Problem 29 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\underset{\rightarrow}{x}\) and \(y\) be two \(\underset{\rightarrow}{\rightarrow}\) non-zero vectors such that \(|x+y|=|x|\) and \(2 x+\lambda y\) is perpendicular to \(y\), then the value of \(\lambda\) is \(\quad\) [NA Sep. 06, 2020 (II)]

Step-by-Step Solution

Answer

\(\lambda = 1\)

Step 1: Understand the Magnitude Condition

Given \(|\mathbf{x} + \mathbf{y}| = |\mathbf{x}|\), interpret this in terms of vector operations. By squaring both sides, \(|\mathbf{x} + \mathbf{y}|^2 = |\mathbf{x}|^2\), we have: \[ (\mathbf{x} + \mathbf{y}) \cdot (\mathbf{x} + \mathbf{y}) = \mathbf{x} \cdot \mathbf{x} \] Which simplifies to: \[ \mathbf{x} \cdot \mathbf{x} + 2 \mathbf{x} \cdot \mathbf{y} + \mathbf{y} \cdot \mathbf{y} = \mathbf{x} \cdot \mathbf{x} \]This leads to: \[ 2 \mathbf{x} \cdot \mathbf{y} + \mathbf{y} \cdot \mathbf{y} = 0 \]So, \(\mathbf{y} \cdot \mathbf{y} = -2 \mathbf{x} \cdot \mathbf{y}\).

Step 2: Interpret the Perpendicularity Condition

The problem states that \(2 \mathbf{x} + \lambda \mathbf{y}\) is perpendicular to \(\mathbf{y}\). The dot product equals zero when vectors are perpendicular:\[ (2 \mathbf{x} + \lambda \mathbf{y}) \cdot \mathbf{y} = 0 \]Expanding the dot product, we have: \[ 2 \mathbf{x} \cdot \mathbf{y} + \lambda (\mathbf{y} \cdot \mathbf{y}) = 0 \]

Step 3: Substitute and Solve for \(\lambda\)

From Step 1, we know \(\mathbf{y} \cdot \mathbf{y} = -2 \mathbf{x} \cdot \mathbf{y}\). Substituting this into the equation from Step 2, we get:\[ 2 \mathbf{x} \cdot \mathbf{y} + \lambda (-2 \mathbf{x} \cdot \mathbf{y}) = 0 \]Factor out \(2 \mathbf{x} \cdot \mathbf{y}\) (since it's non-zero):\[ 2 \mathbf{x} \cdot \mathbf{y} (1 - \lambda) = 0 \]This implies \(1 - \lambda = 0\), hence \(\lambda = 1\).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector calculus. This operation combines two vectors and returns a scalar value. The formula for the dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by:

• \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\theta \)

where \(|\mathbf{a}|\) and \(|\mathbf{b}|\) are the magnitudes of the vectors, and \(\theta\) is the angle between them. If the vectors are expressed with components, then it is computed as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

This operation reveals crucial relationships between vectors. For instance, the dot product is useful in determining if vectors are perpendicular. If \( \mathbf{a} \cdot \mathbf{b} = 0 \), the vectors are orthogonal and form a right angle. The dot product is also integral in projecting one vector onto another. The exercise applies this property to show perpendicularity after adjusting the vectors with a parameter \( \lambda \).

Vector Magnitude

The magnitude of a vector, often referred to as its "length", quantifies its size without direction. For a vector \( \mathbf{v} \) with components \( (v_1, v_2, v_3) \), its magnitude is computed via the Pythagorean Theorem as:

• \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)

Magnitude is like the length of an arrow in space from its starting point to its endpoint. It is a crucial concept in normalization, where a vector is scaled to have a magnitude of one, keeping its direction unchanged. In the exercise, comparing \(|\mathbf{x} + \mathbf{y}| = |\mathbf{x}|\) instantly indicates that the vector \(\mathbf{y} \) must counterbalance \(\mathbf{x} \) so precise that the overall length remains the same, suggesting certain conditions on their orientations. This relationship leads directly into further vector properties when solving for \( \lambda\).

Perpendicular Vectors

When two vectors are perpendicular, their dot product equals zero. This relationship is crucial not only in geometry but also in physics and engineering applications. Perpendicular vectors indicate no projection of one onto the other, meaning they meet at a right angle.

• Condition: \( \mathbf{a} \cdot \mathbf{b} = 0 \)

In the exercise, the vector \(2\mathbf{x} + \lambda\mathbf{y}\) was said to be perpendicular to \(\mathbf{y}\), which helps determine the value of \(\lambda\). Starting from the condition that their dot product is zero, and substituting known values, allows the calculative simplification to solve for unknowns, like \(\lambda\). This strategy effectively utilizes the geometric meaning of perpendicular vectors in vector equations to derive solutions that adhere to these conditions.