If asinα+bsinβ+csinγ=k,then minimum value of sin²α+sin²β+sin²γ isa. k²/(a²+b²+c²)b. 1/(a²+b²+c²)c. k/√3d. none of these

Hi Menka,

Let's see how to solve such questions involving sum of product of terms, using simple Vectors.

Let "X" and "Y" be two vectors given by

X = ai + bj + ck, and

Y = sin(alpha)i + sin(beta)j + sin(gamma)k

Now dot product of X and Y is |X|*|Y|*cos(X^Y).

Also X.Y is the given expression, which is equal to k.

So k = |X|*|Y|*cos(X^Y)

ie k = root(a^2 + b^2 + c^2)*root(sin^2alpha + sin^2beta + sin^2gamma)*cos(X^Y)

Squaring both sides, will give

sin^2alpha + sin^2beta + sin^2gamma = k^2/(a^2 +b^2 +c^2)*cos^2(X^Y)

Now, LHS is minimum, when denominator of RHS is maximum. ie when cos^2(X^Y) = 1.

So Min Value of the expression would be (A).

Hope that was helpful.

All the best,

Regards,

Ashwin (IIT Madras)