Three points having position vectors `bara,barb and barc` will be collinear if

Options

A. `lamdabara+mubarb=(lamda+mu)barc`
B. `[bara barb barc]=0`
C. `bara times barb+barb times barc+barc times bara=0`
D. `bar a times barc=barb`

Answer

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

### Step 1: Understanding Collinearity
Three points are collinear if the vector from one point to another is a scalar multiple of the vector from the first point to the third point. This means that the vectors can be expressed in terms of each other.

### Step 2: Expressing the Position Vectors
Assume the points corresponding to the position vectors are \(A\), \(B\), and \(C\). The position vectors can be expressed as:
- \(\vec{A} = \vec{a}\)
- \(\vec{B} = \vec{b}\)
- \(\vec{C} = \vec{c}\)

### Step 3: Using the Concept of Ratios
If \(C\) divides the line segment \(AB\) in the ratio \( \mu : \lambda \), we can express \(\vec{c}\) as:
\[ \vec{c} = \frac{\mu \vec{b} + \lambda \vec{a}}{\mu + \lambda} \]
This indicates that \(\vec{c}\) is a linear combination of \(\vec{a}\) and \(\vec{b}\).

### Step 4: Rearranging the Equation
Multiplying through by \((\mu + \lambda)\) gives:
\[ (\mu + \lambda) \vec{c} = \mu \vec{b} + \lambda \vec{a} \]
This shows that the position vector \(\vec{c}\) can be expressed as a combination of \(\vec{a}\) and \(\vec{b}\), confirming that they are collinear.

### Step 5: Using the Cross Product
Another method to check for collinearity is to use the cross product. The vectors \(\vec{AB} = \vec{b} - \vec{a}\) and \(\vec{AC} = \vec{c} - \vec{a}\) are collinear if:
\[ (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \vec{0} \]
This means that the area of the triangle formed by the points \(A\), \(B\), and \(C\) is zero.

### Step 6: Conclusion
Thus, the three points are collinear if:
1. \(\vec{c} = \frac{\mu \vec{b} + \lambda \vec{a}}{\mu + \lambda}\) for some scalars \(\mu\) and \(\lambda\).
2. The cross product condition \((\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \vec{0}\) holds.