Finding angle between lines with parametric equations

[escobarro]

\(\displaystyle \mbox{16. Find the an}\)\(\displaystyle \mbox{gle between the lines:}\)

. . .\(\displaystyle \mbox{(a) }\, x\, -\, 3\, =\, 2\, -\, y,\, z\, =\, 1,\, \mbox{ and }\, x\, =\, -3,\, y\, +\, 2\, =\, z\, -\, 5\)

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My work:

\(\displaystyle \mbox{Let }\, x\, -\, 3\, =\, t.\)

$\displaystyle \left(\begin{array}{c}x\\y\\z \end{array}\right)\, =\, \left(\begin{array}{c}3\\2\\1 \end{array}\right)\, +\, t\, \left(\begin{array}{c}1\\-1\\0 \end{array}\right)$

\(\displaystyle \mbox{Let }\, y\, + \,2 \,=\, s.\)

$\displaystyle \left(\begin{array}{c}x\\y\\z \end{array}\right)\, =\, \left(\begin{array}{c}-3\\-2\\5 \end{array}\right)\, +\, s\, \left(\begin{array}{c}0\\1\\1 \end{array}\right)$

\(\displaystyle \left(\begin{array}{c}1\\-1\\0 \end{array}\right)\, \cdot\, \left(\begin{array}{c}0\\1\\1 \end{array}\right)\, =\, \sqrt{\strut 1^2\, +\, (-1)^2\,}\, \times\, \sqrt{\strut 1^2\, +\, 1^2\, }\, \cos(\theta)\)

\(\displaystyle -1\, =\, \sqrt{\strut 2\,}\, \times\, \sqrt{\strut 2\,}\, \cos(\theta)\)

$\displaystyle \cos(\theta)\, =\, -\dfrac{1}{2}$

$\displaystyle \theta\, =\, \dfrac{2\pi}{3}$

So, the answer I get is 2pi/3 but the answer says pi/3
are we just supposed to pick the acute angle complement?