Question

If $l,m,n$ are d.c's of vector $\overline{OP}$ then maximum value of $lmn$ is

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{1}{3\sqrt{3}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{1}{3\sqrt{3}}$

Let the $3$ number be $l^2,m^2,n^2$

we know that

$AM\ge GM$

i.e. Arithemetic Mean $\ge$ Geometric Mean

$\therefore \frac{l^2+m^2+n^2}{3}\ge \sqrt[3]{3(lmn{)}^2}$

$(lmn{)}^{\frac{2}{3}}\le \frac{1}{3}$ $(l^2+m^2+n^2=1)$

$lmn\le \frac{1}{3\sqrt{3}}$

$\therefore$ option C is the correct answer