Let O be the origin and let PQR be an arbitrary triangle. The point S is such that OP⋅OQ+OR⋅OS = OR⋅

Q: Wait, why do we assume the origin O is arbitrary? Does the choice of origin change the result?

That is a brilliant question! In vector geometry, the beauty lies in the fact that the result is independent of the origin. Whether you place O at a vertex, the centroid, or somewhere in deep space, the relative positions of P, Q, R, and S remain invariant. The vector algebra naturally accounts for this, so the result holds true regardless of where you place your origin.

Q: How do I know which terms to group together? Is there a trick?

The trick is to look for commonality. When you see an expression like $\vec{OP} \cdot \vec{OQ} - \vec{OR} \cdot \vec{OP}$, notice that $\vec{OP}$ is a common factor. By grouping terms that share a common vector, you create the opportunity to factor, which is the gateway to revealing the geometric relationship hidden in the algebra.

Q: Does this proof work for any triangle, or only specific ones like equilateral triangles?

It works for any triangle! The beauty of vector algebra is that it generalizes geometric properties. We did not assume any specific angles or side lengths for $\triangle PQR$. Because the derivation holds for any arbitrary vectors $\vec{OP}, \vec{OQ},$ and $\vec{OR}$, the conclusion that S is the orthocenter is universally true for all triangles.

Question

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that $\vec{OP}\cdot\vec{OQ}+\vec{OR}\cdot\vec{OS} = \vec{OR}\cdot\vec{OP}+\vec{OQ}\cdot\vec{OS} = \vec{OQ}\cdot\vec{OR}+\vec{OP}\cdot\vec{OS}$ Then the triangle PQR has S as its

Options:

(A) centroid

(B) circumcentre

(C) incentre

(D) orthocenter

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D space with a triangle $PQR$ floating in front of you. We place an arbitrary origin $O$ somewhere in this space, defining the position vectors $\vec{OP}$, $\vec{OQ}$, $\vec{OR}$, and $\vec{OS}$.

These vectors serve as our compass, defining the location of every point relative to our origin. We are given the following symmetric relation:

$$\vec{OP} \cdot \vec{OQ} + \vec{OR} \cdot \vec{OS} = \vec{OR} \cdot \vec{OP} + \vec{OQ} \cdot \vec{OS}$$

At first glance, this appears to be a jumble of terms. However, this symmetry is the signature of a problem designed to be factored.

### The Algebraic Dance

To bring order to this chaos, we group the terms by moving those involving $\vec{OP}$ to one side and those involving $\vec{OS}$ to the other. By rearranging the equation, we obtain:

$$\vec{OP} \cdot \vec{OQ} - \vec{OR} \cdot \vec{OP} = \vec{OQ} \cdot \vec{OS} - \vec{OR} \cdot \vec{OS}$$

Notice that $\vec{OP}$ is common on the left side, while $\vec{OS}$ is common on the right. Factoring these out, we get:

$$\vec{OP} \cdot (\vec{OQ} - \vec{OR}) = \vec{OS} \cdot (\vec{OQ} - \vec{OR})$$

Bringing all terms to one side leads us to a beautiful, elegant expression:

$$(\vec{OP} - \vec{OS}) \cdot (\vec{OQ} - \vec{OR}) = 0$$

### The Geometric Revelation

This is the moment of truth. Recalling the triangle law of vector addition, we know that $\vec{OP} - \vec{OS}$ is simply the vector $\vec{SP}$, and $\vec{OQ} - \vec{OR}$ is the vector $\vec{RQ}$.

Consequently, our equation simplifies to:

$$\vec{SP} \cdot \vec{RQ} = 0$$

In vector algebra, when the dot product of two non-zero vectors is zero, the vectors are perpendicular. This proves that the line segment $SP$ is perpendicular to the side $QR$.

We have demonstrated that $S$ lies on the altitude from $P$ to $QR$. By repeating this process for the remaining components of the triangle, we find that $\vec{SR} \cdot \vec{QP} = 0$, confirming that $SR$ is perpendicular to $PQ$.

### The Identity of S

We have established that $S$ lies on the altitude from $P$ and the altitude from $R$. By definition, the intersection of the altitudes of a triangle is the **Orthocenter**.

You have successfully navigated the algebra to reveal a fundamental geometric truth. Never fear the complexity of an equation; look for the structure, trust the vector algebra, and the geometry will reveal itself.

Animated Solution for Mathematics - Vector Algebra: Let O be the origin and let PQR be an arbitrary triangle. The point S is such that $O P \cdot O Q + O R \cdot O S = O R \cdot O P + O Q \cdot O S = O Q \cdot O R + O P \cdot O S$ Then the triangle PQR has S as its

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Visualized Solution

Triangle $P QR$ and Origin $O$

Position Vectors

The Given Vector Relation

Rearranging Terms

Factoring Out Common Vectors

Forming a Single Equation

Vector Differences as Directed Line Segments

First Perpendicularity Condition

Analyzing the Second Part of the Equation

Rearranging and Factoring Again

Second Set of Directed Segments

Second Perpendicularity Condition

Conclusion: The Orthocenter

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Algebraic Dance

The Geometric Revelation

The Identity of S

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