Position vectors of \(\vec A\,\&\,\vec B\) are \(\hat i+\hat j + \hat k\) and \(2\hat i+5\hat j\) respectively.
\(\therefore\) Direction vector of line AB is \(\vec a=(2\hat i+5\hat j)-(\hat i+\hat j+\hat k)\)
⇒ \(\vec a=\hat i+4\hat j-\hat k\)
Position vectors of \(\vec C\,\&\,\vec D\) are \(3\hat i+2\hat j - 3\hat k\) and \(\hat i-6\hat j - \hat k\) respectively.
\(\therefore\) Direction vector of line CD is
\(\vec b=(\hat i-6\hat j-\hat k)-(3\hat i+2\hat j-3\hat k)\)
⇒ \(\vec b=-2\hat i-8\hat j+2\hat k\)
Let angle between line AB and CD is θ.
\(\therefore\) cos θ = \(\frac{\vec a.\vec b}{|\vec a||\vec b|}\)
⇒ cos θ = \(\left|\frac{(\hat i+4\hat j-\hat k).(-2\hat i-8\hat j+2\hat k)}{\sqrt{1^2+4^2+(-1)^2}\sqrt{(-2)^2+(-8)^2+2^2}}\right|\)
\(=\left|\frac{-2-32-2}{\sqrt{18}\sqrt{72}}\right|\)
\(=\left|\frac{-36}{\sqrt{18}\sqrt{18}.\sqrt4}\right|\)
\(=\frac{36}{18\times2}\) = 1 = cos 0°
\(\therefore\) θ = 0°
Hence, lines/vectors \(\vec {AB}\) and \(\vec {CD}\) are collinear.
