If $ | \vec{a} | = 3 $, $ | \vec{b} | = 2 $, then find $ (3\vec{a} - 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) $.
In a GP 9, 3, $ \frac{1}{3} $, $ \frac{1}{9} $, … find the 25th term.
The point of intersection of the lines \(\frac{x-3}{2} = \frac{y-2}{2} = \frac{z-6}{1}\) and \(\frac{x-2}{3} = \frac{y-4}{2} = \frac{z-1}{3}\) is:
If $ | \vec{a} | = 3 $, $ | \vec{b} | = 2 $, then find $ (3\vec{a} - 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) $.