If a, b and c are unit vectors such that `a+2b+2c=0," then "abs(a times c)` is equal to

Options

A. `1/4`
B. `sqrt(15)/16`
C. `15/16`
D. `sqrt(15)/4`

Answer

To solve the problem, we need to find the magnitude of the cross product \( | \mathbf{a} \times \mathbf{c} | \) given that \( \mathbf{a} + 2\mathbf{b} + 2\mathbf{c} = 0 \) and that \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are unit vectors.

### Step-by-step Solution:

1. **Rearranging the Equation**:
From the given equation \( \mathbf{a} + 2\mathbf{b} + 2\mathbf{c} = 0 \), we can rearrange it to express \( \mathbf{a} \):
\[ \mathbf{a} = -2\mathbf{b} - 2\mathbf{c} \]

**Hint**: Rearranging the equation helps isolate one vector in terms of the others.

2. **Taking Magnitudes**:
Since \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are unit vectors, we know \( |\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = 1 \). Taking the magnitude of both sides:
\[ |\mathbf{a}|^2 = |-2\mathbf{b} - 2\mathbf{c}|^2 \]
This simplifies to:
\[ 1 = |2\mathbf{b} + 2\mathbf{c}|^2 \]

**Hint**: Use the property of magnitudes to simplify the expression.

3. **Expanding the Magnitude**:
Now, we expand the right-hand side:
\[ |2\mathbf{b} + 2\mathbf{c}|^2 = 4|\mathbf{b} + \mathbf{c}|^2 = 4(|\mathbf{b}|^2 + |\mathbf{c}|^2 + 2\mathbf{b} \cdot \mathbf{c}) \]
Since \( |\mathbf{b}|^2 = |\mathbf{c}|^2 = 1 \):
\[ 1 = 4(1 + 1 + 2\mathbf{b} \cdot \mathbf{c}) = 4(2 + 2\mathbf{b} \cdot \mathbf{c}) \]

**Hint**: Remember to apply the dot product properties correctly.

4. **Solving for \( \mathbf{b} \cdot \mathbf{c} \)**:
We simplify the equation:
\[ 1 = 8 + 8\mathbf{b} \cdot \mathbf{c} \]
Rearranging gives:
\[ 8\mathbf{b} \cdot \mathbf{c} = 1 - 8 \implies \mathbf{b} \cdot \mathbf{c} = -\frac{7}{8} \]

**Hint**: Isolate the dot product to find its value.

5. **Using the Cross Product Formula**:
We know the relationship between the dot product and the cross product:
\[ |\mathbf{a} \times \mathbf{c}|^2 = |\mathbf{a}|^2 |\mathbf{c}|^2 - (\mathbf{a} \cdot \mathbf{c})^2 \]
Substituting \( |\mathbf{a}|^2 = 1 \) and \( |\mathbf{c}|^2 = 1 \):
\[ |\mathbf{a} \times \mathbf{c}|^2 = 1 - (\mathbf{a} \cdot \mathbf{c})^2 \]

**Hint**: Use the identity involving dot and cross products.

6. **Finding \( \mathbf{a} \cdot \mathbf{c} \)**:
We can find \( \mathbf{a} \cdot \mathbf{c} \) using the rearranged equation:
\[ \mathbf{a} \cdot \mathbf{c} = \mathbf{a} \cdot (-\frac{1}{4}\mathbf{b}) = -\frac{1}{4}(\mathbf{b} \cdot \mathbf{c}) = -\frac{1}{4} \left(-\frac{7}{8}\right) = \frac{7}{32} \]

**Hint**: Substitute known values to find the dot product.

7. **Calculating the Magnitude**:
Now substituting back:
\[ |\mathbf{a} \times \mathbf{c}|^2 = 1 - \left(\frac{7}{32}\right)^2 \]
\[ |\mathbf{a} \times \mathbf{c}|^2 = 1 - \frac{49}{1024} = \frac{1024 - 49}{1024} = \frac{975}{1024} \]
Taking the square root:
\[ |\mathbf{a} \times \mathbf{c}| = \sqrt{\frac{975}{1024}} = \frac{\sqrt{975}}{32} \]

**Hint**: Simplify the final expression carefully.

### Final Answer:
Thus, the magnitude \( |\mathbf{a} \times \mathbf{c}| \) is \( \frac{\sqrt{975}}{32} \).