उत्तर: (1)

समाधान:

सूत्र: वेक्टर के अदिश गुणनफल के गुण , वेक्टर के क्रॉस गुणनफल के गुण

$ \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k} $

$ \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k} $

$\lbrace (\vec{a}+\vec{b}) \times (\vec{a} \times \vec{b})\rbrace \times (\vec{a}-\vec{b}) = 8 \hat{i}+40 \hat{j}-24 \hat{k}$

$\Rightarrow \lbrace (\vec{a} - \vec{b}) \cdot (\vec{a}+\vec{b}) \rbrace \cdot (\vec{a} \times \vec{b}) - \lbrace (\vec{a} -\vec{b}) \cdot (\vec{a} +\vec{b}) \rbrace (\vec{a}+\vec{b}) = 8\hat{i}-40 \hat{j}-24 \hat{k}$

$\Rightarrow (|a|^2 - |b|^2)(\vec{a} \times \vec{b}) = 8(\hat{i}-5 \hat{j}-3\hat{k})$

$ \Rightarrow 8(\vec{a} \times \vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$

$\Rightarrow (\vec{a} \times \vec{b}) = (\hat{i}-5\hat{j}-3 \hat{k})$

अब, $\vec{a} \times \vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2\end{vmatrix} $

$=(4-3 \lambda) \hat{i}-(2 \lambda+3) \hat{j}+(-\lambda^{2}-2) \hat{k}$

$\Rightarrow \lambda=1$

$\therefore \overrightarrow{{}a}=\hat{i}+2 \hat{j}-3 \hat{k}$

$\overrightarrow{{}b}=\hat{i}-\hat{j}+2 \hat{k}$

$\Rightarrow \vec{a}+\vec{b}=2 \hat{i}+\hat{j}-\hat{k}, \vec{a}-\vec{b}=3 \hat{j}-5 \hat{k}$

$\Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 0 & 3 & -5\end{vmatrix} =2 \hat{i}+10 \hat{j}+6 \hat{k}$

$\therefore$ आवश्यक उत्तर $=4+100+36=140$