Question

Vector product of three vectors $\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}$ of the type $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})$ is known as the 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 $\overrightarrow{b}\text{ if, }\overrightarrow{a}\bot \overrightarrow{b}$
C) Coplanar with $\overrightarrow{a}\text{ and }\overrightarrow{b}$
D) Normal to the plane containing $\overrightarrow{a}\text{ and }\overrightarrow{b}$.

Answer 1

We need to understand the working of the dot products or the scalar multiplication and the vector or cross products in order to understand and then find the actual relation and the conversions involved between the two types to solve this problem.

Complete step by step solution:
We are given the relation between the vector triple product and the scalar multiplication.

The vector product can be easily calculated using the given relation which involves the dot products and normal multiplication.

In order to solve this, we need to know that the dot product becomes zero, when the vectors are perpendicular to each other and will be the square of the magnitudes when both are the same.

So, let us consider the given situation. It is given that –
$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}$

But,
$\overrightarrow{a}=\overrightarrow{c}$
Therefore, the above relation can be rewritten as –

& \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{a})=(\overrightarrow{a}.\overrightarrow{a})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{a} \\\ & \therefore \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{a})={{a}^{2}}\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{a} \\\ \end{aligned}$$ Now, we can see that the triple product doesn’t become zero in all cases. When, $$\overrightarrow{a}\bot \overrightarrow{b}$$, the product becomes – $$\begin{aligned} & \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{a})={{a}^{2}}\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{a} \\\ & \therefore \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{a})={{a}^{2}}\overrightarrow{b}-0 \\\ \end{aligned}$$ i.e., the vector triple product will be vector parallel to the vector ‘b’ with a magnitude of $${{a}^{2}}$$. Now, in this case, the vector product will be coplanar with the vectors $$\overrightarrow{a}\text{ and }\overrightarrow{b}$$, as they are a linear combination. This rules out the chance of this vector product being normal to $$\overrightarrow{a}\text{ and }\overrightarrow{b}$$. This is the required solution. **The correct answers are option B and C.** **Note:** The vector triple product can only exist if the vector which is perpendicular to the vectors ‘b’ and ‘c’ has a component that is perpendicular to the vector ‘a’. In all other cases, the triple vector product will turn out to be zero and will therefore produce a null vector.