Question

If $\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1$ and $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$ then value of $\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}$ is.

• A. $0$
• B. $11$
• C. $9$
• D. $1$

Answer 1

The correct option is D $1$

let $\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1...\left(1\right)$

$\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0...\left(2\right)$

Squaring both sides of equation (1) we get

${[p/a+q/b+r/c]}^2=(1{)}^2$

$\Rightarrow (p/a{)}^2+(q/b{)}^2+(r/c{)}^2+2(pq/ab+qr/bc+pr/ac)=1$

$\Rightarrow (p/a{)}^2+(q/b{)}^2+(r/c{)}^2+2\left[\frac{pqr}{abc}(\frac{c}{r}+\frac{a}{p}+\frac{b}{q})\right]=1$

$\Rightarrow {\left(p/a\right)}^2+{\left(q/b\right)}^2+{\left(r/c\right)}^2+2[\frac{pqr}{abc}\times 0]=1$

[since $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0]$

$\Rightarrow {\left(p/a\right)}^2+{\left(q//b\right)}^2+{\left(r/c\right)}^2=1$