Problem 20 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

\(A B C\) is a triangle, right angled at \(A\). The resultant of the forces acting along \(\overline{A B}, \overline{B C}\) with magnitudes \(\frac{1}{A B}\) and \(\frac{1}{A C}\) respectively is the force along \(\overline{A D}\), where \(D\) is the foot of the perpendicular from \(A\) onto \(B C\). The magnitude of the resultant is (a) \(\frac{A B^{2}+A C^{2}}{(A B)^{2}(A C)^{2}}\) (b) \(\frac{(A B)(A C)}{A B+A C}\) (c) \(\frac{1}{A B}+\frac{1}{A C}\) (d) \(\frac{1}{A D}\)

Step-by-Step Solution

Answer

The magnitude of the resultant force is option (a): \(\frac{AB^2 + AC^2}{(AB)^2(AC)^2}\).

Step 1: Understand the problem

We have a right-angled triangle \(ABC\), with \(\angle A = 90^\circ\). Forces of magnitudes \(\frac{1}{AB}\) and \(\frac{1}{AC}\) act along \(\overline{AB}\) and \(\overline{AC}\) respectively. We need to find the resultant magnitude of these forces along \(\overline{AD}\), where \(D\) is the foot of the perpendicular from \(A\) onto \(\overline{BC}\).

Step 2: Apply Pythagorean Theorem

Since \(\triangle ABC\) is a right triangle with \(\angle A = 90^\circ\), we apply the Pythagorean Theorem: \[ AB^2 + AC^2 = BC^2. \] This will help understand the relationship between the sides of the triangle as it relates to the directions of the forces.

Step 3: Identify force components

The force along \(\overline{AB}\) is \(\frac{1}{AB}\) and along \(\overline{AC}\) is \(\frac{1}{AC}\). These are perpendicular vectors since \(\angle BAC = 90^\circ\).

Step 4: Calculate the resultant force magnitude

Since the forces act at right angles to each other, use the Pythagorean theorem for vectors: \[ R = \sqrt{\left(\frac{1}{AB}\right)^2 + \left(\frac{1}{AC}\right)^2}. \] Simplifying gives \[ R = \sqrt{\frac{1}{AB^2} + \frac{1}{AC^2}}. \]

Step 5: Identifying the expression among options

Look for the choice that matches \(\sqrt{\frac{1}{AB^2} + \frac{1}{AC^2}}\). To match this with one of the given answers, consider multiplying the expression by \(AB^2\) and \(AC^2\) to adjust for potential equivalent forms.

Step 6: Choice verification

The correct choice corresponds to option (a): \[ \frac{AB^2 + AC^2}{(AB)^2(AC)^2}. \] This simplifies from the resultant force expression, considering \[ \frac{1}{AB^2} + \frac{1}{AC^2} = \frac{AB^2 + AC^2}{(AB)^2(AC)^2}. \]

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a critical concept when dealing with right-angled triangles. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, this can be expressed as:\[ c^2 = a^2 + b^2 \]where \( c \) represents the hypotenuse, and \( a \) and \( b \) are the other two sides. In the context of the exercise, triangle \( ABC \) is a right-angled triangle, allowing us to use this theorem to understand the relationship between its sides. Specifically, with \( \angle A = 90^\circ \), we use the sides \( AB \) and \( AC \) to find \( BC \):\[ BC^2 = AB^2 + AC^2 \] This sets the stage for the next process of analyzing forces acting at right angles.

Forces in Right-Angled Triangles

When forces are applied along the sides of a right-angled triangle, it becomes essential to understand how these forces interact. In triangle \( ABC \), we have two forces acting along the sides \( \overline{AB} \) and \( \overline{AC} \). The magnitudes of these forces are \( \frac{1}{AB} \) and \( \frac{1}{AC} \) respectively.These forces are perpendicular to one another because \( \angle BAC = 90^\circ \). This perpendicular relationship is crucial in determining the resultant force—a single force that represents the cumulative effect of these two forces acting together.In a right-angle context like this, you can't just add the forces' magnitudes directly. Instead, use vector methods to find the resultant force. This involves understanding their directions and magnitudes as perpendicular vectors.

Vector Addition of Forces

Vector addition is a method used to find the resultant force when two or more forces act at a point. In the case of the right-angled triangle \( ABC \), vector addition is very convenient, as the forces are perpendicular to each other. To determine the resultant force magnitude, start by expressing each vector:- The force along \( \overline{AB} \) is \( \frac{1}{AB} \)- The force along \( \overline{AC} \) is \( \frac{1}{AC} \)Being perpendicular vectors, their resultant \( R \) is given by the Pythagorean theorem for forces:\[ R = \sqrt{\left(\frac{1}{AB}\right)^2 + \left(\frac{1}{AC}\right)^2} \]This simplifies to:\[ R = \sqrt{\frac{1}{AB^2} + \frac{1}{AC^2}} \]Further simplification involves matching this expression with the options provided in the exercise. The correct choice turns out to be option (a):\[ \frac{AB^2 + AC^2}{(AB)^2(AC)^2} \]This equation encapsulates the essence of vector addition in a right-angled triangle. It visualizes how individual forces, when combined, result in a new force with a specific magnitude.