TITLE: Constructuing vector equation question
QUESTION [1 upvotes]: 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.

REPLY [0 votes]: $r=(1-\lambda)i+(3-\lambda)j$=$xi+yj$ .....(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
{"set_name": "stack_exchange", "score": 1, "question_id": 647450}