If $$A = \left( {p,\,q,\,r} \right)$$    and $$B = \left( {p',\,q',\,r'} \right)$$    are two points on the line $$\lambda x = \mu y = \nu z$$    such that $$OA = a,\,OB = b$$    then $$pp' + qq' + rr'$$    is equal to :

Solution :
$$\eqalign{ & {\text{Here, }}\lambda p = \mu q = \nu r{\text{ and }}\lambda p' = \mu q' = \nu r' \cr & {\text{So, }}\frac{{p'}}{p} = \frac{{q'}}{q} = \frac{{r'}}{r} = \frac{{\sqrt {p{'^2} + q{'^2} + r{'^2}} }}{{\sqrt {{p^2} + {q^2} + {r^2}} }} = \frac{{\sqrt {{b^2}} }}{{\sqrt {{a^2}} }} = \frac{b}{a} \cr & \therefore \,pp' + qq' + rr' = \frac{b}{a}\left( {{p^2} + {q^2} + {r^2}} \right) = \frac{b}{a}.{a^2} = ab \cr} $$