Solveeit Logo

Question

Question: If l, m, n are the direction cosines of vector if \[l=\dfrac{1}{2}\], then the maximum value of lmn ...

If l, m, n are the direction cosines of vector if l=12l=\dfrac{1}{2}, then the maximum value of lmn is
(a)14\dfrac{1}{4}
(b)38\dfrac{3}{8}
(c)12\dfrac{1}{2}
(d)316\dfrac{3}{16}

Explanation

Solution

Hint: The direction cosine are given by the relation l2+m2+n2=1{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1, put value of l in it. For terms m and n the arithmetic mean & geometric mean are of relation A.M \ge G.M, thus get max value for mn according to the relation and find max value of lmn.

Complete step-by-step answer:
It is said that l, m and n are the direction cosines of a vector.
We have been given that, l=12l=\dfrac{1}{2}.
We need to find the maximum value of lmn.
The direction cosines of a line parallel to any coordinate axis are equal to the direction cosine of the corresponding axis.
The direction cosines are associated by the relation,
l2+m2+n2=1\Rightarrow {{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1
Now let us put l=12l=\dfrac{1}{2} in the above expression. Thus we get,
(12)2+m2+n2=1\Rightarrow {{\left( \dfrac{1}{2} \right)}^{2}}+{{m}^{2}}+{{n}^{2}}=1

& \Rightarrow {{m}^{2}}+{{n}^{2}}=1-\dfrac{1}{4}=\dfrac{4-1}{4} \\\ & \therefore {{m}^{2}}+{{n}^{2}}=\dfrac{3}{4} \\\ \end{aligned}$$ We are considering Arithmetic mean and geometric mean. The arithmetic mean is always greater than geometric mean i.e. A.M $$\ge $$ G.M. We know that for two numbers m and n, Arithmetic mean, A.M = $$\dfrac{{{m}^{2}}+{{n}^{2}}}{2}$$. Geometric mean, G.M = $$\sqrt{{{m}^{2}}+{{n}^{2}}}$$ = mn. $$\therefore $$ A.M $$\ge $$ G.M $$\Rightarrow \dfrac{{{m}^{2}}+{{n}^{2}}}{2}\ge \sqrt{{{m}^{2}}+{{n}^{2}}}$$ Put, $${{m}^{2}}+{{n}^{2}}$$ = $$\dfrac{3}{4}$$. i.e. We get $$\dfrac{3}{4\times 2}\ge \sqrt{{{m}^{2}}{{n}^{2}}}$$ Hence we got $$\dfrac{3}{8}\ge mn$$. Hence we got the maximum value of mn as $$\dfrac{3}{8}$$. What we need is the maximum value of lmn. $$\therefore $$ Maximum value of lmn = l $$\times $$ maximum value of mn Maximum value of lmn = $$\dfrac{1}{2}\times \dfrac{3}{8}=\dfrac{3}{16}$$. Hence we got the maximum value of lmn = $$\dfrac{3}{16}$$. $$\therefore $$ Option (d) is the correct answer. Note: The direction cosines of the x, y and z axis are given as (1, 0, 0), (0, 1, 0) and (0, 0, 1). In case of parallel lines, the direction cosines are always the same. A line can have two sets of direction cosine according to the direction.