Let the position vectors of points A and B be | vedclass

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(C) Using the section formula,the position vector of point $P$ is given by:$\overrightarrow{OP} = \frac{\lambda(2\hat{i}+\hat{j}+3\hat{k}) + 1(\hat{i}

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(C) Using the section formula,the position vector of point $P$ is given by:$\overrightarrow{OP} = \frac{\lambda(2\hat{i}+\hat{j}+3\hat{k}) + 1(\hat{i}+\hat{j}+\hat{k})}{\lambda+1} = \frac{2\lambda+1}{\lambda+1}\hat{i} + \frac{\lambda+1}{\lambda+1}\hat{j} + \frac{3\lambda+1}{\lambda+1}\hat{k}$Now,calculate $\overrightarrow{OB} \cdot \overrightarrow{OP}$:$\overrightarrow{OB} \cdot \overrightarrow{OP} = (2\hat{i}+\hat{j}+3\hat{k}) \cdot \left( \frac{2\lambda+1}{\lambda+1}\hat{i} + \hat{j} + \frac{3\lambda+1}{\lambda+1}\hat{k} ight) = \frac{4\lambda+2 + \lambda+1 + 9\lambda+3}{\lambda+1} = \frac{14\lambda+6}{\lambda+1}$Next,calculate $\overrightarrow{OA} 	imes \overrightarrow{OP}$:$\overrightarrow{OA} 	imes \overrightarrow{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ \frac{2\lambda+1}{\lambda+1} & 1 & \frac{3\lambda+1}{\lambda+1} \end{vmatrix} = \left( \frac{3\lambda+1}{\lambda+1} - 1 ight)\hat{i} - \left( \frac{3\lambda+1}{\lambda+1} - \frac{2\lambda+1}{\lambda+1} ight)\hat{j} + \left( 1 - \frac{2\lambda+1}{\lambda+1} ight)\hat{k}$$= \frac{2\lambda}{\lambda+1}\hat{i} - \frac{\lambda}{\lambda+1}\hat{j} - \frac{\lambda}{\lambda+1}\hat{k}$$|\overrightarrow{OA} 	imes \overrightarrow{OP}|^{2} = \frac{4\lambda^{2} + \lambda^{2} + \lambda^{2}}{(\lambda+1)^{2}} = \frac{6\lambda^{2}}{(\lambda+1)^{2}}$Substituting these into the given equation:$\frac{14\lambda+6}{\lambda+1} - 3 \left( \frac{6\lambda^{2}}{(\lambda+1)^{2}} ight) = 6$Multiply by $(\lambda+1)^{2}$:$(14\lambda+6)(\lambda+1) - 18\lambda^{2} = 6(\lambda+1)^{2}$$14\lambda^{2} + 20\lambda + 6 - 18\lambda^{2} = 6(\lambda^{2} + 2\lambda + 1)$$-4\lambda^{2} + 20\lambda + 6 = 6\lambda^{2} + 12\lambda + 6$$10\lambda^{2} - 8\lambda = 0$Since $\lambda > 0$,we have $10\lambda = 8 \Rightarrow \lambda = 0.8$.

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