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MathMedAdv 2024 • Paper 1

Coplanar Vectors and Point on Plane Parameter Value

Let OP‾=α−1αi^+j^+k^\overline{OP} = \frac{\alpha - 1}{\alpha}\hat{i} + \hat{j} + \hat{k}, OQ‾=i^+β−1βj^+k^\overline{OQ} = \hat{i} + \frac{\beta - 1}{\beta}\hat{j} + \hat{k} and OR‾=i^+j^+12k^\overline{OR} = \hat{i} + \hat{j} + \frac{1}{2}\hat{k} be three vectors, where α,β∈R−{0}\alpha, \beta \in \mathbb{R} - \{0\} and OO denotes the origin. If (OP‾×OQ‾)⋅OR‾=0(\overline{OP} \times \overline{OQ}) \cdot \overline{OR} = 0 and the point (α,β,2)(\alpha, \beta, 2) lies on the plane 3x+3y−z+l=03x + 3y - z + l = 0, then the value of ll is _______.

Official Numerical Answer5

Topics & Concepts

Vector AlgebraScalar Triple ProductPlane in 3D

Step-by-Step Solution

To determine the value of ll, we set the scalar triple product of the vectors OP‾\overline{OP}, OQ‾\overline{OQ}, and OR‾\overline{OR} to zero to represent their coplanarity: ∣1−1α1111−1β11112∣=0\begin{vmatrix} 1 - \frac{1}{\alpha} & 1 & 1 \\ 1 & 1 - \frac{1}{\beta} & 1 \\ 1 & 1 & \frac{1}{2} \end{vmatrix} = 0

Evaluating the determinant yields the relation α+β=−1\alpha + \beta = -1.

Since the point (α,β,2)(\alpha, \beta, 2) lies on the plane 3x+3y−z+l=03x + 3y - z + l = 0, substituting these coordinates into the equation gives: 3(α+β)−2+l=03(\alpha + \beta) - 2 + l = 0

Substituting α+β=−1\alpha + \beta = -1 into the plane equation results in 3(−1)−2+l=03(-1) - 2 + l = 0, which solves to l=5l = 5.