Q.If $\vec{a}$, $\vec{b}$, $\vec{c}$ determine the vertices of a triangle, show that $\dfrac{1}{2}\left[\vec{b}\times\vec{c}+\vec{c}\times\vec{a}+\vec{a}\times\vec{b}\right]$ gives the vector area of the triangle. Hence deduce the condition that the three points $\vec{a}$, $\vec{b}$, $\vec{c}$ are collinear. Also find the unit vector normal to the plane of the triangle.

Concept: Collinearity Condition — The vector area of a triangle is half the magnitude of the cross product of two side vectors; if the area is zero, the points are collinear.

Step 1: Express sides as vectors

Let the position vectors of vertices be $\vec{a}, \vec{b}, \vec{c}$. Two side vectors from $\vec{a}$ are $\vec{b} - \vec{a}$ and $\vec{c} - \vec{a}$.

Step 2: Vector area of triangle

The vector area is

Step 3: Expand the cross product

Since $\vec{a} \times \vec{a} = 0$ and $-\vec{b} \times \vec{a} = \vec{a} \times \vec{b}$, $-\vec{a} \times \vec{c} = \vec{c} \times \vec{a}$, we get

Thus

Step 4: Collinearity and unit normal

If $\vec{a}, \vec{b}, \vec{c}$ are collinear, the triangle area is zero, so

The unit vector normal to the plane is …

Why this works — the collinearity condition

The area of a triangle is fundamentally a geometric quantity, but in vector form it becomes elegant. If you have two sides of a triangle as vectors, say $\vec{AB} = \vec{b} - \vec{a}$ and $\vec{AC} = \vec{c} - \vec{a}$, then the magnitude of their cross product gives twice the area. The direction of that cross product is perpendicular to the plane of the triangle — that's the vector area.

The expression given in the problem is a clever symmetric form of that same idea. Instead of picking one vertex as the "origin" and subtracting, it cycles through all three vertices. This symmetry is what makes it powerful: if the three points are collinear, the triangle collapses to a line, its area becomes zero, and the entire vector sum must vanish.

Step-by-step derivation

1. Start with the standard vector area formula

For a triangle with vertices at position vectors $\vec{a}, \vec{b}, \vec{c}$, take $\vec{a}$ as the reference point. The two edge vectors from $\vec{a}$ are:

The vector area (a vector whose magnitude equals the area and whose direction is normal to the plane) is:

2. Expand the cross product

Using the distributive property of the cross product:

Since $\vec{a}\times\vec{a} = \vec{0}$ and $\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}$, $\vec{a}\times\vec{c} = -\vec{c}\times\vec{a}$, we get:

3. Therefore the vector area is:

This is exactly the expression we needed to show.

4. Condition for collinearity

Three points are collinear if and only if the triangle they form has zero area. Since the vector area $\vec{A}$ has magnitude equal to the area, collinearity means:

That is: …

Method: Vector Area of a Triangle from Position Vectors

Use this whenever a triangle is given by the position vectors of its three vertices and you must find its (vector) area, test the points for collinearity, or find the normal to the plane.

Steps

Step 1: Build two edge vectors from a common vertex

Pick one vertex as the base and subtract to get two sides, e.g. $\vec{b}-\vec{a}$ and $\vec{c}-\vec{a}$. Everything rests on the fact that the cross product of two edges encodes both the area (its magnitude) and the plane's orientation (its direction).

Step 2: Write the vector area as half a cross product

Step 3: Expand to the symmetric (cyclic) form …

Common Mistakes

Mistake 1: Thinking collinearity means $\vec{b}\times\vec{c}=\vec{0}$ alone

Why it's wrong: that only says $\vec{b}$ and $\vec{c}$ are parallel as position vectors from the origin, not that the three points lie on one line. Correct approach: require the full cyclic sum $\vec{b}\times\vec{c}+\vec{c}\times\vec{a}+\vec{a}\times\vec{b}=\vec{0}$ (equivalently, the triangle's area is zero).

Mistake 2: Sign errors when expanding $(\vec{b}-\vec{a})\times(\vec{c}-\vec{a})$ …