Q.(a) Find the angle between the lines
(x-2)/2 = (y-1)/5 = (z+3)/(-3) and (x+2)/(-1) = (y-4)/8 = (z-5)/4 (Scores : 2)

Use the direction-ratio dot-product formula for the angle between two lines, and the skew-lines shortest-distance formula involving the cross product of the two direction vectors.

1. Angle between the lines $\dfrac{x-2}{2}=\dfrac{y-1}{5}=\dfrac{z+3}{-3}$ has direction ratios $\vec{d_1}=(2,5,-3)$. $\dfrac{x+2}{-1}=\dfrac{y-4}{8}=\dfrac{z-5}{4}$ has direction ratios $\vec{d_2}=(-1,8,4)$. $\cos\theta = \dfrac{\vec{d_1}\cdot\vec{d_2}}{|\vec{d_1}||\vec{d_2}|}$ $\vec{d_1}\cdot\vec{d_2} = 2(-1)+5(8)+(-3)(4) = -2+40-12=26$ $|\vec{d_1}| = \sqrt{4+25+9}=\sqrt{38}$, $|\vec{d_2}| = \sqrt{1+64+16}=\sqrt{81}=9$ $\cos\theta = \dfrac{26}{9\sqrt{38}}$, so $\theta = \cos^{-1}\!\left(\dfrac{26}{9\sqrt{38}}\right)$
2. Shortest distance between the skew lines $\vec r_1 = (\hat i+2\hat j+3\hat k)+\lambda(\hat i-3\hat j+2\hat k)$, so $\vec{a_1}=(1,2,3)$, $\vec{d_1}=(1,-3,2)$ $\vec r_2 = (4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)$, so $\vec{a_2}=(4,5,6)$, $\vec{d_2}=(2,3,1)$ Shortest distance formula: $d = \dfrac{\left|(\vec{a_2}-\vec{a_1})\cdot(\vec{d_1}\times\vec{d_2})\right|}{|\vec{d_1}\times\vec{d_2}|}$ $\vec{a_2}-\vec{a_1} = (3,3,3)$ …