Question

$[\overrightarrow{a}=3\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k}]$, $[\overrightarrow{b}=2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}]$, then $[\overrightarrow{a}\times (\overrightarrow{a}\cdot \overrightarrow{b})=]$
A. $[3\overrightarrow{a}]$
B. $[3\sqrt{14}]$
C. $0$
D. None of these

Answer 1

Hint: In this question, the dot and cross products of vectors are applied to find the required vector expression. The dot product is said to be a scalar product and the cross product is said to be a skew product or vector product. By using appropriate formulae, the required vector product is calculated.

Formula Used:
The dot product of two vectors is
$\overrightarrow{a}\cdot \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos (\overrightarrow{a},\overrightarrow{b})$
The cross-product of two vectors is
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\overrightarrow{a},\overrightarrow{b})\overrightarrow{n}$
Scalar triple product of three vectors:
We have the vectors $[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$ as
$[\begin{align}
  & \overrightarrow{a}={{a}_{1}}\overrightarrow{i}-{{a}_{2}}\overrightarrow{j}+{{a}_{3}}\overrightarrow{k} \\
 & \overrightarrow{b}={{b}_{1}}\overrightarrow{i}+{{b}_{2}}\overrightarrow{j}-{{b}_{3}}\overrightarrow{k} \\
 & \overrightarrow{c}={{c}_{1}}\overrightarrow{i}+{{c}_{2}}\overrightarrow{j}-{{c}_{3}}\overrightarrow{k} \\
\end{align}]$
Then, the triple product is calculated by,
$[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|]$
In vector triple product is cross and dot products are interchangeable. I.e.,
$[\begin{align}
  & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b} \\
 & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}]=[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}] \\
 & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]=-[\overrightarrow{c}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=-[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{b}] \\
\end{align}]$
Important vector identities for solving vector equations are:
$[\overrightarrow{a}\times \overrightarrow{a}=0]$
$[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0]$
$[\begin{align}
  & \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
 & \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
 & \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
 & \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}]$


Complete step by step solution: It is given that,
$[\begin{align}
  & \overrightarrow{a}=3\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k} \\
 & \overrightarrow{b}=2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k} \\
\end{align}]$
Then,
$[\overrightarrow{a}\times (\overrightarrow{a}\cdot \overrightarrow{b})=[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]]$
$\begin{align}
  & \Rightarrow [\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=\left| \begin{matrix}
   3 & -1 & 2 \\
   3 & -1 & 2 \\
   2 & 1 & -1 \\
\end{matrix} \right| \\
 & \text{ }=3(1-2)+1(-3-4)+2(3+2) \\
 & \text{ }=3(-1)-7+2(5) \\
 & \text{ }=-3-7+10 \\
 & \text{ }=0 \\
\end{align}$

Thus, Option (C) is correct.

Note: Here we may go wrong with the vector identities and scalar triple product. Here are the simple formulae used for solving the given vector. By applying appropriate vector products, the given vector equation is evaluated