Question

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

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

Answer 1

Answer: (C)

$1$

$\overrightarrow{e}=l\hat{i}+m\hat{j}+n\hat{k}$
$\left|\overrightarrow{e}\right|=\sqrt{l^2+m^2+n^2}$
$1=l^2+m^2+n^2$....(1)
Now the expression

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

$\Rightarrow 2\sum l^2-2\sum lm\ge 0$

Using (1) we get $\sum lm\le 1$