Three points with position vectors `vec(a), vec(b), vec(c ) ` will be collinear if there exist scalars x, y, z such that

Options

A. `x vec(a) + y vec(b)=z vec(c )`
B. `x vec(a) + y vec(b)+z vec(c )=0`
C. `x vec(a) + y vec(b)+z vec(c )=0, " where " x+y+z=0`
D. `x vec(a) + y vec(b)=vec(c ).`

Explanation

To determine the condition under which three points with position vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are collinear, we can follow these steps:

### Step 1: Define the vectors
Let the position vectors of points A, B, and C be represented as:
- \(\vec{A} = \vec{a}\)
- \(\vec{B} = \vec{b}\)
- \(\vec{C} = \vec{c}\)

### Step 2: Use the collinearity condition
Three points A, B, and C are collinear if the vector \(\vec{AB}\) is a scalar multiple of the vector \(\vec{AC}\). This can be expressed mathematically as:
\[ \vec{AB} = \lambda \vec{AC} \]
for some scalar \(\lambda\).

### Step 3: Express the vectors
The vectors \(\vec{AB}\) and \(\vec{AC}\) can be expressed as:
\[ \vec{AB} = \vec{B} - \vec{A} = \vec{b} - \vec{a} \]
\[ \vec{AC} = \vec{C} - \vec{A} = \vec{c} - \vec{a} \]

### Step 4: Substitute into the collinearity condition
Substituting these expressions into the collinearity condition gives:
\[ \vec{b} - \vec{a} = \lambda (\vec{c} - \vec{a}) \]

### Step 5: Rearranging the equation
Rearranging the equation, we have:
\[ \vec{b} - \vec{a} = \lambda \vec{c} - \lambda \vec{a} \]
This can be rewritten as:
\[ \vec{b} + (-\lambda + 1) \vec{a} - \lambda \vec{c} = \vec{0} \]

### Step 6: Group the terms
Grouping the terms, we can express this as:
\[ (1 - \lambda) \vec{a} + \vec{b} - \lambda \vec{c} = \vec{0} \]

### Step 7: Identify coefficients
From the above equation, we can identify coefficients for \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\):
- Coefficient of \(\vec{a}\) is \(1 - \lambda\)
- Coefficient of \(\vec{b}\) is \(1\)
- Coefficient of \(\vec{c}\) is \(-\lambda\)

### Step 8: Set up the condition for collinearity
For the vectors to be collinear, the sum of the coefficients must equal zero:
\[ (1 - \lambda) + 1 - \lambda = 0 \]
This simplifies to:
\[ 2 - 2\lambda = 0 \]
Thus, we have:
\[ \lambda = 1 \]

### Step 9: Final condition
Substituting \(\lambda = 1\) back into the coefficients gives:
\[ X = 1 - \lambda = 0, \quad Y = 1, \quad Z = -\lambda = -1 \]
This leads to the final condition for collinearity:
\[ X + Y + Z = 0 \]

### Conclusion
Thus, the three points with position vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are collinear if there exist scalars \(x\), \(y\), and \(z\) such that:
\[ x + y + z = 0 \]