Let a=6i+9j+12k,b= i+11j-2k and c be vectors such that a c=a b.

Question

Let $\vec{a}=6\hat{i}+9\hat{j}+12\hat{k},\vec{b}=\alpha \hat{i}+11\hat{j}-2\hat{k}$ and $\vec{c}$ be vectors such that $\vec{a}	imes \vec{c}=\vec{a}	imes \vec{b}$. If

$\vec{a}\cdot \vec{c}=-12,\vec{c}\cdot (\hat{i}-2\hat{j}+\hat{k})=5$, then $\vec{c}\cdot (\hat{i}+\hat{j}+\hat{k})$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 11

Accepted Answer

Let $\vec{c}={c}_{1}\hat{i}+{c}_{2}\hat{j}+{c}_{3}\hat{k}$

Now, $\vec{a}\cdot \vec{c}=-12$

$\Rightarrow 6{c}_{1}+9{c}_{2}+12{c}_{3}=-12$ ..............(i)

Also, $\vec{c}\cdot (\hat{i}-2\hat{j}+\hat{k})=5$

$\Rightarrow {c}_{1}-2{c}_{2}+{c}_{3}=5$ ................(ii)

$\begin{gathered} \vec{a}	imes \vec{c}=\vec{a}	imes \vec{b} \ \Rightarrow \vec{a}	imes (\vec{c}-\vec{b})=0 \ \Rightarrow \vec{a}(\vec{c}-\vec{b}) \ \Rightarrow \vec{a}=\lambda (\vec{c}-\vec{b}) \ \Rightarrow 6\hat{i}+9\hat{j}+12\hat{k}=\lambda ({c}_{1}-\alpha )\hat{i}+\lambda ({c}_{2}-11)\hat{j}+\lambda ({c}_{3}+2)\hat{k} \end{gathered}$

On comparing, we get

${c}_{1}=\frac{6}{\lambda }+\alpha ,{c}_{2}=\frac{9}{\lambda }+11,{c}_{3}=\frac{12}{\lambda }-2$

Put there values in (ii), we get

$\begin{gathered} \frac{6}{\lambda }+\alpha -\frac{18}{\lambda }-22+\frac{12}{\lambda }-2=5 \ \Rightarrow \alpha =29 \end{gathered}$

From (i) and values of ${c}_{1},{c}_{2},{c}_{3}$, and $\alpha $ we have

$\begin{gathered} 6(\frac{6}{\lambda }+29)+9(\frac{9}{\lambda }+11)+12(\frac{12}{\lambda }-2)=-12 \ \Rightarrow \frac{261}{\lambda }=-261\Rightarrow \lambda =-1 \end{gathered}$

$\begin{gathered} {c}_{1}=23,{c}_{2}=2,{c}_{3}=-14 \ 	herefore \vec{c}\cdot (\hat{i}+\hat{j}+\hat{k})=(23\hat{i}+2\hat{j}+-14\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k}) \ =23+2-14=11 \end{gathered}$

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