If two out to the three vectors , `veca, vecb , vecc` are unit vectors such that ` veca + vecb + vecc =0 and 2(veca.vecb + vecb .vecc + vecc.veca) +3=0` then the length of the third vector is

Options

A. 3
B. 2
C. 1
D. 0

Explanation

To solve the problem step by step, we will analyze the given conditions and apply vector algebra.

### Step 1: Understand the given conditions
We have three vectors \( \vec{a}, \vec{b}, \vec{c} \). It is given that two of these vectors are unit vectors, which means their magnitudes are 1. We can assume \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \). The third vector \( \vec{c} \) is what we need to find the length of.

### Step 2: Use the first condition
The first condition states:
\[ \vec{a} + \vec{b} + \vec{c} = 0 \]
From this, we can express \( \vec{c} \) in terms of \( \vec{a} \) and \( \vec{b} \):
\[ \vec{c} = -(\vec{a} + \vec{b}) \]

### Step 3: Square both sides
To find the magnitude of \( \vec{c} \), we square both sides:
\[ |\vec{c}|^2 = |-(\vec{a} + \vec{b})|^2 = |\vec{a} + \vec{b}|^2 \]

### Step 4: Expand the squared magnitude
Using the formula for the magnitude of a sum of vectors:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} \]
Since \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \):
\[ |\vec{c}|^2 = 1^2 + 1^2 + 2\vec{a} \cdot \vec{b} = 1 + 1 + 2\vec{a} \cdot \vec{b} = 2 + 2\vec{a} \cdot \vec{b} \]

### Step 5: Use the second condition
The second condition given is:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) + 3 = 0 \]
Rearranging gives:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3 \]
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2} \]

### Step 6: Substitute \( \vec{c} \)
We know \( \vec{c} = -(\vec{a} + \vec{b}) \). Therefore:
\[ \vec{b} \cdot \vec{c} = \vec{b} \cdot (-\vec{a} - \vec{b}) = -\vec{b} \cdot \vec{a} - |\vec{b}|^2 = -\vec{a} \cdot \vec{b} - 1 \]
\[ \vec{c} \cdot \vec{a} = (-\vec{a} - \vec{b}) \cdot \vec{a} = -|\vec{a}|^2 - \vec{b} \cdot \vec{a} = -1 - \vec{a} \cdot \vec{b} \]

### Step 7: Substitute back into the equation
Now substituting into the dot product equation:
\[ \vec{a} \cdot \vec{b} + (-\vec{a} \cdot \vec{b} - 1) + (-1 - \vec{a} \cdot \vec{b}) = -\frac{3}{2} \]
This simplifies to:
\[ \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{b} - 1 - 1 = -\frac{3}{2} \]
\[ -2 = -\frac{3}{2} \]
This indicates that:
\[ \vec{a} \cdot \vec{b} = -\frac{1}{2} \]

### Step 8: Substitute back to find \( |\vec{c}|^2 \)
Now substituting \( \vec{a} \cdot \vec{b} \) back into the equation for \( |\vec{c}|^2 \):
\[ |\vec{c}|^2 = 2 + 2(-\frac{1}{2}) = 2 - 1 = 1 \]
Thus, the magnitude of \( \vec{c} \) is:
\[ |\vec{c}| = \sqrt{1} = 1 \]

### Final Answer
The length of the third vector \( \vec{c} \) is \( 1 \).

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