There are two column vectors `X=((x),(1))` and `((1,4),(5,2))` X is parallel to X. If `theta` is the angle between them, the `tan theta ` is

Options

A. 3
B. 5
C. 7
D. 9

Answer

To solve the problem, we need to find the value of \( \tan \theta \) where \( \theta \) is the angle between the two column vectors \( X = \begin{pmatrix} x \\ 1 \end{pmatrix} \) and \( A = \begin{pmatrix} 1 & 4 \\ 5 & 2 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Understanding the Condition for Parallel Vectors**:
Two vectors \( A \) and \( B \) are parallel if there exists a scalar \( \lambda \) such that:
\[ A = \lambda B \]
In this case, we can express the vector \( A \) as a matrix multiplying the vector \( X \).

2. **Setting Up the Equation**:
We can express the relationship as:
\[ \begin{pmatrix} 1 & 4 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} x \\ 1 \end{pmatrix} = \lambda \begin{pmatrix} x \\ 1 \end{pmatrix} \]
This expands to:
\[ \begin{pmatrix} 1 \cdot x + 4 \cdot 1 \\ 5 \cdot x + 2 \cdot 1 \end{pmatrix} = \begin{pmatrix} \lambda x \\ \lambda \end{pmatrix} \]

3. **Equating Components**:
From the above equation, we can derive two equations:
\[ x + 4 = \lambda x \quad \text{(1)} \]
\[ 5x + 2 = \lambda \quad \text{(2)} \]

4. **Solving for \( \lambda \)**:
From equation (1):
\[ \lambda = \frac{x + 4}{x} \quad \text{(3)} \]
Substitute equation (3) into equation (2):
\[ 5x + 2 = \frac{x + 4}{x} \]
Multiply both sides by \( x \) to eliminate the fraction:
\[ 5x^2 + 2x = x + 4 \]
Rearranging gives:
\[ 5x^2 + 2x - x - 4 = 0 \implies 5x^2 + x - 4 = 0 \]

5. **Applying the Quadratic Formula**:
We can solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5, b = 1, c = -4 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 5 \cdot (-4)}}{2 \cdot 5} = \frac{-1 \pm \sqrt{1 + 80}}{10} = \frac{-1 \pm 9}{10} \]
This gives us two solutions:
\[ x = \frac{8}{10} = \frac{4}{5} \quad \text{and} \quad x = \frac{-10}{10} = -1 \]

6. **Finding the Values of \( \tan \theta \)**:
We have two values for \( x \): \( x = \frac{4}{5} \) and \( x = -1 \). We will use these to find \( \tan \theta \) using the formula:
\[ \tan \theta = \frac{a_1 b_2 - b_1 a_2}{a_1 a_2 + b_1 b_2} \]
Here, \( a_1 = x \), \( b_1 = 1 \), \( a_2 = 4 \), \( b_2 = 5 \).

For \( x = \frac{4}{5} \):
\[ \tan \theta = \frac{\left(\frac{4}{5}\right) \cdot 5 - 1 \cdot 4}{\left(\frac{4}{5}\right) \cdot 4 + 1 \cdot 5} \]
Simplifying:
\[ = \frac{4 - 4}{\frac{16}{5} + 5} = \frac{0}{\frac{16 + 25}{5}} = 0 \]

For \( x = -1 \):
\[ \tan \theta = \frac{-1 \cdot 5 - 1 \cdot 4}{-1 \cdot 4 + 1 \cdot 5} = \frac{-5 - 4}{-4 + 5} = \frac{-9}{1} = -9 \]

### Final Result:
Thus, the value of \( \tan \theta \) can be either \( 0 \) or \( -9 \) depending on the value of \( x \).