Question

If $\overline{e}=l\overline{i}+m\overline{j}+n\overline{k}$ is a unit vector, the maximum value of $lm+mn+nl$ is

• A. $-\frac{1}{2}$
• B. $0$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $-\frac{1}{2}$

Given $\overrightarrow{e}$ is unit vector,

$\therefore l^2+m^2+n^2=1\dots \dots \left(1\right)$

We know that the square of any real number is always greater than zero.

$\Rightarrow (l+m+n{)}^2\ge 0$

$\Rightarrow l^2+m^2+n^2+2(lm+mn+nl)\ge 0$

From $\left(1\right)$, we get

$lm+nm+nl\ge -\frac{1}{2}$