Question

E=li+mj+nk is a unit vector and l=1/3then the maximum value of lmn is

Answer 1

Given:-

$E = l\vec{i} +m\vec{j} +n\vec{k}$   is a unit vector and,

$l = \frac{1}{3}$

To find : maximum value of $lmn$

Solution:-

We know that, for a unit vector it's magnitude = 1.

$\therefore |E| = 1$

⇒$\sqrt{l^{2} +m^{2} +n^{2} } =1$

Squaring both sides we get,

⇒$l^{2} +m^{2} +n^{2} =1$

⇒$\frac{1}{9} + m^{2} +n^{2} =1$

⇒$m^{2} +n^{2} =\frac{8}{9}$      ---------------------------(1)

Since we know that Arithmetic Mean (AM) of a set of non-negative numbers is always greater than or equal to their Geometric Mean (GM),

We can write as follows for the numbers $m^{2}$ and $n^{2}$ :-

$AM\geq GM$

⇒ $\frac{m^{2}+n^{2} }{2} \geq ( m^{2} n^{2} )^{\frac{1}{2} }$

⇒ $\frac{\frac{8}{9} }{2} \geq mn$ -------------------(putting value of $m^{2} +n^{2}$ from equation (1))

⇒$mn\leq \frac{4}{9}$ ---(This is maximum value of $mn$)

Multiplying both sides by $l$ to get maximum value of $lmn$ :-

⇒ $lmn\leq \frac{4}{9}\times\frac{1}{3} \leq \frac{4}{27}$

$\therefore$ The maximum value of $lmn$ is $\frac{4}{27}$.