Question

Let $\overrightarrow{c}$be a vector perpendicular to the vectors $a=i+j-k$and $b=i+2j+k$. If $c·(i+j+3k)=8$, then the value of $c·(axb)$ is equal to

Answer 1

Step 1: Calculating the value of $(axb)$ and $c$ vectors

$\overrightarrow{c}·(\overrightarrow{a}x\overrightarrow{b})=\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]$

$$\begin{gathered} (\overrightarrow{a}x\overrightarrow{b})=\left|\begin{array}{ccc}i& j& k\\ 1& 1& -1\\ 1& 2& 1\end{array}\right| \\ \Rightarrow (\overrightarrow{a}x\overrightarrow{b})=i(1+2)-j(1+1)+k(2-1) \\ \Rightarrow (\overrightarrow{a}x\overrightarrow{b})=3i-2j+k \\ \Rightarrow (\overrightarrow{a}x\overrightarrow{b})=(3,-2,1) \end{gathered}$$

$$\begin{gathered} \overrightarrow{c}\perp \overrightarrow{a},\overrightarrow{c}\perp \overrightarrow{b}\Rightarrow C\parallel (\overrightarrow{a}\times \overrightarrow{b}) \\ \Rightarrow \overrightarrow{c}=\lambda (\overrightarrow{a}\times \overrightarrow{b}) \\ \Rightarrow \overrightarrow{c}=\lambda (3i-2j+k) \end{gathered}$$

so,

$$\begin{gathered} \overrightarrow{c}(i+j+3k)=8 \\ \Rightarrow 3\lambda -2\lambda +3\lambda =8 \\ \Rightarrow 4\lambda =8 \\ \Rightarrow \lambda =\frac{8}{4} \\ \Rightarrow \lambda =2 \end{gathered}$$

Step 2: Calculating the value of $\overrightarrow{c}.(\overrightarrow{a}x\overrightarrow{b})$

$$\begin{gathered} \overrightarrow{c}=6i-4j+2k \\ \overrightarrow{c}·(\overrightarrow{a}\times \overrightarrow{b})=\left|\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right| \\ \Rightarrow \overrightarrow{c}·(\overrightarrow{a}\times \overrightarrow{b})=\left|\begin{array}{ccc}6& -4& 2\\ 1& 1& -1\\ 1& 2& 1\end{array}\right| \\ \Rightarrow \overrightarrow{c}·(\overrightarrow{a}\times \overrightarrow{b})=6(1+2)+4(1+1)+2(2-1) \\ \Rightarrow \overrightarrow{c}·(\overrightarrow{a}\times \overrightarrow{b})=6\left(3\right)+4\left(2\right)+2\left(1\right) \\ \Rightarrow \overrightarrow{c}·(\overrightarrow{a}\times \overrightarrow{b})=18+8+2 \\ \Rightarrow \overrightarrow{c}·(\overrightarrow{a}\times \overrightarrow{b})=28 \end{gathered}$$

Hence, the value of $c·(axb)$ is $28$.