyle w = z(t) = u + \{(p - u) \cdot T\}T = p - \{(p - u) \cdot N\}N$ have to be the projection of $\displaystyle p$?

I also found the answer.

Because: $\displaystyle [z(t) - p] \cdot T = 0$ only for $\displaystyle w = z(t) = u + \{(p - u) \cdot T\}T = p - \{(p - u) \cdot N \} N$