Question

If $a,b,c$ are the $p^{th},q^{th},r^{th}$ terms of an $HP$ and $\overrightarrow{u}=(q-r)\hat{i}+(r-p)\hat{j}+(p-q)\hat{k}$, $\overrightarrow{v}=\frac{\hat{i}}{a}+\frac{\hat{j}}{b}+\frac{\hat{k}}{c}$ then

• A. $\overrightarrow{u},\overrightarrow{v}$ are parallel vectors
• B. $\overrightarrow{u},\overrightarrow{v}$ are orthogonal vectors
• C. $\overrightarrow{u}\cdot \overrightarrow{v}=1$
• D. $\overrightarrow{u}\times \overrightarrow{v}=\hat{i}+\hat{j}+\hat{k}$

Answer 1

The correct option is B $\overrightarrow{u},\overrightarrow{v}$ are orthogonal vectors

$a,b,c$ are in H.P. so $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in $A.P$.

$\frac{1}{a}=A+(p-1)D,\frac{1}{b}=A+(q-1)D,\frac{1}{c}=A+(r-1)D$

Now, $p-q=\frac{1}{aD}-\frac{1}{bD};q-r=\frac{1}{bD}-\frac{1}{cD};r-p=\frac{1}{cD}-\frac{1}{aD}$

Now if keenly observed, we find that the terms are interchanged and then written in the question, implying that if a dot product is taken, the terms will cancel out to give zero and thus imply that the vectors are orthogonal vectors.