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

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,

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$$