Correct Answer - Option 4 :
\(\frac{1}{14}\)
Concept:
If \(x\vec{a}\ +\ y\vec{b}\ +\ z\vec{c}\ + k\vec{d}\ =\ 0\) where \(\vec{a},\ \vec{b},\ \vec{c},\ \vec{d}\) are coplanar vector then
x + y + z + k should be zero
Dot product of two vectors is defined as:
\({\rm{\vec A}}{\rm{.\vec B = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times cos}}\;{\rm{\theta }}\)
If \(⃗ a = {a_1}̂ i \ + \ {a_2}̂ j\ +\ {a_3}̂ k\)
Then \(|a|\ =\ \sqrt{a_1^2\ +\ a_2^2\ +\ a_3^2}\)
Calculation
Given that,
\((sin\ A) \vec{a}\ +\ (2sin\ 2B) \vec{b}\ +\ (3sin\ 3C) \vec{c}\ -\ \vec{d}\ =\ 0\)
where, \(\vec{a},\ \vec{b},\ \vec{c}\ and\ \vec{d}\) are coplanar vector then
sin A + 2sin 2B + 3sin 3C - 1 = 0
⇒ 1.sin A + 2.sin 2B + 3.sin 3C = 1
This can be written as dot product of two vector
\(⇒ (\hat{i}\ +\ 2\hat{j}\ +\ \hat{k}).[(sin A) \hat{i}+ (sin 2B)\hat{j} + (sin 3C)\hat{k}] = 1 \)
∵ \({\rm{\vec A}}{\rm{.\vec B = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times cos}}\;{\rm{\theta }}\)
\(⇒ \sqrt{1^2\ +\ 2^2\ +\ 3^2}.\sqrt{(sin^2 A) + (sin^2 2B)+ (sin^2 3C)} = 1 \)
\(⇒\ \sqrt{(sin^2 A) + (sin^2 2B)+ (sin^2 3C)} = \ \frac{1 }{ \sqrt{14}}\)
\(⇒\ {(sin^2 A) + (sin^2 2B)+ (sin^2 3C)} = \ \frac{1}{{14}}\)
Hence. option 4 is correct.
