If $\vec{b}$ and $\vec{c}$ are any two non-collinear unit vectors and $\vec{a}$ is any vector, then

Question

If $\vec{b}$ and $\vec{c}$ are any two non-collinear unit vectors and $\vec{a}$ is any vector, then $(\vec{a} \cdot \vec{b}) \vec{b}+(\vec{a} \cdot \vec{c}) \vec{c}+\frac{\vec{a} \cdot(\vec{b} 	imes \vec{c})}{|\vec{b} 	imes \vec{c}|}(\vec{b} 	imes \vec{c})= \_\_\_\_$

Professor PC Thomas Classes · Answer

Since $\vec{b}$ and $\vec{c}$ are non-collinear, therefore $\vec{b} 	imes \vec{c}$ is normal to their plane. Thus $\vec{b}, \vec{c}$ and $\vec{b} 	imes \vec{c}$ are three non- coplanar vectors. Now any vector $\vec{r}$ can be expressed in terms of three non-coplanar vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ as $\vec{r}=x \vec{\alpha}+y \vec{\beta}+z \vec{\gamma}$\ where $x, y, z$ are projections of $r$ in the direction of $\vec{\alpha} \cdot \vec{\beta}+\vec{\gamma}=\frac{\vec{r} \cdot \vec{\alpha}}{|\vec{\alpha}|}=\vec{r} \cdot \hat{\alpha}$\ $y=\vec{r} \cdot \hat{\beta}, z=\vec{r} \cdot \hat{\gamma}$\ In our problem $\hat{\alpha}=\vec{b}, \hat{\beta}=\vec{c}, \hat{\gamma}=\frac{\vec{b} 	imes \vec{c}}{|\vec{b} 	imes \vec{c}|}$ where $\vec{r}=\vec{a}$. Hence the given expression is $\vec{a}$.\