Question 121

$$\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1$$ and $$\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$$ where p,q,r and a,b,c are non-zero then the value of $$\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}$$ is


Given , $$\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1$$
Squaring on both sides gives, $$(\frac{p}{a}+\frac{q}{b}+\frac{r}{c})^2=1^2$$
$$\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}$$ + 2$$(\frac{p}{a}\frac{q}{b}+\frac{q}{b}\frac{r}{c}+\frac{r}{c}\frac{p}{a})$$ = 1 ........equ(1)
Also given that $$\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$$
Solving this , we get aqr+ bpr +cpq = 0
Divide this with abc on both sides, we get $$\frac{aqr}{abc}+\frac{bpr}{abc}+\frac{cpq}{abc}=0$$
i.e. $$\frac{qr}{bc}+\frac{pr}{ac}+\frac{pq}{ab}=0$$ . Substituting this in equ(1) .
We get , $$\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}$$ = 1

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