Let a=2i+5j-k,b=2i-2j+2k and c be three vectors such that (c+i) (a+b+i)=a (c+i).

Question

Let $\vec{a}=2\hat{i}+5\hat{j}-\hat{k},\vec{b}=2\hat{i}-2\hat{j}+2\hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c}+\hat{i})	imes (\vec{a}+\vec{b}+\hat{i})=\vec{a}	imes (\vec{c}+\hat{i})$. If $\vec{a}\cdot \vec{c}=-29$, then $\vec{c}\cdot (-2\hat{i}+\hat{j}+\hat{k})$ is equal to:

Accepted Answer

$\begin{gathered} (\vec{c}+\hat{i})	imes (\vec{a}+\vec{b}+\hat{i}+\vec{a})=0 \ \Rightarrow \, \vec{c}+\hat{i}=\lambda (\vec{a}+\vec{b}+\hat{i}+\vec{a}) \ =\lambda (2\vec{a}+\vec{b}+\hat{i}) \ \, =\lambda (7\hat{i}+8\hat{j}) \ \Rightarrow \, \vec{c}=(7\lambda -1)\hat{i}+8\lambda \hat{j} \ \, \vec{c}\cdot \vec{a}=-29 \ \Rightarrow \, 14\lambda -2+40\lambda =-29 \ \Rightarrow \, 54\lambda =-27 \ \Rightarrow \, \lambda =-\frac{1}{2} \end{gathered}$

$\begin{gathered} 	herefore \, \vec{c}=(\frac{-7}{2}-1)\hat{i}-4\hat{j}=\frac{-9}{2}\hat{i}-4\hat{j} \ \vec{c}\cdot (-2\hat{i}+\hat{j}+\hat{k})=9-4=5 \end{gathered}$

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