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Let a,b,c denote the lengths of the side...

Let a,b,c denote the lengths of the sides of a triangle such that
` ( a-b) vecu + ( b-c) vecv + ( c-a) (vecu xx vecv) = vec0`
For any two non-collinear vectors ` vecu and vecu`,then the triangle is

A

right angled

B

equilateral

C

isoscels

D

obtuse angled

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the given vector equation and its implications on the triangle formed by the sides \( a, b, c \). ### Step-by-step Solution: 1. **Understand the Given Equation**: The equation given is: \[ (a - b) \vec{u} + (b - c) \vec{v} + (c - a)(\vec{u} \times \vec{v}) = \vec{0} \] This equation involves three terms, each multiplied by a vector. 2. **Recognize the Implications of Non-Collinearity**: Since \( \vec{u} \) and \( \vec{v} \) are non-collinear vectors, it implies that \( \vec{u} \), \( \vec{v} \), and \( \vec{u} \times \vec{v} \) are non-coplanar. This means they do not lie in the same plane. 3. **Analyze the Vector Equation**: For the sum of these three vectors to equal the zero vector, and since they are non-coplanar, each coefficient of the vectors must individually equal zero. Thus, we can set up the following equations: \[ a - b = 0 \] \[ b - c = 0 \] \[ c - a = 0 \] 4. **Solve the Equations**: From \( a - b = 0 \), we get: \[ a = b \] From \( b - c = 0 \), we get: \[ b = c \] From \( c - a = 0 \), we get: \[ c = a \] 5. **Conclude the Relationship Between Sides**: Since \( a = b \), \( b = c \), and \( c = a \), we conclude that: \[ a = b = c \] This means all sides of the triangle are equal. 6. **Identify the Type of Triangle**: A triangle with all sides equal is known as an equilateral triangle. ### Final Conclusion: The triangle is equilateral.

To solve the problem, we need to analyze the given vector equation and its implications on the triangle formed by the sides \( a, b, c \). ### Step-by-step Solution: 1. **Understand the Given Equation**: The equation given is: \[ (a - b) \vec{u} + (b - c) \vec{v} + (c - a)(\vec{u} \times \vec{v}) = \vec{0} ...
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