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If \vec{a}, \vec{b}, \vec{c} are three non-zero vectors and \hat{n} is a unit vector perpendicular to \hat{c} such that \vec{a}=\alpha \vec{b}-\hat{n} (a\neq 0)  and \vec{b} \cdot \vec{c}=12, then |\vec{c} \times(\vec{a} \times \vec{b})| is equal to :

Option: 1

9


Option: 2

15


Option: 3

6


Option: 4

12


Answers (1)

best_answer

\begin{aligned} & \overrightarrow{\mathrm{a}}=\alpha \overrightarrow{\mathrm{b}}-\hat{\mathrm{n}}, \vec{b} \cdot \overrightarrow{\mathrm{c}}=12 \\ & \overrightarrow{\mathrm{c}} \times(\vec{a} \times \vec{b})=(\vec{c} \cdot \vec{b}) \vec{a}-(\vec{c} \cdot \vec{a}) \vec{b} \\ & \overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=12 \vec{a}-(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}) \\ & \because \overrightarrow{\mathrm{a}}=\alpha \overrightarrow{\mathrm{b}}-\mathrm{n} \\ & \vec{c} \cdot \vec{a}=\alpha \overrightarrow{\mathrm{c}} \cdot \vec{b}-\overrightarrow{\mathrm{c}} \cdot \mathrm{n} \end{aligned}------------(i)

\overrightarrow{\vec{c}} \cdot \vec{a}=12 \alpha

Equation (2) put in equation (1)-------------------------(ii)

\begin{aligned} & \overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=12 \overrightarrow{\mathrm{a}}-12 \alpha \overrightarrow{\mathrm{b}} \\ & |\overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|=12|\overrightarrow{\mathrm{a}}-\alpha \overrightarrow{\mathrm{b}}| \quad[\because \overrightarrow{\mathrm{a}}-\alpha \overrightarrow{\mathrm{b}}=-\mathrm{n} \text { then }|\overrightarrow{\mathrm{a}}-\alpha \overrightarrow{\mathrm{b}}|=1] \\ & \Rightarrow|\overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|=12 \end{aligned}

|\overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|=12

 

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