If p,q are two-collinear vectors such that `(b-c)ptimesq+(c-a)p+(a-b)q=0` where a, b, c are lengths of sides of a triangle, then the triangle is

Options

A. right angled
B. obtuse
C. equilateral
D. right angled isosceles triangle

Answer

To solve the problem, we start with the equation given:

\[ (b - c) \mathbf{p} \times \mathbf{q} + (c - a) \mathbf{p} + (a - b) \mathbf{q} = 0 \]

where \( a, b, c \) are the lengths of the sides of a triangle, and \( \mathbf{p} \) and \( \mathbf{q} \) are two collinear vectors.

### Step 1: Analyze the Cross Product
Since \( \mathbf{p} \) and \( \mathbf{q} \) are collinear, the cross product \( \mathbf{p} \times \mathbf{q} \) is equal to the zero vector. Therefore, we can simplify the equation:

\[ (b - c) \mathbf{0} + (c - a) \mathbf{p} + (a - b) \mathbf{q} = 0 \]

This simplifies to:

\[ (c - a) \mathbf{p} + (a - b) \mathbf{q} = 0 \]

### Step 2: Rearranging the Equation
We can rearrange the equation to isolate one vector:

\[ (c - a) \mathbf{p} = -(a - b) \mathbf{q} \]

### Step 3: Analyzing the Coefficients
Since \( \mathbf{p} \) and \( \mathbf{q} \) are non-zero vectors, for the equation to hold, the coefficients must also satisfy certain conditions. This implies that:

\[ \frac{c - a}{a - b} = -\frac{\|\mathbf{q}\|}{\|\mathbf{p}\|} \quad \text{(1)} \]

### Step 4: Setting Conditions
From the equation (1), we can derive three cases based on the equality of the sides of the triangle:

1. If \( c - a = 0 \) then \( c = a \).
2. If \( a - b = 0 \) then \( a = b \).
3. If \( b - c = 0 \) then \( b = c \).

### Step 5: Conclusion
If all three conditions hold true, we have:

\[ a = b = c \]

This means that all sides of the triangle are equal, which defines an equilateral triangle.

### Final Answer
Thus, the triangle is **equilateral**.

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