Problem 50 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The values of a, for which points \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) with position vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}\) and \(a \hat{i}-3 \hat{\jmath}+\hat{k}\) respectively are the vertices of a right angled triangle with \(C=\frac{\pi}{2}\) are \([2006]\) (a) 2 and 1 (b) \(-2\) and \(-1\) (c) \(-2\) and 1 (d) 2 and \(-1\)

Step-by-Step Solution

Answer

The values of \( a \) are 1 and 2. Option (a) is correct.

Step 1: Determine Vector AB

Find the vector from point A to B, denoted as \( \overrightarrow{AB} \). Use the position vectors of A \(2 \hat{i} - \hat{j} + \hat{k} \) and B \(\hat{i} - 3 \hat{j} - 5 \hat{k} \): \[ \overrightarrow{AB} = (\hat{i} - 3 \hat{j} - 5 \hat{k}) - (2 \hat{i} - \hat{j} + \hat{k}) = -\hat{i} - 2 \hat{j} - 6 \hat{k}. \]

Step 2: Determine Vector BC

Find the vector from point B to C, denoted as \( \overrightarrow{BC} \). Use the position vectors of B \(\hat{i} - 3 \hat{j} - 5 \hat{k} \) and C \(a \hat{i} - 3 \hat{j} + \hat{k} \): \[ \overrightarrow{BC} = (a \hat{i} - 3 \hat{j} + \hat{k}) - (\hat{i} - 3 \hat{j} - 5 \hat{k}) = (a-1)\hat{i} + 6\hat{k}. \]

Step 3: Determine Vector CA

Find the vector from point C to A, denoted as \( \overrightarrow{CA} \). Use the position vectors of C \(a \hat{i} - 3 \hat{j} + \hat{k} \) and A \(2 \hat{i} - \hat{j} + \hat{k} \): \[ \overrightarrow{CA} = (2 \hat{i} - \hat{j} + \hat{k}) - (a \hat{i} - 3 \hat{j} + \hat{k}) = (2-a)\hat{i} + 2 \hat{j}. \]

Step 4: Apply Right Triangle Condition

For the points A, B, and C to form a right-angled triangle with the right angle at C, the dot product \( \overrightarrow{BC} \cdot \overrightarrow{CA} \) should be zero: \[ \overrightarrow{BC} \cdot \overrightarrow{CA} = [(a-1)\hat{i} + 6\hat{k}] \cdot [(2-a)\hat{i} + 2\hat{j}] = (a-1)(2-a) + 0 + 0 = (a-1)(2-a). \] Simplifying: \[ (a-1)(2-a) = 0. \]

Step 5: Solve the Equation

Solve the equation obtained in the previous step: \[ (a-1)(2-a) = 0. \] This gives the potential solutions for \( a \):1. \( a - 1 = 0 \Rightarrow a = 1 \)2. \( 2 - a = 0 \Rightarrow a = 2 \)Thus, the solutions for \( a \) are 1 and 2.

Key Concepts

Vector Calculation Made Simple

Vectors are essential in understanding different geometric relationships, including those in triangles. In this exercise, we look at the position vectors for the points \( A \), \( B \), and \( C \), which allow us to calculate vectors \( \overrightarrow{AB} \), \( \overrightarrow{BC} \), and \( \overrightarrow{CA} \). This process involves subtracting coordinates of these points:

• \( \overrightarrow{AB} = (\hat{i} - 3 \hat{j} - 5 \hat{k}) - (2 \hat{i} - \hat{j} + \hat{k}) = -\hat{i} - 2 \hat{j} - 6 \hat{k} \)
• \( \overrightarrow{BC} = (a \hat{i} - 3 \hat{j} + \hat{k}) - (\hat{i} - 3 \hat{j} - 5 \hat{k}) = (a-1)\hat{i} + 6\hat{k} \)
• \( \overrightarrow{CA} = (2 \hat{i} - \hat{j} + \hat{k}) - (a \hat{i} - 3 \hat{j} + \hat{k}) = (2-a)\hat{i} + 2\hat{j} \)

Each vector has direction and magnitude that helps in further calculations. Understanding each part of vector creation and operation makes tackling complex geometry problems straightforward.

Exploring the Dot Product

The dot product is a way to multiply two vectors and results in a scalar value. It's particularly useful to determine the angle between vectors. For right-angled triangles, the dot product of the vectors forming the right angle is zero. In this case, calculating \( \overrightarrow{BC} \cdot \overrightarrow{CA} \) gives an insight:

• Calculate the dot product: \( [(a-1)\hat{i} + 6\hat{k}] \cdot [(2-a)\hat{i} + 2\hat{j}] \)
• This simplifies to \( (a-1)(2-a) + 0 + 0 = (a-1)(2-a) \)

The zero result indicates orthogonality, proving the vectors are at right angles. Thus, finding when the product equals zero confirms the condition for the triangle being right-angled at a certain vertex. The clever arrangement of terms and understanding dot product properties are key here.

Crafting a Mathematical Proof

A mathematical proof is about constructing a logical argument. In the context of this problem, we are proving that the given points form a right-angled triangle when \( a \) satisfies certain conditions. The proof unfolds through:

• Defining the vectors \( \overrightarrow{BC} \) and \( \overrightarrow{CA} \) as calculated earlier.
• Using the dot product condition for orthogonality. This is essential in proving that angle at point \( C \) is right.
• Solving the equation \( (a-1)(2-a) = 0 \) leads us to find possible values for \( a \).
• We conclude through possible solutions, \( a = 1 \) and \( a = 2 \).

Through linear algebra and dot product properties, mathematical proof helps us confirm true statements. Mastering these fundamental concepts ensures accurate solutions to geometric problems.