Question

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

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 $\overrightarrow{a}\cdot \frac{\overrightarrow{b}\times \overrightarrow{c}}{|\overrightarrow{b}\times \overrightarrow{c}|}=\overrightarrow{a}\cdot \hat{k}=a_3$
$\therefore (\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{b}+(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{c}+\frac{\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})}{|\overrightarrow{b}\times \overrightarrow{c}{|}^2}(\overrightarrow{b}\times \overrightarrow{c})$
$=a_1\overrightarrow{b}+a_2\overrightarrow{c}+a_3\frac{(\overrightarrow{b}\times \overrightarrow{c})}{|\overrightarrow{b}\times \overrightarrow{c}|}=a_1\hat{i}+a_{2}j+a_3\hat{k}=\overrightarrow{a}$