Consider three vectors `A=hati +hatj-2hatk`,
`B=hati +hatj+2hatk and 2hati -3hatj +4hatk` A vector X of the from `alphaA +betaB (alpha and beta"are numbers")` is perpendicular to C .The ratio of `alpha and beta` is

Options

A. `1:1`
B. `2:1`
C. `-1:1`
D. `3:1`

Explanation

To solve the problem, we need to find the ratio of the coefficients \( \alpha \) and \( \beta \) such that the vector \( \mathbf{X} = \alpha \mathbf{A} + \beta \mathbf{B} \) is perpendicular to the vector \( \mathbf{C} \).

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Given vectors:
\[ \mathbf{A} = \hat{i} + \hat{j} - 2\hat{k} \]
\[ \mathbf{B} = \hat{i} + \hat{j} + 2\hat{k} \]
\[ \mathbf{C} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]

2. **Express Vector \( \mathbf{X} \)**:
- The vector \( \mathbf{X} \) can be expressed as:
\[ \mathbf{X} = \alpha \mathbf{A} + \beta \mathbf{B} = \alpha(\hat{i} + \hat{j} - 2\hat{k}) + \beta(\hat{i} + \hat{j} + 2\hat{k}) \]
- Expanding this gives:
\[ \mathbf{X} = (\alpha + \beta)\hat{i} + (\alpha + \beta)\hat{j} + (-2\alpha + 2\beta)\hat{k} \]

3. **Dot Product Condition for Perpendicularity**:
- For \( \mathbf{X} \) to be perpendicular to \( \mathbf{C} \), the dot product must be zero:
\[ \mathbf{X} \cdot \mathbf{C} = 0 \]
- Calculating the dot product:
\[ \mathbf{X} \cdot \mathbf{C} = (\alpha + \beta)(2) + (\alpha + \beta)(-3) + (-2\alpha + 2\beta)(4) \]
- Simplifying this:
\[ = 2(\alpha + \beta) - 3(\alpha + \beta) + 4(-2\alpha + 2\beta) \]
\[ = (2 - 3)(\alpha + \beta) + 4(-2\alpha + 2\beta) \]
\[ = -1(\alpha + \beta) + 8\beta - 8\alpha \]
\[ = -\alpha - \beta + 8\beta - 8\alpha \]
\[ = -9\alpha + 7\beta \]

4. **Set the Dot Product to Zero**:
- Setting the equation to zero:
\[ -9\alpha + 7\beta = 0 \]
- Rearranging gives:
\[ 9\alpha = 7\beta \]
- Thus, the ratio \( \frac{\alpha}{\beta} \) is:
\[ \frac{\alpha}{\beta} = \frac{7}{9} \]

5. **Finding the Ratio**:
- To express the ratio \( \alpha : \beta \):
\[ \alpha : \beta = 7 : 9 \]

### Final Answer:
The ratio of \( \alpha \) to \( \beta \) is \( 7 : 9 \).