Question

If origin is the ortho-center of an equilateral triangle whose vertices are \[ \vec{a},\vec{b},\vec{c}, \] then

• A. \(\vec{a}+\vec{b}=\vec{c}\)
• B. \(\vec{a}+\vec{b}=-\vec{c}\)
• C. \(|\vec{a}|^2=|\vec{b}|^2=|\vec{c}|^2\)
• D. \(\vec{a}-\vec{b}=\vec{c}\)

Hint:

In an equilateral triangle, \[ \text{orthocenter}=\text{centroid}=\text{circumcenter} \] Hence, if the common center is the origin, then \[ \vec{a}+\vec{b}+\vec{c}=\vec{0}. \]

Answer 1

The Correct Option is B

Step 1: Use the property of an equilateral triangle.
In an equilateral triangle, the orthocenter, centroid and circumcenter all coincide.
Given that the origin is the orthocenter, it is also the centroid.

Step 2: Apply the centroid condition.
If the position vectors of the vertices are \[ \vec{a},\vec{b},\vec{c}, \] then the centroid is \[ \frac{\vec{a}+\vec{b}+\vec{c}}{3} \] Since the centroid is at the origin, \[ \frac{\vec{a}+\vec{b}+\vec{c}}{3}=\vec{0} \] Multiplying by \(3\), \[ \vec{a}+\vec{b}+\vec{c}=\vec{0} \] Therefore, \[ \vec{a}+\vec{b}=-\vec{c} \]

Step 3: Final conclusion.
Hence, \[ \boxed{\vec{a}+\vec{b}=-\vec{c}} \]