I've a task to find the distance in $E^4$ between:
$L = [1,2,-1,4] + \text{lin}((1,2,-1,0))$
and
$M = [2,3,1,5] + \text{lin}((2,1,0,2))$
My efforts to find the correct solution:
Let
$\alpha_{1}=(1,2,-1,0) , \alpha_{2}=(2,1,0,2)$
$p_{0}=[2,3,1,5]$
$p\in R^{4} p\in L$
$p=(1+t,2+2t,-1-t,4)$
$\overrightarrow{p_{0}p}=(t-1,2t-1,2t,-1)$
$f(t)= d(p,M)$
Then (W is a determinant of grammian matrix)
$f(t)=\sqrt{\frac{W(\alpha_{2},\overrightarrow{p_{0}p})}{W(\alpha_{2})}}$
$<\alpha_{2},\alpha_{2}>=3$
$W(\alpha_{2})=3$
$W(\alpha_{2},\overrightarrow{p_{0}p})=\det\left(\begin{array}{cc} <\alpha_{2},\alpha_{2}> & <\alpha_{2},\overrightarrow{p_{0}p}>\\ <\overrightarrow{p_{0}p},\alpha_{2}> & <\overrightarrow{p_{0}p},\overrightarrow{p_{0}p}> \end{array}\right)$
$<\overrightarrow{p_{0}p},\overrightarrow{p_{0}p}>=\sqrt{(t-1)^{2}+(2t-1)^{2}+4t^{2}+1}=\sqrt{t^{2}-2t+1-4t^{2}+1+4t^{2}+1}=\sqrt{t^{2}-2t+1}=\sqrt{(t-1)^{2}}=t-1$
$<\alpha_{2},\overrightarrow{p_{0}p}>=<\overrightarrow{p_{0}p},\alpha_{2}>=\sqrt{2t-2+2t-1-2}=\sqrt{4t-5}$
$W(\alpha_{2},\overrightarrow{p_{0}p})=\det\left(\begin{array}{cc} 3 & \sqrt{4t-5}\\ \sqrt{4t-5} & \mid t-1\mid \end{array}\right)=\\3\cdot\mid t-1\mid-\mid4t-5\mid=\begin{cases} -3t+3+4t-5 & \iff t<1\\ 3t-3+4t-5 & \iff t\in[1,\frac{5}{4})\\ 3t-3-4t+5 & \iff t>\frac{5}{4} \end{cases}=\begin{cases} t-2 & \iff t<1\\ 7t-8 & \iff t\in[1,\frac{5}{4})\\ -t+2 & \iff t\geq\frac{5}{4} \end{cases}$
From that I noticed, that distance can be negative, and I don't know how to fix it
Could you help me to point, where I made an error in reasoning?
Thanks in advance for all advices!