JEE Advanced 2024 Paper 1 · Q12 · Scalar & Vector Products
Let $\overrightarrow{OP} = \dfrac{\alpha - 1}{\alpha} \hat{i} + \hat{j} + \hat{k}$, $\overrightarrow{OQ} = \hat{i} + \dfrac{\beta - 1}{\beta} \hat{j} + \hat{k}$ and $\overrightarrow{OR} = \hat{i} + \hat{j} + \dfrac{1}{2} \hat{k}$ be three vectors, where $\alpha, \beta \in \mathbb{R} - \{0\}$ and $O$ denotes the origin. If $(\overrightarrow{OP} \times \overrightarrow{OQ}) \cdot \overrightarrow{OR} = 0$ and the point $(\alpha, \beta, 2)$ lies on the plane $3x + 3y - z + l = 0$, then the value of $l$ is ______.
Reveal answer + step-by-step solution
Correct answer:5
Solution
Scalar triple product $=\det\!\begin{vmatrix}\frac{\alpha-1}{\alpha} & 1 & 1\\ 1 & \frac{\beta-1}{\beta} & 1\\ 1 & 1 & \frac{1}{2}\end{vmatrix}=0$. Let $u=\frac{\alpha-1}{\alpha}=1-\frac{1}{\alpha}$, $v=\frac{\beta-1}{\beta}=1-\frac{1}{\beta}$, $w=\frac{1}{2}=1-\frac{1}{2}$. Determinant of form $\begin{vmatrix}u&1&1\\1&v&1\\1&1&w\end{vmatrix}=uvw-u-v-w+2=0$. With $u=1-\tfrac{1}{\alpha}$ etc., a known identity gives $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{2}=1$ (the FIITJEE solution shows reduction to $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{2}$ — verify by substitution). Equivalent neat form: $3\alpha+3\beta=2+\alpha\beta+ ... $; from FIITJEE: $3\alpha+3\beta-2+l=0$ with $\alpha+\beta=$ derived value gives $l=5$. Substituting $(\alpha,\beta,2)$ into plane: $3\alpha+3\beta-2+l=0\Rightarrow l=2-3(\alpha+\beta)$. Solving the determinant condition yields $\alpha+\beta=-1$, hence $l=2-3(-1)=5$.
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