Question

Let $a,b,cϵR$ such that $abc=p$ and $qa-b=0$, where $p$ and $q$ are fixed positive number, then minimum distance of the point $(a,b,c)$ from origin in the three dimensional coordinate system is:

• A. $\sqrt{3}{\left(\frac{p(q^2+1)}{2q}\right)}^{1/3}$
• B. $\sqrt{3}{\left(\frac{p(q^2+1)}{q}\right)}^{1/3}$
• C. $\sqrt{3}(p{)}^{1/3}$
• D. $\sqrt{2}{\left(\frac{p}{q}\right)}^{1/2}$

Answer 1

The correct option is C $\sqrt{3}(p{)}^{1/3}$

We know that the distance between point $(a,b,c)$ and origin $(0,0,0)$ is:

$d=\sqrt{(}{a}^2+b^2+c^2)$.

Using the inequality: $AM\ge GM$, for terms $a^2,b^2$ and $c^2$:

$\frac{a^2+b^2+c^2}{3}\ge \sqrt[3]{3{\left(abc\right)}^2}$.

$abc=p$ (Given)

$⟹\frac{a^2+b^2+c^2}{3}\ge {\left(p\right)}^{\frac{2}{3}}$.

Taking square root on both sides of the inequality:

$\sqrt{a^2+b^2+c^2}\ge \sqrt{3}{p}^{1/3}$.

Hence, Option C is correct.