Problem 51 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\bar{u}, \bar{v}, \bar{w}\) be such that \(|\bar{u}|=1,|\bar{v}|=2,|\bar{w}|=3\). If the projection \(\bar{v}\) along \(\bar{u}\) is equal to that of \(\bar{w}\) along \(\bar{u}\) and \(\bar{v}, \bar{w}\) are perpendicular to each other then \(|\bar{u}-\bar{v}+\bar{w}|\) equals (a) 14 (b) \(\sqrt{7}\) (c) \(\sqrt{14}\) (d) 2

Step-by-Step Solution

Answer

The magnitude \(|\bar{u} - \bar{v} + \bar{w}|\) equals \(\sqrt{14}\). (c) is correct.

Step 1: Calculate Projection of Vectors

The projection of vector \( \bar{v} \) along \( \bar{u} \) is given by \( \text{proj}_u v = \frac{\bar{v} \cdot \bar{u}}{\bar{u} \cdot \bar{u}} \bar{u} \). The magnitude of this projection is \( \frac{\bar{v} \cdot \bar{u}}{1} = \bar{v} \cdot \bar{u} \), where \( \bar{u} \) is a unit vector. Similarly, the magnitude of the projection of \( \bar{w} \) along \( \bar{u} \) is \( \bar{w} \cdot \bar{u} \). Given that these two projections are equal, we have \( \bar{v} \cdot \bar{u} = \bar{w} \cdot \bar{u} \).

Step 2: Determine Condition from Perpendicularity

Since \( \bar{v} \) and \( \bar{w} \) are perpendicular, \( \bar{v} \cdot \bar{w} = 0 \). This is a key condition that will help us find the desired magnitude.

Step 3: Setup Equation for Magnitude Using Dot Product

The magnitude of the vector \( \bar{u} - \bar{v} + \bar{w} \) is given by \( |\bar{u} - \bar{v} + \bar{w}| = \sqrt{(\bar{u} - \bar{v} + \bar{w}) \cdot (\bar{u} - \bar{v} + \bar{w})} \). Expanding the dot product, we have:\[|\bar{u} - \bar{v} + \bar{w}| = \sqrt{\bar{u} \cdot \bar{u} - 2(\bar{u} \cdot \bar{v}) + \bar{v} \cdot \bar{v} - 2(\bar{v} \cdot \bar{w}) + \bar{w} \cdot \bar{w} + 2(\bar{w} \cdot \bar{u})}\]

Step 4: Simplify the Expression

Given \( \bar{v} \cdot \bar{w} = 0 \) and \( \bar{u} \cdot \bar{v} = \bar{u} \cdot \bar{w} \) which we denote as \( x \), the expression simplifies to:\[|\bar{u} - \bar{v} + \bar{w}| = \sqrt{1 - 2x + 4 + 9 + 2x} = \sqrt{14}\]Therefore, \(|\bar{u} - \bar{v} + \bar{w}| = \sqrt{14}\).

Step 5: Conclude the Result

The correct value for \(|\bar{u} - \bar{v} + \bar{w}|\) is \(\sqrt{14}\). Therefore, the answer is option (c).

Key Concepts

Dot Product

In vector mathematics, the dot product is a foundational operation that provides significant insights into the relationship between two vectors. The dot product, also known as the scalar product, is a way to multiply two vectors and results in a scalar value. If we have two vectors \(\bar{a} \) and \(\bar{b} \), their dot product is calculated as follows:

• \( \bar{a} \cdot \bar{b} = |\bar{a}| \, |\bar{b}| \, \cos(\theta) \)
• Where \(|\bar{a}|\) and \(|\bar{b}|\) are the magnitudes of the vectors, and \(\theta\) is the angle between them.
• If \(\bar{a} = (a_1, a_2, a_3)\) and \(\bar{b} = (b_1, b_2, b_3)\), then \(\bar{a} \cdot \bar{b} = a_1b_1 + a_2b_2 + a_3b_3\).

The outcome of the dot product is highly informative:

• If it equals zero, the vectors are perpendicular.
• If positive, the vectors are in the same general direction.
• If negative, they are in opposite directions.

In the exercise, the equal dot products for \(\bar{v} \cdot \bar{u}\) and \(\bar{w} \cdot \bar{u}\) imply that the projection of these vectors onto \(\bar{u}\) is identical. This lays the groundwork for further calculations to find the magnitude of the resulting vector expression.

Projection of Vectors

Projection of a vector onto another vector represents the shadow or footprint of the first vector onto the line defined by the second vector. Let’s break this down using vector \( \bar{v} \) projected onto vector \( \bar{u} \):

• To calculate this, the formula is: \( \text{proj}_u v = \frac{\bar{v} \cdot \bar{u}}{\bar{u} \cdot \bar{u}} \bar{u} \).
• Since \( \bar{u} \) is a unit vector, the denominator (\( \bar{u} \cdot \bar{u} \)) simplifies to 1, making our task easier.
• This results in: \( \text{proj}_u v = (\bar{v} \cdot \bar{u}) \bar{u} \).

The projection helps illustrate how much of \( \bar{v} \) goes in the direction of \( \bar{u} \). In our problem, knowing that the projection of \( \bar{v} \) along \( \bar{u} \) equals that of \( \bar{w} \) along \( \bar{u} \), we conclude \( \bar{v} \cdot \bar{u} = \bar{w} \cdot \bar{u} \). This key insight is crucial in solving for the magnitude of the resulting vector given its expression.

Perpendicular Vectors

When two vectors are said to be perpendicular, this implies a geometric relationship where they create a 90-degree angle with each other. In mathematical terms, this is expressed by their dot product being zero. Consider vectors \( \bar{v} \) and \( \bar{w} \):

• Mathematically, \( \bar{v} \cdot \bar{w} = 0 \) confirms that they are perpendicular.
• This is due to \( \cos(90^\circ) = 0 \), resulting in zero contribution to the dot product.

Understanding perpendicular vectors are essential in decomposition and orthogonal projections. In the exercise, \( \bar{v} \) and \( \bar{w} \) being perpendicular simplifies the process of calculating the magnitude, as their interaction contributes nothing to the length of the resultant expression. This absence simplifies to eliminate specific terms in the overall calculation, focusing solely on the magnitudes of component vectors and their equivalent projections. Such scenarios illustrate the elegance of vector algebra in reducing complex problems into simpler, solvable parts.