Question

Vector product of three vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ of the type $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})$ is known as vector triple product. It is defined as $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}$ . Vector triple product $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{a})$ is

• A. A null vector
• B. Parallel to $\overrightarrow{b}$ if $\overrightarrow{a}\perp \overrightarrow{b}$
• C. Coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$
• D. Normal to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$

Answer 1

Answer: (B), (C)

The correct options are
B Coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$
C Parallel to $\overrightarrow{b}$ if $\overrightarrow{a}\perp \overrightarrow{b}$

Given

$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}$

When $\overrightarrow{c}=\overrightarrow{a}$, triple product becomes :

$(\overrightarrow{a}.\overrightarrow{a})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{a}$ which is not a null vector always.

Hence, Option A is wrong.

If $\overrightarrow{b}\perp \overrightarrow{a},\text{then}(\overrightarrow{a}.\overrightarrow{b})=0$

The triple product becomes ${\left|\overrightarrow{a}\right|}^2\overrightarrow{b}=\lambda \overrightarrow{b}$

which is parallel to $\overrightarrow{b}$

Hence, Option B is correct.

Since $(\overrightarrow{a}.\overrightarrow{a})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{a}$ is a linear combination of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, the triple product shall be in plane of these vectors.

Hence, Option C is correct and Option D is wrong.