Question

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

• A. –1/2
• B. 0
• C. 1
• D. 3/2

Answer 1

Answer: (C)

1

Answer 2

Since $A.M \geq G.M.$,

$\frac{l^{2} + m^{2}}{2} \geq lm;\frac{m^{2} + n^{2}}{2} \geq mn;\frac{n^{2} + l^{2}}{2} \geq ln.$

Adding these three inequalities,

$l^{2} + m^{2} + n^{2} \geq lm + mn + nl$.

But $l^{2} + m^{2} + n^{2} = 1$

Hence $lm + mn + nl \leq 1.$ Maximum value = 1.