If the position vectors of the points A,B and C be `hati+hatj,hati-hatj` and `ahati+bhatj+chatk` respectively, then the points A,B and C are collinear, if

Options

A. a=b=c=1
B. a=1,b and c are arbitrary scalars
C. ab=c=0
D. c=0,a=1 and b is arbitrary scalars

Explanation

To determine the condition under which the points A, B, and C are collinear, we can use the concept of vectors and the position vectors provided.

**Given:**
- Position vector of point A: **OA** = \( \hat{i} + \hat{j} \)
- Position vector of point B: **OB** = \( \hat{i} - \hat{j} \)
- Position vector of point C: **OC** = \( a\hat{i} + b\hat{j} + c\hat{k} \)

**Step 1: Find the vectors AB and AC.**

The vector **AB** can be calculated as:
\[ \vec{AB} = \vec{OB} - \vec{OA} = (\hat{i} - \hat{j}) - (\hat{i} + \hat{j}) = \hat{i} - \hat{j} - \hat{i} - \hat{j} = -2\hat{j} \]

The vector **AC** can be calculated as:
\[ \vec{AC} = \vec{OC} - \vec{OA} = (a\hat{i} + b\hat{j} + c\hat{k}) - (\hat{i} + \hat{j}) = (a - 1)\hat{i} + (b - 1)\hat{j} + c\hat{k} \]

**Step 2: Establish the condition for collinearity.**

The points A, B, and C are collinear if the vectors **AB** and **AC** are parallel. This can be expressed as:
\[ \vec{AB} = k \cdot \vec{AC} \]
for some scalar \( k \).

Substituting the vectors we found:
\[ -2\hat{j} = k \cdot ((a - 1)\hat{i} + (b - 1)\hat{j} + c\hat{k}) \]

**Step 3: Equate the components.**

From the equation above, we can equate the components:
1. The coefficient of \( \hat{i} \): \( 0 = k(a - 1) \)
2. The coefficient of \( \hat{j} \): \( -2 = k(b - 1) \)
3. The coefficient of \( \hat{k} \): \( 0 = kc \)

**Step 4: Analyze the equations.**

From \( 0 = k(a - 1) \):
- If \( k \neq 0 \), then \( a - 1 = 0 \) which gives \( a = 1 \).
- If \( k = 0 \), then this condition does not provide information about \( a \).

From \( -2 = k(b - 1) \):
- If \( k \neq 0 \), we can solve for \( b \): \( b - 1 = -\frac{2}{k} \) which gives \( b = 1 - \frac{2}{k} \).

From \( 0 = kc \):
- If \( k \neq 0 \), then \( c = 0 \).
- If \( k = 0 \), this condition does not provide information about \( c \).

**Conclusion:**

The points A, B, and C are collinear if:
1. \( a = 1 \) (if \( k \neq 0 \))
2. \( c = 0 \) (if \( k \neq 0 \))
3. \( b \) can be any value depending on \( k \).

Thus, the condition for collinearity can be summarized as:
- \( a = 1 \)
- \( c = 0 \)
- \( b \) is arbitrary.