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Let the position vector S of point A and B be \widehat{i}+\widehat{j}+\widehat{k} and \widehat{2i}-\widehat{j}-\widehat{3k}  respectively. A point 'p'  divides the line segment AB intervally in the ratio \lambda : 1(\lambda ,0) if O is the origin and \overrightarrow{OB}.\overrightarrow{OP}-3\left | \overrightarrow{OA}\times \overrightarrow{OP} \right |^{2}=6 then \lambda is equal to _______________
Option: 1 8
Option: 2 0.8
Option: 3 6
Option: 4 0.6

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best_answer

\overrightarrow{\mathrm{OA}} \times \overline{\mathrm{OP}}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ \frac{2 \lambda+1}{\lambda+1} & 1 & \frac{3 \lambda+1}{\lambda+1} \end{array}\right|=\frac{2 \lambda+1}{\lambda+1} \hat{\mathbf{i}}+\frac{-\lambda}{\lambda+1} \hat{\mathbf{j}}+\frac{-\lambda}{\lambda+1} \hat{\mathbf{k}}=\frac{6 \lambda^{2}+1}{(\lambda+1)^{2}}

\\\Rightarrow \frac{14 \lambda+6}{\lambda+1}-3 \times \frac{\left(6 \lambda^{2}+1\right)}{(\lambda+1)^{2}}=6 \\ \\\Rightarrow 10 \lambda^{2}-8 \lambda=0 \\ \\\Rightarrow \lambda=0, \frac{8}{10}=0.8 \Rightarrow \lambda=0.8

 

Posted by

Suraj Bhandari

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