# [Notes](#fn1): Problem Class, 21 Oct 2020
> Group 11 -- Problem Class, MATH40001/40009 IUM
> Stergios M. Antonakoudis ([e-mail])

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Welcome to IUM Problem Class Group 11. \
Today's class: <u>**Problem Sheet 3 of Part III.**</u>

[^1]: <small> *These notes[^1] are typed up and created during and for the purposes of the discussion in the online session of Problem Class Group 11 for IUM on MS Teams, similarly to the use of a 'whiteboard' in an actual class.* </small>

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## Questions

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### ==Problem 12==
Consider two vectors $u$ and $v$ in $\mathbb{R}^3$ with $u\cdot v = 0$ and $u \neq 0$. Show that the vectors $r$ in $\mathbb{R}^3$ satisfying the equation ==$r\times u=v$== form the line parallel to $u$ and passing through the point $\frac{u \times v}{|u|^2}$.

### ==Solution 12==

First, a few thoughts -- I expect the point $\frac{u \times v}{|u|^2}$ to satisfy the equation $r\times u=v$. Is this true?

Check: ==$v = (\frac{u \times v}{|u|^2}) \times u$==.
Proof: $(u \times v) \times u = (u \cdot u) v + (v \cdot u) u = |u|^2v$, because $u\cdot v =0$. Since $u \neq 0$, we can divide both sides by $|u|^2$. *QED*

Using the above we can re-write the equation:
$r\times u=v \iff r\times u = (\frac{u \times v}{|u|^2}) \times u \iff r\times u - (\frac{u \times v}{|u|^2}) \times u =0$
$(r - (\frac{u \times v}{|u|^2})) \times u =0$. Since $u \neq 0$, we conclude that
$(r - (\frac{u \times v}{|u|^2})) \times u =0 \iff r - (\frac{u \times v}{|u|^2}) = \lambda u$, for some scalar $\lambda$. I.e. It is the equation of a line parallel to $u$ through the given point.

**QED**

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### ==Problem 17==
Let $\mathcal{L}$ a line in $\mathbb{R}^3$ going through the points $(1,1,1)$ and $(1,2,3)$.
(a) Find $v,w \in \mathbb{R}^3$ such that $\mathcal{L}=\{x \in \mathbb{R}^3\;\;:\;\;v\times x = w \}$.
(b) Find the distance from $\mathcal{L}$ to the origin $O$, and the point on $\mathcal{L}$ which is closest to $O$.
### ==Solution 17==

(a) The line $\mathcal{L}$ goes through the point $p=(1,1,1)$ in the direction of $v=(1,2,3)-(1,1,1)=(0,1,2)$, we know that

$x \in \mathcal{L} \iff \exists \lambda \in \mathbb{R}\;\;:\;\; x-p = \lambda v \iff v \times (x-p) =0$.
$v \times x = v\times p = (-1,2,-1)$.
In particular, $v=(0,1,2)$ and $w=(-1,2,-1)$.

(b) The point on the line that closest to the origin in orthogonal to the direction of the line, in this case $v$, but it is also orthogonal to the vector $w$ (because $w = v \times x$, for any $x \in \mathcal{L}$ !).
So the point on the line closest to the origin is of the form $\lambda (v \times w)$, but it has to satisfy $v \times x = w$; hence, $\lambda v \times (v \times w)=w$. We can re-write (as earlier/above) as $\lambda (-(v\cdot v)w) =w$, so we conclude that $\lambda = - 1/|v|^2$. Hence, the point on the line is $-(v \times w)/|v|^2$ and the distance is $|v \times w |/|v|^2 = \sqrt{6/5}$.

**QED**