Constructuing vector equation question I'm working on a three-part vector problem where a given straight line has a vector equation of 
$$r = ( i + 3 j ) + \lambda( -i - j )$$
which I rearranged to get
$$r  = (1 - \lambda) i + (3 - \lambda) j$$
For the first part I had to show  the point (1,2) does not lie on this line, so I did
1 - $\lambda$ = 1 so $\lambda$ = 0 and
3 - $\lambda$ = 2 so $\lambda$ = 1
so $(1,2)$ does not lie on the line r, is this right? And if someone could give a more detailed explanation as to why/why not or a better way to set it out I'd love to hear it.
For the second part I have to construct a vector equation for a line that does go through  the point (1,2) and is perpendicular to r, so I  figured I'd try 
$$s = ( i + 2 j ) + (\mu)( i - j )$$
$$s = (1 +\mu) i + (2 - \mu)j$$
Does this look like it would make s and r vectors perpendicular? Also I know that if the dot product of the two equations is 0 then they are perpendicular, but how would I go about multiplying the lines r and s? Do I have to substitute values in for lambda and mu, and if so which ones?
Finally for the third part I have to determine the point of intersection between the two lines r and s, how would I be able to do this? Any help would be much appreciated, thanks.
 A: Your first part is just fine.
You will not be finding the dot product of the lines $r$ and $s$ in the second part, but rather, of their "direction vectors," namely: $-i-j$ and $i-j.$ Note that $\lambda$ and $\mu$ don't come into play at all here. They are simply parameters that generate the lines from the starting vectors by multiples of the direction vectors.
As for the third part, try solving the system of equations $$1-\lambda=1+\mu\\3-\lambda=2-\mu$$ for $\lambda$ and $\mu,$ then confirm that this gets you the same point on both $r$ and $s.$
A: The dot product of two vectors is zero when they are perpendicular; and when dealing with lines you should be looking at the direction vectors.  For your original line $r$, the direction vector was $-i-j$, and for your line $s$, the direction vector is $i-j$; so those are the things you should dot.
An equation for a line needs an independent variable, which is what your $\lambda$ and $\mu$ are.  (You might be accustomed to seeing $t$ there, but any variable will do.)
To find the intersection point, set the i-components equal and the j-components equal and see what you get. (Your variables are $\lambda$ and $\mu$, as mentioned above.) Plugging them in to their respective equations should give one (i.e. the same) point.
A: $r=(1-\lambda)i+(3-\lambda)j$=$xi+yj$ .....(let)
The point (1,2) does not lie on the line if $x= 1$ & $y=2$ or $(1-\lambda)=1$ & &$(3-\lambda)=2$ or $\lambda=0$ & also $\lambda=1$ both the values of $\lambda$
are not same hence point (1,2) can't be on the line 
for second part taking dot product of r & s zero then we find
[$(1-\lambda)i+(3-\lambda)j$][$(1+\mu)i+(2-\mu)j$]=0
$(1-\lambda)$ $(1+\mu)$+$(3-\lambda)$ $(2-\mu)$=0
$7-2\mu-3\lambda=0$
$\mu=\left(\frac{7-3\lambda}{2}\right)$ so $(1+\mu)=1+\left(\frac{7-3\lambda}{2}\right)$=$\left(\frac{9-3\lambda}{2}\right)$  & $ (2-\mu)=\left(\frac{3(\lambda-1)}{2}\right)$
using this in 's' we get
$s=\left(\frac{9-3\lambda}{2}\right)i+\left(\frac{3(\lambda-1)}{2}\right)j$
now for the point of intersection equating the coefficients of $i$ & $j$ of $r$ & $s$ we get two equations 
$\left(\frac{9-3\lambda}{2}\right)=(1-\lambda)$ 
on solving we get $\lambda=7$
& $\left(\frac{3(\lambda-1}{2}\right)=(3-\lambda)$
on solving we get $\lambda=\left(\frac{9}{5}\right)$
Hence the point of intersection is $(7,\frac{9}{5})$
IF ANY CORRECTION MOST WELCOME
