Question

If $\overrightarrow e = l\overrightarrow i + m\overrightarrow j + n\overrightarrow k$is a unit vector, then the maximum value of $lm + mn + nl$ is
A)$\dfrac{-1}{2}$
B) 0
C) 1
D) None of these

Answer 1

Hint: First we will calculate the magnitude of the given vector and put it equal to 1 and then apply the relation between A.M and G.M of three numbers to get the desired answer.
The relation between AM and GM is given by:
$A.M \geqslant G.M$

Complete step by step solution:
The given vector is :
$\overrightarrow e = l\overrightarrow i + m\overrightarrow j + n\overrightarrow k$
The magnitude of the vector $\overrightarrow v = a\overrightarrow i + b\overrightarrow j + c\overrightarrow k$ is given by:
$|\overrightarrow v | = \sqrt {{a^2} + {b^2} + {c^2}}$
Applying this formula for given vector we get:
$|\overrightarrow e | = \sqrt {{l^2} + {m^2} + {n^2}}$
Since the magnitude of given vector is equal to 1 therefore,

$$\sqrt {{l^2} + {m^2} + {n^2}} = 1 \\\ {l^2} + {m^2} + {n^2} = 1.........\left( 1 \right) \\\$$

Now calculating the Arithmetic mean of ${l^2},{m^2},{n^2}$ we get:
$A.M = \dfrac{{{l^2} + {m^2} + {n^2}}}{3}$
Now calculating Geometric mean of ${l^2},{m^2},{n^2}$ we get:
$G.M = \sqrt[3]{{{l^2}{m^2}{n^2}}}$
Now since $A.M \geqslant G.M$ therefore,
$\dfrac{{{l^2} + {m^2} + {n^2}}}{3} \geqslant \sqrt[3]{{{l^2}{m^2}{n^2}}}$
According to equation 1 we get:
$\dfrac{1}{3} \geqslant \sqrt[3]{{{l^2}{m^2}{n^2}}}$ (relation 1)
Now applying the relation of AM and GM for we get:

$$A.M = \dfrac{{lm + mn + nl}}{3} \\\ G.M = \sqrt[3]{{\left( {lm} \right)\left( {mn} \right)\left( {nl} \right)}} \\\ G.M = \sqrt[3]{{{l^2}{m^2}{n^2}}} \\\$$

And since $A.M \geqslant G.M$ therefore,
$\dfrac{{lm + mn + nl}}{3} \geqslant \sqrt[3]{{{l^2}{m^2}{n^2}}}$ (relation 2)
Comparing relation 1 and relation 2 we get:

$$\dfrac{{lm + mn + nl}}{3} \leqslant \dfrac{1}{3} \\\ lm + mn + nl \leqslant 1 \\\$$

Therefore the maximum value of $lm + mn + nl$ is 1.
Hence option (C) is the correct option.

Note:
The arithmetic mean of numbers is always greater than their geometric mean.
Arithmetic mean of $a1,a2,......an$ is:
$AM = \dfrac{{a1 + a2...... + an}}{n}$
Geometric mean of $a1,a2,......an$ is:
$GM = \sqrt[n]{{a1a2......an}}$