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. (a)1
B. (b)`(1)/(8)`
C. (c)`(-16)/(11)`
D. (d)`(-11)/(16)`

Explanation

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

### Step-by-step Solution:

1. **Set up the linear dependence condition**:
The vector \( \vec{a} \) is linearly dependent on the vectors \( \vec{v_1} = \hat{i} + 2\hat{j} \) and \( \vec{v_2} = \hat{j} - 2\hat{k} \). This means there exist scalars \( m \) and \( l \) such that:
\[ a\hat{i} + b\hat{j} + \frac{1}{2}\hat{k} = m(\hat{i} + 2\hat{j}) + l(\hat{j} - 2\hat{k}) \]

2. **Expand the right-hand side**:
\[ m(\hat{i} + 2\hat{j}) + l(\hat{j} - 2\hat{k}) = m\hat{i} + (2m + l)\hat{j} - 2l\hat{k} \]

3. **Equate coefficients**:
From the equation \( a\hat{i} + b\hat{j} + \frac{1}{2}\hat{k} = m\hat{i} + (2m + l)\hat{j} - 2l\hat{k} \), we can equate coefficients:
- For \( \hat{i} \): \( a = m \)
- For \( \hat{j} \): \( b = 2m + l \)
- For \( \hat{k} \): \( \frac{1}{2} = -2l \)

4. **Solve for \( l \)**:
From \( \frac{1}{2} = -2l \), we find:
\[ l = -\frac{1}{4} \]

5. **Substitute \( l \) into the equation for \( b \)**:
Substitute \( l \) into \( b = 2m + l \):
\[ b = 2m - \frac{1}{4} \]

6. **Substitute \( m \) into the equation for \( a \)**:
Since \( a = m \), we can express \( b \) in terms of \( a \):
\[ b = 2a - \frac{1}{4} \]

7. **Use the unit vector condition**:
The vector \( \vec{a} \) is a unit vector, so:
\[ a^2 + b^2 + \left(\frac{1}{2}\right)^2 = 1 \]
Substitute \( b \):
\[ a^2 + \left(2a - \frac{1}{4}\right)^2 + \frac{1}{4} = 1 \]

8. **Expand and simplify**:
\[ a^2 + (4a^2 - a + \frac{1}{16}) + \frac{1}{4} = 1 \]
Combine terms:
\[ 5a^2 - a + \frac{1}{16} + \frac{4}{16} = 1 \]
\[ 5a^2 - a + \frac{5}{16} = 1 \]
Multiply through by 16 to eliminate the fraction:
\[ 80a^2 - 16a + 5 = 16 \]
\[ 80a^2 - 16a - 11 = 0 \]

9. **Find the roots using the quadratic formula**:
The roots \( a_1 \) and \( a_2 \) are given by:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 80 \cdot (-11)}}{2 \cdot 80} \]
Calculate the discriminant:
\[ 256 + 3520 = 3776 \]
Thus,
\[ a = \frac{16 \pm \sqrt{3776}}{160} \]

10. **Calculate \( \frac{1}{a_1} + \frac{1}{a_2} \)**:
Using the property of roots:
\[ \frac{1}{a_1} + \frac{1}{a_2} = \frac{a_1 + a_2}{a_1 a_2} \]
From Vieta's formulas:
\[ a_1 + a_2 = \frac{16}{80} = \frac{1}{5}, \quad a_1 a_2 = \frac{-11}{80} \]
Therefore,
\[ \frac{1}{a_1} + \frac{1}{a_2} = \frac{\frac{1}{5}}{\frac{-11}{80}} = \frac{80}{5 \cdot -11} = -\frac{16}{11} \]

### Final Answer:
\[ \frac{1}{a_1} + \frac{1}{a_2} = -\frac{16}{11} \]