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Step 1: Identify the Given Vectors and Ratios
Let
\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}
and
\overrightarrow{b} = 2\hat{i} + \hat{j} + 3\hat{k}.
Points A and B have position vectors \overrightarrow{a} and \overrightarrow{b} respectively. A point P divides AB internally in the ratio \lambda :1 (with \lambda > 0 ). Hence,
\overrightarrow{OP} = \frac{\overrightarrow{a} + \lambda \,\overrightarrow{b}}{1 + \lambda}.
Step 2: Express the Condition Involving Dot and Cross Products
We are given:
\overrightarrow{OB} \cdot \overrightarrow{OP} \;-\; 3 \Big|\overrightarrow{OA} \times \overrightarrow{OP}\Big|^2 \;=\; 6.
Since \overrightarrow{OB} = \overrightarrow{b} and \overrightarrow{OA} = \overrightarrow{a}, substitute:
\overrightarrow{OP} = \dfrac{\overrightarrow{a} + \lambda \,\overrightarrow{b}}{1 + \lambda}.
Step 3: Substitute and Simplify the Dot Product
Thus,
\overrightarrow{OB} \cdot \overrightarrow{OP}
= \overrightarrow{b} \cdot \left(\dfrac{\overrightarrow{a} + \lambda \,\overrightarrow{b}}{1 + \lambda}\right)
= \dfrac{\overrightarrow{b} \cdot \overrightarrow{a} + \lambda(\overrightarrow{b} \cdot \overrightarrow{b})}{1 + \lambda}.
Compute each dot product:
\overrightarrow{b} \cdot \overrightarrow{a}
= (2\hat{i} + \hat{j} + 3\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})
= 2 + 1 + 3
= 6.
\overrightarrow{b} \cdot \overrightarrow{b}
= (2\hat{i} + \hat{j} + 3\hat{k}) \cdot (2\hat{i} + \hat{j} + 3\hat{k})
= 4 + 1 + 9
= 14.
Step 4: Substitute and Simplify the Cross Product Term
Next, consider
\Big|\overrightarrow{OA} \times \overrightarrow{OP}\Big|^2.
First, note that:
\overrightarrow{OA} \times \overrightarrow{OP}
= \overrightarrow{a} \times \left(\dfrac{\overrightarrow{a} + \lambda \,\overrightarrow{b}}{1 + \lambda}\right)
= \dfrac{\overrightarrow{a} \times \overrightarrow{a} + \lambda\big(\overrightarrow{a} \times \overrightarrow{b}\big)}{1 + \lambda}.
But \overrightarrow{a} \times \overrightarrow{a} = \overrightarrow{0}. Hence we need the magnitude of
\dfrac{\lambda(\overrightarrow{a} \times \overrightarrow{b})}{1 + \lambda}.
Compute
\overrightarrow{a} \times \overrightarrow{b}.
\overrightarrow{a} \times \overrightarrow{b}
= \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
2 & 1 & 3
\end{vmatrix}
= (3 - 1)\,\hat{i} \;-\; (3 - 2)\,\hat{j} \;+\; (1 - 2)\,\hat{k}
= 2\hat{i} - \hat{j} - \hat{k}.
Then
\Big|\overrightarrow{a} \times \overrightarrow{b}\Big|
= \sqrt{2^2 + (-1)^2 + (-1)^2}
= \sqrt{6}.
Therefore,
\Big|\overrightarrow{OA} \times \overrightarrow{OP}\Big|^2
= \left|\dfrac{\lambda \,(\overrightarrow{a} \times \overrightarrow{b})}{1 + \lambda}\right|^2
= \dfrac{\lambda^2 \,\big|\overrightarrow{a} \times \overrightarrow{b}\big|^2}{(1 + \lambda)^2}
= \dfrac{\lambda^2 \cdot 6}{(1 + \lambda)^2}.
Step 5: Form the Final Equation and Solve for \lambda
Putting these results into the given condition:
\dfrac{6 + 14\,\lambda}{1 + \lambda}
\;-\; 3 \times \dfrac{\lambda^2 \cdot 6}{(1 + \lambda)^2}
= 6.
Multiply through by (1 + \lambda)^2 to clear denominators:
(6 + 14\lambda)(1 + \lambda)
\;-\;
18\,\lambda^2
=
6 \,(1 + \lambda)^2.
Expand each side:
Left side: (6 + 14\lambda)(1 + \lambda) - 18\lambda^2 = (6 + 14\lambda)(1 + \lambda) - 18\lambda^2.
Right side: 6(1 + \lambda)^2 = 6(1 + 2\lambda + \lambda^2).
Simplifying gives a quadratic in \lambda . After arranging terms, one obtains:
10\lambda^2 - 8\lambda = 0
\quad\Longrightarrow\quad
\lambda\,(10\lambda - 8) = 0.
Since \lambda > 0, we discard \lambda = 0 and get
\lambda = \dfrac{8}{10} = 0.8.
Step 6: Conclude the Result
Hence, the value of \lambda that satisfies the given condition is
\lambda = 0.8.
