Search any question & find its solution
Question:
Answered & Verified by Expert
If $\left|\begin{array}{ccc}\mathrm{p} & \mathrm{q}-\mathrm{y} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0,$ then the value of
$\frac{p}{x}+\frac{q}{y}+\frac{r}{z}$ is
Options:
$\frac{p}{x}+\frac{q}{y}+\frac{r}{z}$ is
Solution:
1065 Upvotes
Verified Answer
The correct answer is:
2
$\left|\begin{array}{ccc}\mathrm{p} & \mathrm{q}-\mathrm{y} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0$
Apply $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3},$ we get
$\left|\begin{array}{ccc}\mathrm{x} & 0 & -\mathrm{z} \\ 0 & \mathrm{y} & -\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0$
$\Rightarrow \mathrm{x}[\mathrm{yr}+\mathrm{z}(\mathrm{q}-\mathrm{y})]-\mathrm{z}[0-\mathrm{y}(\mathrm{p}-\mathrm{x})]=0$
[Expansion along first row $]$
$\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2$
Apply $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3},$ we get
$\left|\begin{array}{ccc}\mathrm{x} & 0 & -\mathrm{z} \\ 0 & \mathrm{y} & -\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0$
$\Rightarrow \mathrm{x}[\mathrm{yr}+\mathrm{z}(\mathrm{q}-\mathrm{y})]-\mathrm{z}[0-\mathrm{y}(\mathrm{p}-\mathrm{x})]=0$
[Expansion along first row $]$
$\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.