Let $\overrightarrow{OP}=\frac{\alpha-1}{\alpha}\hat{i}+\hat{j}+\hat{k},\overrightarrow{OQ}=\hat{i}+\frac{\beta-1}{\beta}\hat{j}+\hat{k}$ and $\overrightarrow{OR}=\hat{i}+\hat{j}+\frac{1}{2}\hat{k}$ be three vector where α, β ∈ R - {0} and 0 denotes the origin. If $(\overrightarrow{OP}\times\overrightarrow{OQ}).\overrightarrow{OR}=0$ and the point (α, β, 2) lies on the plane 3x + 3y - z + l = 0, then the value of l is _______.

Question

Let $\overrightarrow{OP}=\frac{\alpha-1}{\alpha}\hat{i}+\hat{j}+\hat{k},\overrightarrow{OQ}=\hat{i}+\frac{\beta-1}{\beta}\hat{j}+\hat{k}$  and  $\overrightarrow{OR}=\hat{i}+\hat{j}+\frac{1}{2}\hat{k}$  be three vector where α, β ∈ R - {0} and 0 denotes the origin. If  $(\overrightarrow{OP}\times\overrightarrow{OQ}).\overrightarrow{OR}=0$  and the point (α, β, 2) lies on the plane 3x + 3y - z + l = 0, then the value of l is _______.

cdquestions Admin · Accepted Answer

Vector Expressions  Given vectors: $\overrightarrow{OP} = \left( \frac{\alpha - 1}{\alpha} \right) i + j + k$ $\overrightarrow{OQ} = i + \left( \frac{\beta - 1}{\beta} \right) j + k$ $\overrightarrow{OR} = i + j + \frac{1}{2} k$ Cross Product $ \overrightarrow{OP} \times \overrightarrow{OQ} $ Computing the determinant: $$ \overrightarrow{OP} \times \overrightarrow{OQ} = \begin{vmatrix} i & j & k \ \frac{\alpha - 1}{\alpha} & 1 & 1 \ 1 & \frac{\beta - 1}{\beta} & 1 \end{vmatrix} $$ Expanding along the first row: $$ \overrightarrow{OP} \times \overrightarrow{OQ} = i \left( 1 - \frac{\beta - 1}{\beta} \right) - j \left( \frac{\alpha - 1}{\alpha} - 1 \right) + k \left( \frac{\alpha - 1}{\alpha} \cdot \frac{\beta - 1}{\beta} - 1 \right) $$ Dot Product with $ \overrightarrow{OR} $ $$ (\overrightarrow{OP} \times \overrightarrow{OQ}) \cdot \overrightarrow{OR} = 0 $$ Solving for $ \alpha, \beta $. Substituting $ (\alpha, \beta, 2) $ into the Plane Equation $$ 3\alpha + 3\beta - 2 + l = 0 $$ Solving for $ l $: $$ l = 5 $$ Final Answer: 5

List-I	List-II
(A) 4î − 2ĵ − 4k̂	(I) A vector perpendicular to both î + 2ĵ + k̂ and 2î + 2ĵ + 3k̂
(B) 4î − 4ĵ + 2k̂	(II) Direction ratios are −2, 1, 2
(C) 2î − 4ĵ + 4k̂	(III) Angle with the vector î − 2ĵ − k̂ is cos⁻¹(1/√6)
(D) 4î − ĵ − 2k̂	(IV) Dot product with −2î + ĵ + 3k̂ is 10

Correct Answer: 5

Solution and Explanation

Vector Expressions

Cross Product \( \overrightarrow{OP} \times \overrightarrow{OQ} \)

Dot Product with \( \overrightarrow{OR} \)

Substituting \( (\alpha, \beta, 2) \) into the Plane Equation

Final Answer: 5

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