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Let alpha , beta , gamma , delta be rea...

Let `alpha , beta , gamma , delta ` be real numbers such that `alpha^(2) +beta^(2) +gamma^(2) != 0 and alpha +gamma = 1 ` . Supose the point `(3,2,-1)` is the mirror image of the point `(1,0,-1)` with respect to the planet `alphax + beta y + gammaz = delta` . Then which of the following statements is/are TRUE ?

A

`alpha + beta = 2`

B

`delta- gamma = 3`

C

`delta + beta = 4`

D

`alpha + beta + gamma = delta`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the values of the parameters \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\) based on the given conditions, and then verify which statements are true. ### Step-by-step Solution: 1. **Understand the Given Conditions**: - We have the condition: \(\alpha^2 + \beta^2 + \gamma^2 \neq 0\) and \(\alpha + \gamma = 1\). - The points \(A(1, 0, -1)\) and \(C(3, 2, -1)\) are mirror images with respect to the plane defined by the equation \(\alpha x + \beta y + \gamma z = \delta\). 2. **Find the Midpoint**: - The midpoint \(B\) between points \(A\) and \(C\) can be calculated as: \[ B = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) = \left(\frac{1 + 3}{2}, \frac{0 + 2}{2}, \frac{-1 + (-1)}{2}\right) = (2, 1, -1) \] 3. **Substitute Point B into the Plane Equation**: - Since point \(B\) lies on the plane, we substitute its coordinates into the plane equation: \[ \alpha(2) + \beta(1) + \gamma(-1) = \delta \] This simplifies to: \[ 2\alpha + \beta - \gamma = \delta \quad \text{(Equation 1)} \] 4. **Find Direction Ratios**: - The direction vector from \(A\) to \(C\) is given by: \[ \text{Direction} = (3 - 1, 2 - 0, -1 - (-1)) = (2, 2, 0) \] - The normal vector of the plane is given by \((\alpha, \beta, \gamma)\). Thus, we can set up the following ratios: \[ \frac{\alpha}{2} = \frac{\beta}{2} = \frac{\gamma}{0} = \lambda \] - From this, we can express: \[ \alpha = 2\lambda, \quad \beta = 2\lambda, \quad \gamma = 0 \] 5. **Substituting into the Condition**: - Using the condition \(\alpha + \gamma = 1\): \[ 2\lambda + 0 = 1 \implies \lambda = \frac{1}{2} \] - Therefore: \[ \alpha = 2 \cdot \frac{1}{2} = 1, \quad \beta = 2 \cdot \frac{1}{2} = 1, \quad \gamma = 0 \] 6. **Finding Delta**: - Substitute \(\alpha\), \(\beta\), and \(\gamma\) into Equation 1: \[ 2(1) + 1 - 0 = \delta \implies \delta = 3 \] 7. **Final Values**: - We have found: \[ \alpha = 1, \quad \beta = 1, \quad \gamma = 0, \quad \delta = 3 \] 8. **Check the Statements**: - Now we check the provided statements: 1. \(\alpha + \beta = 2\) (True) 2. \(\delta - \gamma = 3\) (True) 3. \(\delta + \beta = 4\) (True) 4. \(\alpha + \beta + \gamma = 2\) (False) ### Conclusion: The true statements are: - \(\alpha + \beta = 2\) - \(\delta - \gamma = 3\) - \(\delta + \beta = 4\)
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