ABC is a triangle, right angled at A. The resultant of the forces acting along AB, BC with magnitude

Q: Wait, why are the forces perpendicular? I thought they were along $AB$ and $BC$?

That is a common point of confusion! The problem states the forces act along $AB$ and $AC$. Since the triangle is right-angled at $A$, the sides $AB$ and $AC$ are perpendicular. This makes the vector addition straightforward using the Pythagorean theorem.

Q: Can we use the parallelogram law of vector addition here?

Absolutely! The parallelogram law is the fundamental way to add vectors. Since the angle between the two forces is $90^\circ$, the parallelogram formed is actually a rectangle, and the diagonal (the resultant) is simply the hypotenuse of the right triangle formed by the two force vectors.

Q: Why does the area formula help us here?

The area formula is a powerful geometric bridge. By expressing the area of $\triangle ABC$ as $\frac{1}{2} \cdot AB \cdot AC$ and also as $\frac{1}{2} \cdot BC \cdot AD$, we create an algebraic link between the legs ($AB, AC$), the hypotenuse ($BC$), and the altitude ($AD$). This allows us to substitute the product $AB \cdot AC$ with $BC \cdot AD$, which is the key to simplifying the final expression.

Question

ABC is a triangle, right angled at A. The resultant of the forces acting along $\vec{AB}, \vec{BC}$ with magnitudes $\frac{1}{AB}$ and $\frac{1}{AC}$ respectively is the force along $\vec{AD}$, where D is the foot of the perpendicular from A onto BC. The magnitude of the resultant is

Options:

(A) $\frac{AB^2 + AC^2}{(AB)^2(AC)^2}$

(B) $\frac{(AB)(AC)}{AB + AC}$

(C) $\frac{1}{AB} + \frac{1}{AC}$

(D) $\frac{1}{AD}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at vertex $A$ of a right-angled triangle $ABC$. You have two forces, $F_1 = \frac{1}{AB}$ acting along the leg $AB$, and $F_2 = \frac{1}{AC}$ acting along the leg $AC$.

Because the triangle is right-angled at $A$, these two forces are perfectly orthogonal. We are dealing with a simple, clean, right-angled vector addition.

### The Vector Dance

When two forces are perpendicular, the magnitude of the resultant $R$ is given by the Pythagorean theorem for vectors:

$$R = \sqrt{F_1^2 + F_2^2}$$

Substituting our given magnitudes, we get:

$$R = \sqrt{\left(\frac{1}{AB}ight)^2 + \left(\frac{1}{AC}ight)^2}$$

Simplifying this by taking the common denominator, we obtain:

$$R = \sqrt{\frac{AC^2 + AB^2}{AB^2 \cdot AC^2}}$$

### The Pythagorean Bridge

In our right-angled triangle $ABC$, the Pythagorean theorem states that $AB^2 + AC^2 = BC^2$. We can replace the numerator with $BC^2$:

$$R = \sqrt{\frac{BC^2}{AB^2 \cdot AC^2}}$$

Taking the square root, we find:

$$R = \frac{BC}{AB \cdot AC}$$

### The Geometric Masterstroke

To simplify the denominator, we relate the product $AB \cdot AC$ to the altitude $AD$ dropped from $A$ to the hypotenuse $BC$. The area of $	riangle ABC$ can be expressed in two ways:

$$	ext{Area} = \frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} \cdot BC \cdot AD$$

Equating these two expressions, we get:

$$AB \cdot AC = BC \cdot AD$$

### The Final Calculation

Now, substitute this relation back into our expression for the resultant:

$$R = \frac{BC}{BC \cdot AD}$$

The $BC$ terms in the numerator and denominator cancel out perfectly. We are left with the elegant result:

**$$R = \frac{1}{AD}$$**

The resultant of these two forces is simply the reciprocal of the altitude. This result demonstrates how geometry and physics intertwine to provide an elegant, simple solution.

Select Answer:

Visualized Solution

Visualizing the Geometry

Defining the Force Vectors

Resultant of Orthogonal Forces

Simplifying the Expression

Applying Pythagoras Theorem

Removing the Square Root

Relating to Altitude $A D$

Equating the Areas

Final Substitution

The Sigma Insight: Scalar and Vector Quantities

Analyzing the Setup

The Vector Dance

The Pythagorean Bridge

The Geometric Masterstroke

The Final Calculation

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