If a, b, c are the pth, qth and rth terms of a HP, then the vectors $\vec u = a^{-1}\hat i+b^{-1}\hat j+c^{-1}\hat k$ and $\vec v = (q − r) \hat i+ (r− p) \hat j + (p − q)\hat k$
Answer & explanation
Correct answer: option 2
Let A be the first term and D be the common difference of the corresponding AP. Then,
$\frac{1}{a}=A+ (p-1) D,\frac{1}{b}= A + (q-1) D$ and, $\frac{1}{c}= A + (r−1) D$
$⇒a^{-1} (q-r) +b^{-1} (r-p) + c^{-1} (p −q) = 0$
$⇒\vec u.\vec v=0⇒\vec u⊥\vec v$
Hence, $\vec u$ and $\vec v$ are orthogonal vectors.