# Show that the following two lines are skew lines: $\overrightarrow{r}=(\overrightarrow{3i}+\overrightarrow{5j}+\overrightarrow{7k})+ t (\overrightarrow{i}-\overrightarrow{2j}+\overrightarrow{k})$ and $\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}) + s (\overrightarrow{7i}-\overrightarrow{6j}+\overrightarrow{7k})$

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• The lines $\overrightarrow r=\overrightarrow a_1+t\overrightarrow u, \: \overrightarrow r=\overrightarrow a_2+s\overrightarrow u$ intersect if $[(\overrightarrow a_2-\overrightarrow a_1)\overrightarrow u\: \overrightarrow v]=0$
$\overrightarrow r=(3\overrightarrow i+5\overrightarrow j+7\overrightarrow k)+t(\overrightarrow i-2\overrightarrow j+\overrightarrow k)$ (i)
$\overrightarrow r = (\overrightarrow i+\overrightarrow j+\overrightarrow k)+s(7\overrightarrow i+6\overrightarrow j+7\overrightarrow k)$ (ii)
(i) Passes through the point with pv $\overrightarrow a_1 = 3\overrightarrow i+5\overrightarrow j+7\overrightarrow k$ and is parallel to $\overrightarrow u=\overrightarrow i-2\overrightarrow j+\overrightarrow k$
(ii) Passes through the point with pv $\overrightarrow a_2=\overrightarrow i+\overrightarrow j+\overrightarrow k$ and is parallel to $\overrightarrow v = 7\overrightarrow i+6\overrightarrow j+7\overrightarrow k$
Since the vectors $\overrightarrow u\: and \overrightarrow v$ are not parallel.
(i) and (ii) are either intersecting or skew lines.
Now $\overrightarrow a_2-\overrightarrow a_1=-2\overrightarrow i-4\overrightarrow j-6\overrightarrow k$
Consider $[\overrightarrow a_2-\overrightarrow a_1\: \: \overrightarrow u\: \overrightarrow v]$
$= \begin{vmatrix} -2 & -4 & -6 \\ 1 & -2 & 1 \\ 7 & 6 & 7 \end{vmatrix} = (-2)(-14-6)-(-4)(7-7)-6(6+14)$
$= 40+0-120=-80 \neq 0$
Since $[\overrightarrow a_2-\overrightarrow a_1\: \: \overrightarrow u\overrightarrow v] \neq 0$ the lines do not intersect.
$\therefore$ they are skew lines.