A, B, C and D, are four points in a plane with position vectors a, b, c and d respectively such that

Q: Why is $\vec{a} - \vec{d}$ equal to $\vec{DA}$? Shouldn't it be $\vec{AD}$?

That is a fantastic question! Remember the definition of a directed vector: the vector $\vec{PQ}$ is defined as the position vector of the terminal point minus the position vector of the initial point, which is $\vec{q} - \vec{p}$. So, $\vec{a} - \vec{d}$ is the vector starting at $D$ and ending at $A$, which is $\vec{DA}$. It is a common trap, but once you master this, you will never look at vector subtraction the same way again!

Q: Does this proof hold if the triangle is obtuse?

Yes, absolutely! The beauty of vector algebra is that it is coordinate-independent and geometry-agnostic. Whether the triangle is acute, right-angled, or obtuse, the definition of the orthocentre as the intersection of altitudes remains the same. The vector equations we derived are universal truths for any triangle in a plane.

Q: Why do we only need two altitudes to identify the orthocentre?

In geometry, the orthocentre is defined as the point of concurrency of all three altitudes. However, since two lines in a plane intersect at a unique point (unless they are parallel, which is not the case for the altitudes of a non-degenerate triangle), finding the intersection of two altitudes is sufficient to uniquely identify the point $D$. The third altitude will naturally pass through this same point $D$.

Question

$A, B, C$ and $D$, are four points in a plane with position vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ respectively such that $(\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) = (\vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a}) = 0$. The point $D$, then, is the ......... of the triangle $ABC$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving a problem; we are uncovering a hidden truth about the architecture of triangles.

We are going to look at a set of vector equations and, through the lens of pure logic, reveal the identity of a mysterious point $D$. Imagine you are standing on a flat plane, a blank canvas.

You have a triangle $ABC$ with vertices defined by position vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$. Somewhere on this plane, there is a point $D$ with position vector $\vec{d}$.

We are given two conditions:
$$(\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) = 0$$
$$(\vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a}) = 0$$

At first glance, these look like abstract algebraic expressions, but I want you to see them as geometric blueprints.

### The Vector Bridge: Decoding the Language

Before we touch the equations, let us master the language of vectors. In the world of JEE physics and mathematics, the most powerful tool you possess is the ability to translate between algebra and geometry.

If you have two points $P$ and $Q$ with position vectors $\vec{p}$ and $\vec{q}$, the directed line segment $\vec{PQ}$ is simply $\vec{q} - \vec{p}$. This is the bridge.

Whenever you see a difference of position vectors, you are looking at a directed line segment. Let us apply this to our first condition: $(\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) = 0$.

The term $(\vec{a} - \vec{d})$ is the vector $\vec{DA}$. The term $(\vec{b} - \vec{c})$ is the vector $\vec{CB}$.

Suddenly, the equation transforms into:
$$\vec{DA} \cdot \vec{CB} = 0$$

Do you see the elegance? We have moved from abstract symbols to a concrete geometric statement.

### The Power of the Dot Product

Now, what does it mean for the dot product of two vectors to be zero? In the realm of Euclidean geometry, the dot product $\vec{u} \cdot \vec{v}$ is defined as $|\vec{u}| |\vec{v}| \cos 	heta$, where $	heta$ is the angle between the vectors.

For this product to be zero, either the magnitude of one of the vectors must be zero (which would make the point trivial) or $\cos 	heta$ must be zero. If $\cos 	heta = 0$, then $	heta = 90^\circ$.

This means the vectors are perpendicular! So, our first condition, $\vec{DA} \cdot \vec{CB} = 0$, tells us that the line segment $AD$ is perpendicular to the side $BC$.

In the language of triangles, a line segment from a vertex perpendicular to the opposite side is called an altitude. Therefore, point $D$ must lie on the altitude dropped from vertex $A$ to the side $BC$.

### The Second Condition: Unveiling the Identity

We have found the first clue. Now, let us apply the same rigorous logic to the second condition: $(\vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a}) = 0$.

Using our vector bridge, $(\vec{b} - \vec{d})$ becomes $\vec{DB}$, and $(\vec{c} - \vec{a})$ becomes $\vec{AC}$. The equation becomes:
$$\vec{DB} \cdot \vec{AC} = 0$$

Just as before, this implies that $\vec{DB} \perp \vec{AC}$. This means the line segment $BD$ is perpendicular to the side $AC$.

Point $D$ must also lie on the altitude dropped from vertex $B$ to the side $AC$.

### The Synthesis: The Orthocentre

We have arrived at the heart of the matter. We know that point $D$ lies on the altitude from $A$ to $BC$, and it also lies on the altitude from $B$ to $AC$.

In any triangle, the point where the altitudes intersect is known as the **orthocentre**. By proving that $D$ lies on two of these altitudes, we have successfully identified $D$ as the **orthocentre of $\Delta ABC$**.

This is the beauty of vector geometry—it allows us to prove complex geometric properties using simple, elegant algebraic manipulations. You have taken a set of abstract vectors and reconstructed the very definition of the orthocentre.

Take a moment to appreciate this. You didn't just solve a problem; you navigated the logical structure of a triangle. Keep this clarity, keep this curiosity, and you will conquer any problem the JEE throws your way.

Animated Solution for Mathematics - Vector Algebra: $A, B, C$ and $D$ , are four points in a plane with position vectors $a, b, c$ and $d$ respectively such that $(a - d) \cdot (b - c) = (b - d) \cdot (c - a) = 0$ . The point $D$ , then, is the ......... of the triangle $A B C$ .

Visualized Solution

Visualizing the Points $A, B, C, D$

Position Vectors and Directed Line Segments

Analyzing the First Condition

Converting to Directed Vectors

Geometric Meaning of Zero Dot Product

Analyzing the Second Condition

Converting the Second Condition

Second Perpendicularity

Identifying Point $D$ as the Orthocentre

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Vector Bridge

The Power of the Dot Product

The Second Condition

The Synthesis

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