Question

If \[\overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k}\], \[\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}\] and \[\overrightarrow{c}=\widehat{i}-\widehat{j}+2\widehat{k}\], then $\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=$
A. \[6~\]
B. \[10\]
C. \[12\]
D. \[24\]

Answer 1

Hint: In the above question, the dot and cross products are applied to find the required vector product.

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
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\overrightarrow{a},\overrightarrow{b})\overrightarrow{n}$

Complete step by step solution: Given that,
\[\overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k}\]
\[\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}\]
\[\overrightarrow{c}=\widehat{i}-\widehat{j}+2\widehat{k}\]
Then,
$\begin{align}
  & \overrightarrow{b}\times \overrightarrow{c}=\left| \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   1 & 2 & 1 \\
   1 & -1 & 2 \\
\end{matrix} \right| \\
 & =\widehat{i}(4+1)-\widehat{j}(2-1)+\widehat{k}(-1-2) \\
 & =5\widehat{i}-\widehat{j}-3\widehat{k} \\
\end{align}$
Then, the required scalar triple product is
\[\begin{align}
  & \overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})=(2\widehat{i}+\widehat{j}-\widehat{k})\cdot (5\widehat{i}-\widehat{j}-3\widehat{k}) \\
 & \text{ }=2\times 5+1\times -1+1\times 3 \\
 & \text{ }=10-1+3 \\
 & \text{ }=12 \\
\end{align}\]

Thus, Option (C) is the correct value.

Additional Information: Important vector identities for solving vector equations are:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
\[\begin{align}
  & \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
 & \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
 & \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
 & \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}\]

Note: Here we may go wrong with the vector identities and product identities (dot and cross).