vectors
The points A and B have position vectors, relative to the origin $O$, given by $\overrightarrow {OA} = i + j+ k$ and $\overrightarrow {OB} = 2i + 3k$. The line $l$ has vector equation $r = 2i -2j -k + \mu(-i +2j +k)$.
$i)$ Show that the line passing through A and B does not intersect $l$.
$ii)$ Show that the length of the perpendicular from A to $l$ is $\frac {1}{\sqrt 2}$.
I did part $i$, how do I do part $ii$?
Keeping it abstract, you have a line $L_1$ given by $\vec P + \lambda \vec v$, and you want to find a line $L_2$ perpendicular to $L_1$ and passing through an external point $\vec E$.
We want to find the point $\vec C$ on $L_1$ closest to $\vec E$, because then $L_2$ will be the line through $\vec C$ and $\vec E$. The vector $\vec{CE}$ will be perpendicular to $L_1$, i.e. perpendicular to $\vec v$.
Finding $\lambda_C$ for $\vec C$: $$\vec{CE}\cdot\vec v= \vec 0\\ (\vec E-(\vec P+\lambda_C \vec v))\cdot \vec v = \vec 0\\ (\vec E-\vec P-\lambda_C \vec v)\cdot \vec v = \vec 0\\ (\vec E-\vec P)\cdot\vec v=\lambda_C \vec v\cdot \vec v\\ \lambda_C = \dfrac{(\vec E-\vec P)\cdot\vec v}{\vec v\cdot \vec v}\\ $$
In your case, $\lambda_C = \dfrac{((4\mathrm i+3\mathrm j)-(2\mathrm i -\mathrm j))\cdot(3\mathrm i+2\mathrm j)}{(3\mathrm i+2\mathrm j)\cdot(3\mathrm i+2\mathrm j)}=\dfrac{(2\mathrm i+4\mathrm j)\cdot(3\mathrm i+2\mathrm j)}{9+4}=\dfrac{6+8}{13}=\dfrac{14}{13}$
Sub $\lambda_C$ into the equation for $L_1$ to get $\vec C$. Then $L_2$ is the line through $\vec C$ and $\vec E$.
Let $\mathbf{OP} = <a,b,c> $.
We know that $AB$ and $\mathbf{OP}$ must be perpendicular, so the dot product of $\mathbf{OP}$ and $<2,2,-2>$ is $0$ ($<2,2,-2>$ plays the role of $\mathbf{v}$ in the diagram below)
$$<2,2,-2> \bullet <a,b,c> = 2a+2b-2c = 0$$ $$ \Rightarrow a+b-c=0$$ We also have that for some value of $\lambda$ (let's just call it $t$) $$<1,2,2> + <2,2,-2>t $$ $$=<1+2t, 2+2t, 2-2t> = <a,b,c>$$ because $P$ lies in the $AB $
Combining the two equations, we get $$1+2t +2 +2t - 2 +2t = 0 \Rightarrow 6t = -1$$ $$\Rightarrow t = -1/6$$
So $\mathbf{OP} = <2/3, 5/3, 7/3> = \frac13 <2,5,7>$
Best Answer
Let $C$ be the point on line $l$ which is the shortest distance from $A$. This is the perpendicular. $C$ has co-ordinates given by $2\vec i-2\vec j-\vec k+\mu_0(-\vec i+2\vec j+\vec k)=(2-\mu_0)\vec i+(2\mu_0-2)\vec j+(\mu_0-1)\vec k$ for some $\mu_0$ which we must determine.
To minimise the distance we minimise the distance between $C$ and $A$:$$\begin{align}|C-A|^2 = & (2-\mu_0-1)^2+(2\mu_0-2-1)^2+(\mu_0-1-1)^2 \\ & = {\mu_0}^2-2\mu_0+1+4{\mu_0}^2-12\mu_0+9+{\mu_0}^2-4\mu_0+4\\ & = 6{\mu_0}^2-18\mu_0+14\\ \end{align}$$ This is a quadratic. To minimise, we differentiate and set the derivative equal to zero, which gives: $$12\mu_0-18=0\implies\mu_0=3/2\\\implies |C-A|^2=6\cdot{\frac32}^2-18\cdot\frac32+14=\frac12\\\implies|C-A|=\frac{1}{\sqrt2}$$