If `a_(1)` and `a_(2)` are two values of a for which the unit vector `aveci + bvecj +1/2veck` is linearly dependent with `veci+2vecj` and `vecj-2veck`, then `1/a_(1)+1/a_(2)` is equal to

Options

A. 1
B. `1/8`
C. `-16/11`
D. `-11/16`

Explanation

To solve the problem, we need to determine the values of \( a_1 \) and \( a_2 \) such that the vector \( \vec{a} = a \hat{i} + b \hat{j} + \frac{1}{2} \hat{k} \) is linearly dependent on the vectors \( \vec{v_1} = \hat{i} + 2\hat{j} \) and \( \vec{v_2} = \hat{j} - 2\hat{k} \).

### Step 1: Set Up the Linear Dependence Condition
The vectors \( \vec{a} \), \( \vec{v_1} \), and \( \vec{v_2} \) are linearly dependent if there exist scalars \( \lambda_1 \) and \( \lambda_2 \) such that:
\[ \vec{a} = \lambda_1 \vec{v_1} + \lambda_2 \vec{v_2} \]

### Step 2: Write the Vectors in Component Form
We can express the vectors as follows:
- \( \vec{a} = a \hat{i} + b \hat{j} + \frac{1}{2} \hat{k} \)
- \( \vec{v_1} = \hat{i} + 2\hat{j} + 0\hat{k} \)
- \( \vec{v_2} = 0\hat{i} + 1\hat{j} - 2\hat{k} \)

### Step 3: Equate Components
From the linear combination, we have:
\[ a \hat{i} + b \hat{j} + \frac{1}{2} \hat{k} = \lambda_1 (\hat{i} + 2\hat{j}) + \lambda_2 (0\hat{i} + \hat{j} - 2\hat{k}) \]
This leads to the following equations by equating coefficients:
1. For \( \hat{i} \): \( a = \lambda_1 \)
2. For \( \hat{j} \): \( b = 2\lambda_1 + \lambda_2 \)
3. For \( \hat{k} \): \( \frac{1}{2} = -2\lambda_2 \)

### Step 4: Solve for \( \lambda_2 \)
From the third equation:
\[ \lambda_2 = -\frac{1}{4} \]

### Step 5: Substitute \( \lambda_2 \) into the Second Equation
Substituting \( \lambda_2 \) into the second equation:
\[ b = 2\lambda_1 - \frac{1}{4} \]

### Step 6: Express \( b \) in Terms of \( a \)
Since \( a = \lambda_1 \), we can write:
\[ b = 2a - \frac{1}{4} \]

### Step 7: Use the Unit Vector Condition
The vector \( \vec{a} \) is a unit vector, which means:
\[ \sqrt{a^2 + b^2 + \left(\frac{1}{2}\right)^2} = 1 \]
Squaring both sides gives:
\[ a^2 + b^2 + \frac{1}{4} = 1 \]
Thus,
\[ a^2 + b^2 = \frac{3}{4} \]

### Step 8: Substitute for \( b \)
Substituting \( b = 2a - \frac{1}{4} \) into the equation:
\[ a^2 + \left(2a - \frac{1}{4}\right)^2 = \frac{3}{4} \]
Expanding the equation:
\[ a^2 + (4a^2 - a + \frac{1}{16}) = \frac{3}{4} \]
Combining terms:
\[ 5a^2 - a + \frac{1}{16} - \frac{12}{16} = 0 \]
This simplifies to:
\[ 5a^2 - a - \frac{11}{16} = 0 \]

### Step 9: Apply the Quadratic Formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( A = 5, B = -1, C = -\frac{11}{16} \):
\[ a = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 5 \cdot \left(-\frac{11}{16}\right)}}{2 \cdot 5} \]
Calculating the discriminant:
\[ 1 + \frac{220}{16} = \frac{236}{16} = \frac{59}{4} \]
Thus,
\[ a = \frac{1 \pm \frac{\sqrt{59}}{2}}{10} = \frac{1 \pm \sqrt{59}}{20} \]

### Step 10: Find \( \frac{1}{a_1} + \frac{1}{a_2} \)
Let \( a_1 = \frac{1 + \sqrt{59}}{20} \) and \( a_2 = \frac{1 - \sqrt{59}}{20} \):
\[ \frac{1}{a_1} + \frac{1}{a_2} = \frac{a_1 + a_2}{a_1 a_2} \]
Calculating \( a_1 + a_2 \) and \( a_1 a_2 \):
\[ a_1 + a_2 = \frac{1 + \sqrt{59} + 1 - \sqrt{59}}{20} = \frac{2}{20} = \frac{1}{10} \]
\[ a_1 a_2 = \frac{(1 + \sqrt{59})(1 - \sqrt{59})}{400} = \frac{1 - 59}{400} = -\frac{58}{400} = -\frac{29}{200} \]
Thus,
\[ \frac{1}{a_1} + \frac{1}{a_2} = \frac{\frac{1}{10}}{-\frac{29}{200}} = -\frac{20}{29} \]

### Final Answer
\[ \frac{1}{a_1} + \frac{1}{a_2} = -\frac{20}{29} \]