Question
If $$\overrightarrow a + \overrightarrow b \bot \overrightarrow a $$
and $$\left| {\overrightarrow b } \right| = \sqrt 2 \left| {\overrightarrow a } \right|$$ then :
A.
$$\left( {2\overrightarrow a + \overrightarrow b } \right)||\overrightarrow b $$
B.
$$\left( {2\overrightarrow a + \overrightarrow b } \right) \bot \overrightarrow b $$
C.
$$\left( {2\overrightarrow a - \overrightarrow b } \right) \bot \overrightarrow b $$
D.
$$\left( {2\overrightarrow a + \overrightarrow b } \right) \bot \overrightarrow a $$
Answer :
$$\left( {2\overrightarrow a + \overrightarrow b } \right) \bot \overrightarrow b $$
Solution :
$$\eqalign{
& {\text{Here, }}\overrightarrow a .\left( {\overrightarrow a + \overrightarrow b } \right) = 0{\text{ or }}{\left| {\overrightarrow a } \right|^2} + \overrightarrow a .\overrightarrow b = 0\,\,\,{\text{or }}\frac{1}{2}{\text{ }}{\left| {\overrightarrow b } \right|^2} + \overrightarrow a .\overrightarrow b = 0 \cr
& \left( {2\overrightarrow a + \overrightarrow b } \right).\overrightarrow b = 2\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow b } \right|^2} = 2\left\{ {\overrightarrow a .\overrightarrow b + \frac{1}{2}{{\left| {\overrightarrow b } \right|}^2}} \right\} = 0 \cr
& \therefore \,\left( {2\overrightarrow a + \overrightarrow b } \right) \bot \overrightarrow b \cr
& \left( {2\overrightarrow a - \overrightarrow b } \right).\overrightarrow b = \,2\overrightarrow a .\overrightarrow b - {\left| {\overrightarrow b } \right|^2} = 2\overrightarrow a .\overrightarrow b + 2\overrightarrow a .\overrightarrow b = 4\overrightarrow a .\overrightarrow b \ne 0 \cr
& \overrightarrow a .\left( {2\overrightarrow a + \overrightarrow b } \right) = 2{\left| {\overrightarrow a } \right|^2} + \overrightarrow a .\overrightarrow b = {\left| {\overrightarrow a } \right|^2} + \left( {{{\left| {\overrightarrow a } \right|}^2} + \overrightarrow a .\overrightarrow b } \right) = {\left| {\overrightarrow a } \right|^2} \ne 0 \cr} $$