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

**x3bnm** · August 31st 2012, 10:21 AM

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

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

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

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

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

Proof:

$L$ can be parameterized as $z(t) = u + tT$ (Vector equation of line). We seek a point $z(t)$ for which $z(t) - p$ is orthogonal to the direction $T$ of the line, i.e. for which
$0 = (p - z(t)) \cdot T = (p - u - tT) \cdot T$ This relation has a unique solution $t = (p - u) \cdot T$ Substituting for $t$ in $z(t)$ gives the first
formula(first proof) for $w$ in the statement of the result.

My question is:

How did the author find the solution $t = (p - u) \cdot T$ from the equation? What method he used for this?
How did he conclude that $w = u + \{(p-u) \cdot T\}T$ ?

And back to the proof.

Substituting $v = p - u$ into:

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

gives the second part of the proof.

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

**x3bnm** · August 31st 2012, 10:46 AM

Don't worry I've found the answers.

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

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

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

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

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

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

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

**x3bnm** · August 31st 2012, 11:47 AM

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

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

**x3bnm** · August 31st 2012, 11:53 AM

Originally Posted by x3bnm

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

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

I also found the answer.

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

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

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