If \(\vec a,\;\vec b\) and \(\vec c\) are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then which one of the following is correct?
Equilateral Triangle Position Vectors and Orthocentre
This problem involves the properties of an equilateral triangle, specifically the relationship between the position vectors of its vertices and the location of its orthocentre.
Let the equilateral triangle be denoted by ABC, with vertices having position vectors \(\vec a\), \(\vec b\), and \(\vec c\) respectively, relative to an origin O.
The question states that the orthocentre of the equilateral triangle is at the origin.
Properties of an Equilateral Triangle
A key property of an equilateral triangle is that its orthocentre, centroid, circumcentre, and incenter all coincide at the same point.
In this specific problem, since the orthocentre is at the origin, it means the centroid of the equilateral triangle is also at the origin.
Centroid of a Triangle
The position vector of the centroid (G) of a triangle with vertices A, B, and C having position vectors \(\vec a\), \(\vec b\), and \(\vec c\) is given by the formula:
\[ \text{Position vector of Centroid (G)} = \frac{\vec a + \vec b + \vec c}{3} \]
Since the centroid is located at the origin, its position vector is the zero vector, \(\vec 0\).
Solving for the Vector Sum
Equating the position vector of the centroid to the zero vector:
\[ \frac{\vec a + \vec b + \vec c}{3} = \vec 0 \]
To find the sum \(\vec a + \vec b + \vec c\), we can multiply both sides of the equation by 3:
\[ 3 \times \left( \frac{\vec a + \vec b + \vec c}{3} \right) = 3 \times \vec 0 \]
\[ \vec a + \vec b + \vec c = \vec 0 \]
Conclusion
Based on the properties of an equilateral triangle and the definition of a centroid, if the orthocentre (and thus the centroid) is at the origin, the sum of the position vectors of the vertices is the zero vector.
Let's check the given options:
- Option 1: \(\vec a + \vec b + \vec c = \vec 0\)
- Option 2: \(\vec a + \vec b + \vec c =\) unit vector
- Option 3: \(\vec a + \vec b = \vec c\)
- Option 4: \(\vec a = \vec b + \vec c\)
Our derived result is \(\vec a + \vec b + \vec c = \vec 0\), which matches Option 1.
Revision Table: Equilateral Triangle Centres
| Centre Type | Description | Location in Equilateral Triangle |
|---|---|---|
| Centroid | Intersection of medians | Coincides with Orthocentre, Circumcentre, Incenter |
| Orthocentre | Intersection of altitudes | Coincides with Centroid, Circumcentre, Incenter |
| Circumcentre | Intersection of perpendicular bisectors of sides | Coincides with Centroid, Orthocentre, Incenter |
| Incenter | Intersection of angle bisectors | Coincides with Centroid, Orthocentre, Circumcentre |
Additional Information: Vector Properties
Position vectors are used to denote the location of a point in space relative to a fixed origin. If a point P has position vector \(\vec p\), it means the vector from the origin O to P is \(\vec{OP} = \vec p\).
The sum of vectors follows the triangle or parallelogram law of addition. Vector subtraction is used to find displacement vectors between two points.
The zero vector, denoted by \(\vec 0\), is a vector with zero magnitude and no specific direction. Adding the zero vector to any vector does not change the vector.
Similar Questions
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