If `veca,vecb,vecc` are three vectors such that `|veca|=2,|vecb|=4,|vecc|=4,vecb.vecc=0,vecb.veca=vecc.veca`, then find the value of `|veca+veca-vecc|`

Options

A. 6
B. `sqrt(6)`
C. `7`
D. `2sqrt(6)`

Answer

To solve the problem, we need to find the value of \( |\vec{a} + \vec{b} - \vec{c}| \) given the conditions on the vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \).

### Step-by-Step Solution:

1. **Given Information:**
- \( |\vec{a}| = 2 \)
- \( |\vec{b}| = 4 \)
- \( |\vec{c}| = 4 \)
- \( \vec{b} \cdot \vec{c} = 0 \) (indicating that \( \vec{b} \) and \( \vec{c} \) are orthogonal)
- \( \vec{b} \cdot \vec{a} = \vec{c} \cdot \vec{a} \)

2. **Finding the Magnitude of \( |\vec{a} + \vec{b} - \vec{c}| \):**
We can express the magnitude squared:
\[ |\vec{a} + \vec{b} - \vec{c}|^2 = (\vec{a} + \vec{b} - \vec{c}) \cdot (\vec{a} + \vec{b} - \vec{c}) \]

3. **Expanding the Dot Product:**
\[ |\vec{a} + \vec{b} - \vec{c}|^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b}) - 2(\vec{a} \cdot \vec{c}) - 2(\vec{b} \cdot \vec{c}) \]

4. **Substituting Known Values:**
- \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 2^2 = 4 \)
- \( \vec{b} \cdot \vec{b} = |\vec{b}|^2 = 4^2 = 16 \)
- \( \vec{c} \cdot \vec{c} = |\vec{c}|^2 = 4^2 = 16 \)
- \( \vec{b} \cdot \vec{c} = 0 \) (since they are orthogonal)
- From the condition \( \vec{b} \cdot \vec{a} = \vec{c} \cdot \vec{a} \), let \( \vec{b} \cdot \vec{a} = k \). Then \( \vec{c} \cdot \vec{a} = k \) as well.

5. **Putting Everything Together:**
\[ |\vec{a} + \vec{b} - \vec{c}|^2 = 4 + 16 + 16 + 2k - 2k - 0 \]
\[ = 4 + 16 + 16 = 36 \]

6. **Taking the Square Root:**
\[ |\vec{a} + \vec{b} - \vec{c}| = \sqrt{36} = 6 \]

### Final Answer:
The value of \( |\vec{a} + \vec{b} - \vec{c}| \) is \( 6 \).