Question

If $\overrightarrow{ b }$ and $\overrightarrow{ c }$ are any two non-collinear unit vectors and $\overrightarrow{ a }$ is any vector, then $ (\overrightarrow{ a } \cdot \overrightarrow{ b }) \overrightarrow{ b }+(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }+\frac{\overrightarrow{ a } \cdot(\overrightarrow{ b } \times \overrightarrow{ c })}{|\overrightarrow{ b } \times \overrightarrow{ c }|^{2}}(\overrightarrow{ b } \times \overrightarrow{ c })=\ldots \ldots $

Answer 1

Let $\hat{ i }$ be a unit vector in the direction of $\overrightarrow{ b }, \hat{ j }$ in the direction of $\overrightarrow{ c }$. Note that $\overrightarrow{ c }=\hat{ j }$ and
$(\overrightarrow{ b } \times \overrightarrow{ c })=|\overrightarrow{ b }||\overrightarrow{ c }| \sin \alpha \hat{ k }=\sin \alpha \hat{ k }$
where, $\hat{ k }$ is a unit vector perpendicular to $\overrightarrow{ b }$ and $\overrightarrow{ c }$.
$\Rightarrow |\overrightarrow{ b } \times \overrightarrow{ c }|=\sin \alpha \Rightarrow \hat{ k }=\frac{\overrightarrow{ b } \times \overrightarrow{ c }}{|\overrightarrow{ b } \times \overrightarrow{ c }|}$
Let $ \overrightarrow{ a }=a_{1} \hat{ i }+a_{2} \hat{ j }+a_{3} \hat{ k }$
Now, $ \overrightarrow{ a } \cdot \overrightarrow{ b }=\overrightarrow{ a } \cdot \overrightarrow{ i }=\hat{ i } \cdot\left(a_{1} \hat{ i }+a_{2} \hat{ j }+a_{9} \hat{ k }\right)=a_{1}$
and $ \overrightarrow{ a } \cdot \overrightarrow{ c }=\overrightarrow{ a } \cdot \hat{ j }=\hat{ j } \cdot\left(a_{1} \hat{ i }+a_{2} \hat{ j }+a_{3} \hat{ k }\right)=a_{2}$
and $ \overrightarrow{ a } \cdot \frac{\overrightarrow{ b } \times \overrightarrow{ c }}{|\overrightarrow{ b } \times \overrightarrow{ c }|}=\overrightarrow{ a } \cdot \hat{ k }=a_{3}$
$\therefore(\overrightarrow{ a } \cdot \overrightarrow{ b }) \overrightarrow{ b }+(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }+\frac{\overrightarrow{ a } \cdot(\overrightarrow{ b } \times \overrightarrow{ c })}{|\overrightarrow{ b } \times \overrightarrow{ c }|^{2}}(\overrightarrow{ b } \times \overrightarrow{ c })$
$=a_{1} \overrightarrow{ b }+a_{2} \overrightarrow{ c }+a_{3} \frac{(\overrightarrow{ b } \times \overrightarrow{ c })}{|\overrightarrow{ b } \times \overrightarrow{ c }|}=a_{1} \hat{ i }+a_{2} j +a_{3} \hat{ k }=\overrightarrow{ a }$