Question
Resolved part of vector $$\overrightarrow a $$ along vector $$\overrightarrow b $$ is $${\overrightarrow a _1}$$ and that perpendicular to $$\overrightarrow b $$ is $${\overrightarrow a _2}$$ then $${\overrightarrow a _1} \times {\overrightarrow a _2}$$ is equal to :
A.
$$\frac{{\left( {\overrightarrow a \times \overrightarrow b } \right) . \overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}}$$
B.
$$\frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}$$
C.
$$\frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)}}{{{{\left| {\overrightarrow b } \right|}^2}}}$$
D.
$$\frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)}}{{\left| {\overrightarrow b \times \overrightarrow a } \right|}}$$
Answer :
$$\frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)}}{{{{\left| {\overrightarrow b } \right|}^2}}}$$
Solution :
$$\eqalign{
& {\overrightarrow a _1} = \left( {\overrightarrow a .\hat b} \right)\hat b = \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}} \cr
& \Rightarrow {\overrightarrow a _2} = \overrightarrow a - {\overrightarrow a _1} = \overrightarrow a - \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}} \cr
& {\text{Thus, }}{\overrightarrow a _1} \times {\overrightarrow a _2} = \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}} \times \left( {\overrightarrow a - \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}}} \right) \cr
& = \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)}}{{{{\left| {\overrightarrow b } \right|}^2}}} \cr} $$