Let a, b, c be three vectors such that each of them are non-collinear, a+b and b+c are collinear with c and a respectively and a+b+c=k. Then (|k|, |k|) lies on

Options

A. `y^(2)=4ax`
B. `x^(2)+y^(2)-ax-by=0`
C. `x^(2)-y^(2)=1`
D. `|x|+|y|=1`

Answer

To solve the problem, we need to analyze the conditions given in the question step by step.

### Step-by-Step Solution:

1. **Understanding the Conditions**:
We have three vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) that are non-collinear. This means that no vector can be expressed as a linear combination of the others.

2. **Collinearity Conditions**:
- The vector \( \mathbf{a} + \mathbf{b} \) is collinear with \( \mathbf{c} \).
- The vector \( \mathbf{b} + \mathbf{c} \) is collinear with \( \mathbf{a} \).

We can express these conditions mathematically:
\[ \mathbf{a} + \mathbf{b} = \lambda \mathbf{c} \quad (1) \]
\[ \mathbf{b} + \mathbf{c} = \mu \mathbf{a} \quad (2) \]
where \( \lambda \) and \( \mu \) are scalars.

3. **Subtracting the Equations**:
We subtract equation (2) from equation (1):
\[ (\mathbf{a} + \mathbf{b}) - (\mathbf{b} + \mathbf{c}) = \lambda \mathbf{c} - \mu \mathbf{a} \]
Simplifying this gives:
\[ \mathbf{a} - \mathbf{c} = \lambda \mathbf{c} - \mu \mathbf{a} \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \mathbf{a} + \mu \mathbf{a} = \lambda \mathbf{c} + \mathbf{c} \]
Factoring out gives:
\[ (1 + \mu) \mathbf{a} = (1 + \lambda) \mathbf{c} \]

5. **Non-Collinearity Implication**:
Since \( \mathbf{a} \) and \( \mathbf{c} \) are non-collinear, the only way for the equation to hold is if both sides equal zero:
\[ 1 + \mu = 0 \quad \text{and} \quad 1 + \lambda = 0 \]
Thus, we find:
\[ \mu = -1 \quad \text{and} \quad \lambda = -1 \]

6. **Substituting Back**:
Substituting \( \lambda \) and \( \mu \) back into equations (1) and (2):
\[ \mathbf{a} + \mathbf{b} = -\mathbf{c} \quad \Rightarrow \quad \mathbf{a} + \mathbf{b} + \mathbf{c} = 0 \]
This implies:
\[ \mathbf{a} + \mathbf{b} + \mathbf{c} = k \quad \Rightarrow \quad k = 0 \]

7. **Magnitude of k**:
The magnitude of \( k \) is:
\[ |k| = 0 \]
Therefore, the point \( (|k|, |k|) \) is \( (0, 0) \).

8. **Conclusion**:
The point \( (0, 0) \) lies on the equations or curves given in the options.