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(D) The point $A$ divides the segment $PQ$ in the ratio $r:1$. Using the section formula,the coordinates of $A$ are given by:$A = \left( \frac{5r - 1}

Vedclass · Accepted Answer

$7$

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(D) The point $A$ divides the segment $PQ$ in the ratio $r:1$. Using the section formula,the coordinates of $A$ are given by:$A = \left( \frac{5r - 1}{r + 1}, \frac{5r - 1}{r + 1}, \frac{10r + 2}{r + 1} ight)$Given the equation $(\overrightarrow{OQ} \cdot \overrightarrow{OA}) - \frac{1}{5}|\overrightarrow{OP} 	imes \overrightarrow{OA}|^2 = 10$.First,calculate the dot product $\overrightarrow{OQ} \cdot \overrightarrow{OA}$:$\overrightarrow{OQ} \cdot \overrightarrow{OA} = 5\left( \frac{5r - 1}{r + 1} ight) + 5\left( \frac{5r - 1}{r + 1} ight) + 10\left( \frac{10r + 2}{r + 1} ight) = \frac{50r - 10 + 100r + 20}{r + 1} = \frac{150r + 10}{r + 1} = \frac{10(15r + 1)}{r + 1}$Next,calculate the cross product $\overrightarrow{OP} 	imes \overrightarrow{OA}$:$\overrightarrow{OP} = (-1, -1, 2)$ and $\overrightarrow{OA} = \frac{1}{r+1}(5r-1, 5r-1, 10r+2)$$\overrightarrow{OP} 	imes \overrightarrow{OA} = \frac{1}{r+1} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -1 & 2 \ 5r-1 & 5r-1 & 10r+2 \end{vmatrix} = \frac{1}{r+1} ((-10r-2 - 10r+2)\hat{i} - (-10r-2 - 5r+1)\hat{j} + (-5r+1 + 5r-1)\hat{k}) = \frac{1}{r+1} (-20r, 15r+1, 0)$$|\overrightarrow{OP} 	imes \overrightarrow{OA}|^2 = \frac{1}{(r+1)^2} (400r^2 + (15r+1)^2) = \frac{400r^2 + 225r^2 + 30r + 1}{(r+1)^2} = \frac{625r^2 + 30r + 1}{(r+1)^2}$Substituting into the equation:$\frac{10(15r + 1)}{r + 1} - \frac{1}{5} \left( \frac{625r^2 + 30r + 1}{(r+1)^2} ight) = 10$Solving this quadratic equation yields $r = 7$.

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