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 _______.

Answer 1

Correct Answer: 5

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

Answer 2

To solve the problem, analyze the given vectors and conditions.

Given:
\[ \overrightarrow{OP} = \frac{\alpha - 1}{\alpha} \hat{i} + \hat{j} + \hat{k} \] \[ \overrightarrow{OQ} = \hat{i} + \frac{\beta - 1}{\beta} \hat{j} + \hat{k} \] \[ \overrightarrow{OR} = \hat{i} + \hat{j} + \frac{1}{2} \hat{k} \] with \(\alpha, \beta \in \mathbb{R} \setminus \{0\}\), and \[ (\overrightarrow{OP} \times \overrightarrow{OQ}) \cdot \overrightarrow{OR} = 0 \] Point \((\alpha, \beta, 2)\) lies on plane \[ 3x + 3y - z + l = 0 \] Find \(l\).

Step 1: Compute \(\overrightarrow{OP} \times \overrightarrow{OQ}\):

\[ \overrightarrow{OP} = \left( \frac{\alpha - 1}{\alpha}, 1, 1 \right), \quad \overrightarrow{OQ} = (1, \frac{\beta - 1}{\beta}, 1) \] Calculate cross product: \[ \overrightarrow{OP} \times \overrightarrow{OQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\alpha - 1}{\alpha} & 1 & 1 \\ 1 & \frac{\beta - 1}{\beta} & 1 \end{vmatrix} \] Calculate each component: \[ \hat{i}: (1)(1) - (1)\left(\frac{\beta - 1}{\beta}\right) = 1 - \frac{\beta - 1}{\beta} = \frac{\beta - (\beta - 1)}{\beta} = \frac{1}{\beta} \] \[ \hat{j}: -\left( \frac{\alpha - 1}{\alpha} \cdot 1 - 1 \cdot 1 \right) = -\left( \frac{\alpha - 1}{\alpha} - 1 \right) = -\left( \frac{\alpha - 1 - \alpha}{\alpha} \right) = -\left( \frac{-1}{\alpha} \right) = \frac{1}{\alpha} \] \[ \hat{k}: \frac{\alpha - 1}{\alpha} \cdot \frac{\beta - 1}{\beta} - 1 \cdot 1 = \frac{(\alpha - 1)(\beta - 1)}{\alpha \beta} - 1 \] So, \[ \overrightarrow{OP} \times \overrightarrow{OQ} = \left( \frac{1}{\beta}, \frac{1}{\alpha}, \frac{(\alpha - 1)(\beta - 1)}{\alpha \beta} - 1 \right) \]

Step 2: Compute the dot product with \(\overrightarrow{OR}\):
\[ (\overrightarrow{OP} \times \overrightarrow{OQ}) \cdot \overrightarrow{OR} = \frac{1}{\beta} \times 1 + \frac{1}{\alpha} \times 1 + \left( \frac{(\alpha - 1)(\beta - 1)}{\alpha \beta} - 1 \right) \times \frac{1}{2} = 0 \] Multiply both sides by \(2 \alpha \beta\) (non-zero since \(\alpha, \beta \neq 0\)) to clear denominators: \[ 2 \alpha + 2 \beta + \alpha \beta \left( (\alpha - 1)(\beta - 1)/(\alpha \beta) - 1 \right) = 0 \] Simplify the term inside the parenthesis: \[ (\alpha - 1)(\beta - 1)/(\alpha \beta) - 1 = \frac{(\alpha - 1)(\beta - 1) - \alpha \beta}{\alpha \beta} \] Multiply by \(\alpha \beta\): \[ (\alpha - 1)(\beta - 1) - \alpha \beta = \alpha \beta - \alpha - \beta + 1 - \alpha \beta = -\alpha - \beta + 1 \] Therefore, the equation becomes: \[ 2 \alpha + 2 \beta + (-\alpha - \beta + 1) = 0 \] \[ (2\alpha - \alpha) + (2\beta - \beta) + 1 = 0 \] \[ \alpha + \beta + 1 = 0 \] \[ \alpha + \beta = -1 \]

Step 3: Use the plane equation:
Given point \((\alpha, \beta, 2)\) lies on plane: \[ 3 \alpha + 3 \beta - 2 + l = 0 \implies 3 (\alpha + \beta) + l = 2 \] Substitute \(\alpha + \beta = -1\): \[ 3 \times (-1) + l = 2 \implies -3 + l = 2 \implies l = 5 \]

Final Answer:
\[ \boxed{5} \]