Question

Find \[\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right),\] if \[\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k},\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k},\overrightarrow{c}=3\widehat{i}+\widehat{j}+2\widehat{k}.\]

Answer 1

Hint: We are asked to find the dot product of vector a with \[\overrightarrow{b}\times \overrightarrow{c}.\] First, we will find the cross product of \[\overrightarrow{b}\times \overrightarrow{c}\] using the formula of \[A={{a}_{1}}i+{{b}_{i}}j+{{c}_{1}}k\] and \[B={{a}_{2}}i+{{b}_{2}}j+{{c}_{2}}k.\] Once we find \[\overrightarrow{A}\times \overrightarrow{B}=\left| \begin{matrix}
   i & j & k \\
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|,\] then using \[\overrightarrow{b}\times \overrightarrow{c}\] we will find the dot product of \[\overrightarrow{b}\times \overrightarrow{c}\] with \[\overrightarrow{a}\] using \[A.B={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}\] and then we will get our required solution.

Complete step by step answer:
We are given \[\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k},\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k},\overrightarrow{c}=3\widehat{i}+\widehat{j}+2\widehat{k}.\] We are asked to find \[\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right).\] We will first find the cross product of \[\overrightarrow{b}\] and \[\overrightarrow{c}\] first and then we will solve further.
We know that for any \[X=x\widehat{i}+y\widehat{j}+z\widehat{k}\] and \[Y=a\widehat{i}+b\widehat{j}+c\widehat{k}\] the cross product is given as,
\[\overrightarrow{X}\times \overrightarrow{Y}=\left| \begin{matrix}
   i & j & k \\
   x & y & z \\
   a & b & c \\
\end{matrix} \right|\]
So, for \[\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k},\overrightarrow{c}=3\widehat{i}+\widehat{j}+2\widehat{k},\] we will have,
\[\overrightarrow{b}\times \overrightarrow{c}=\left| \begin{matrix}
   i & j & k \\
   1 & 2 & 1 \\
   3 & 1 & 2 \\
\end{matrix} \right|\]
Now, we will expand the determinant along the row 1, we will get,
\[\overrightarrow{b}\times \overrightarrow{c}=i\left( 2\times 2-1\times 1 \right)-j\left( 1\times 2-3\times 1 \right)+k\left( 1\times 1-3\times 2 \right)\]
\[\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=i\left( 4-1 \right)-j\left( 2-3 \right)+k\left( 1-6 \right)\]
Simplifying further, we get,
\[\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=3i+j-5k\]
Now, we will find the dot product of \[\overrightarrow{b}\times \overrightarrow{c}\] with \[\overrightarrow{a}.\] We know that for \[\alpha =x\widehat{i}+y\widehat{j}+z\widehat{k}\] and \[\beta =a\widehat{i}+b\widehat{j}+c\widehat{k}.\]
\[\alpha .\beta =x.a+y.b+z.c\]
So for, \[\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}\] and \[\overrightarrow{b}\times \overrightarrow{c}=3i+j-5k,\] we have,
\[\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( 2\widehat{i}+\widehat{j}+3\widehat{k} \right).\left( 3i+j-5k \right)\]
\[\Rightarrow \overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=2.3+1.1+3.\left( -5 \right)\]
Simplifying, we get,
\[\Rightarrow \overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=6+1-15\]
\[\Rightarrow \overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=-8\]

So, we will get \[\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=-8\] as our answer.

Note: We can do this in an alternate method. We know that for any \[X.\left( \overrightarrow{Y}\times \overrightarrow{Z} \right)\] is given as,
\[X.\left( \overrightarrow{Y}\times \overrightarrow{Z} \right)=\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]
So for, \[\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k},\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k},\overrightarrow{c}=3\widehat{i}+\widehat{j}+2\widehat{k},\]
\[a.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left| \begin{matrix}
   2 & 1 & 3 \\
   1 & 2 & 1 \\
   3 & 1 & 2 \\
\end{matrix} \right|\]
Expanding along row 1, we will get,
\[\Rightarrow a.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=2\left( 2\times 2-1\times 1 \right)-1\left( 1\times 2-3\times 1 \right)+3\left( 1\times 1-3\times 2 \right)\]
\[\Rightarrow a.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=2\left( 4-1 \right)-1\left( 2-3 \right)+3\left( 1-6 \right)\]
\[\Rightarrow a.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=2\left( 3 \right)-1\left( -1 \right)+3\left( -5 \right)\]
\[\Rightarrow a.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=6+1-15\]
\[\Rightarrow a.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=-8\]