Question

a, b, c are the pth, qth, rth terms of an H.P. 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. $\overrightarrow{u}.\overrightarrow{v}=1$
• D. $\overrightarrow{u}\times\overrightarrow{v}=\hat{i} +\hat{j} +\hat{k} $

Answer 1

Answer: (B)

$\vec{u}\,.\vec{v}=\frac{q-r}{a}+\frac{r-p}{b}+\frac{p-q}{c}$
Since $\frac{1}{a}=x+\left(p-1\right)y$
$\frac{1}{b}=x+\left(q-1\right)y$
$\frac{1}{c}=x+\left(r-1\right)y$
Putting in $\left(1\right)$, we get
$\vec{u}\,.\vec{v}=\left(q-r\right)\left[x+\left(p-1\right)y\right]+\left(r-p\right)$
$\left[x+\left(q-1\right)y\right]+\left(p-q\right)\left[x+\left(r-1\right)y\right]=0$
$\therefore \vec{u}\,.\vec{v}=0\quad\ldots\left(1\right)$
$\therefore \vec{u}\,.\vec{v}$ are orthogonal vectors