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:
\[GM = \sqrt[n]{{a1a2......an}}\]