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  • Let <img alt="\lambda \in \mathbb{R}, \vec{a}=\lambda \hat{\imath}+2 \hat{\jmath}-3 \hat{k}, \vec{b}=\hat{\imath}-\lambda \hat{\jmath}+2 \hat{k}" src="https://entrancecorner.oncodecogs.com/gif.late

Let \lambda \in \mathbb{R}, \vec{a}=\lambda \hat{\imath}+2 \hat{\jmath}-3 \hat{k}, \vec{b}=\hat{\imath}-\lambda \hat{\jmath}+2 \hat{k}
If ((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{\imath}-40 \hat{\jmath}-24 \hat{k}, then |\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^{2}is equal to

Option: 1

132


Option: 2

136


Option: 3

140


Option: 4

144


Answers (1)

best_answer

((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})-=8 \hat{i}-40 \hat{j}-24 \hat{k}
\Rightarrow(\vec{a} \times(\vec{a} \times \vec{b})+\vec{b} \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})
\Rightarrow((\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}+(\vec{b} \cdot \vec{b}) \vec{a}-(\vec{b} \cdot \vec{a}) \vec{b}) \times(\vec{a}-\vec{b})
\Rightarrow 0-(\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b})-a^{2}(\vec{b} \times \vec{a})+0-\mathrm{b}^{2}(\vec{a} \times \vec{b})-(\vec{a} \cdot \vec{b}) \vec{b} \times \vec{a}=8 \hat{\mathrm{i}}-40 \hat{\mathrm{j}}-24 \hat{\mathrm{k}}
\Rightarrow\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)(\vec{a} \times \overrightarrow{\mathrm{b}})=8 \hat{\mathrm{i}}-40 \hat{\mathrm{j}}-24 \hat{\mathrm{k}}

\left(\left(\lambda^{2}+4+9\right)-\left(1+\lambda^{2}+4\right)\right)(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})
8(\vec{a} \times \vec{b})=8(\hat{i}-5 \hat{j}-3 \hat{k})

\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2 \end{array}\right|=\hat{i}-5 \hat{j}-3 \hat{k}

\hat{\mathrm{i}}(4-3 \lambda)-\hat{\mathrm{j}}(2 \lambda+3)+\hat{\mathrm{k}}\left(-\lambda^{2}-2\right)=\hat{\mathrm{i}}-5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}
\Rightarrow 4-3 \lambda=1 \quad 2 \lambda+3=5 \quad-\lambda^{2}-2=-3
3 \lambda=3 \quad \quad                                                            \lambda^{2}=1  
\lambda=1 \quad                                                                \lambda=1 \quad

|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^{2}
\Rightarrow |-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}|^{2}=|2(\vec{a} \times \vec{b})|^{2}=4(1+25+9)=140
 

Posted by

avinash.dongre

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