Problem 83 - A Complete Resource Book in Mathematics for JEE Main 2017

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 [2004] (A) 2 (B) \(\sqrt{7}\) (C) \(\sqrt{14}\) (D) 14

Step-by-Step Solution

Answer

(C) \(\sqrt{14}\)

Step 1: Understanding Vector Projections

The projection of vector \(\bar{v}\) on \(\bar{u}\) is given by \(\text{proj}_{\bar{u}} \bar{v} = \left( \frac{\bar{v} \cdot \bar{u}}{|\bar{u}|^2} \right) \bar{u}\). Similarly, the projection of \(\bar{w}\) on \(\bar{u}\) is \(\text{proj}_{\bar{u}} \bar{w} = \left( \frac{\bar{w} \cdot \bar{u}}{|\bar{u}|^2} \right) \bar{u}\). Given that these two projections are equal, we have \(\frac{\bar{v} \cdot \bar{u}}{|\bar{u}|^2} = \frac{\bar{w} \cdot \bar{u}}{|\bar{u}|^2}\).

Step 2: Equating Dot Products

Since \(|\bar{u}| = 1\), the equation simplifies to \(\bar{v} \cdot \bar{u} = \bar{w} \cdot \bar{u}\). This means the scalar products of \(\bar{v}\) and \(\bar{w}\) with \(\bar{u}\) are the same.

Step 3: Using Orthogonality

It is given that \(\bar{v}\) and \(\bar{w}\) are perpendicular. Thus, \(\bar{v} \cdot \bar{w} = 0\). This will be useful later.

Step 4: Evaluating \\(|\bar{u}-\bar{v}+\bar{w}|^2\\)

The magnitude we need is related to the square of the vector sum: \[ |\bar{u} - \bar{v} + \bar{w}|^2 = (\bar{u} - \bar{v} + \bar{w}) \cdot (\bar{u} - \bar{v} + \bar{w}) \]},

Step 5: Expanding Using Dot Product

Expand the expression:\[ (\bar{u} - \bar{v} + \bar{w}) \cdot (\bar{u} - \bar{v} + \bar{w}) = \bar{u} \cdot \bar{u} - 2 \bar{u} \cdot \bar{v} + 2 \bar{u} \cdot \bar{w} + \bar{v} \cdot \bar{v} - 2 \bar{v} \cdot \bar{w} + \bar{w} \cdot \bar{w} \]

Step 6: Substituting Known Values

Since \(\bar{v} \cdot \bar{w} = 0\) (orthogonality), \(|\bar{u}| = 1\), \(|\bar{v}| = 2\), and \(|\bar{w}| = 3\), as well as \(\bar{u} \cdot \bar{v} = \bar{u} \cdot \bar{w}\), we substitute these into our expanded expression:\[ = 1 - 2(\bar{u} \cdot \bar{v}) + 2(\bar{u} \cdot \bar{w}) + 4 + 0 + 9 \]

Step 7: Solving the Final Equation

Combine and simplify the equation:\[ = 1 + 4 + 9 = 14 \]Hence, \(|\bar{u}-\bar{v}+\bar{w}|^2 = 14\), so \(|\bar{u}-\bar{v}+\bar{w}| = \sqrt{14}\).

Step 8: Selecting the Correct Answer

The magnitude \(|\bar{u}-\bar{v}+\bar{w}|\) has been calculated as \(\sqrt{14}\), which corresponds to option (C).

Key Concepts

Vector Projection

To project one vector onto another means to "drop" one vector onto the line defined by another, creating a shadow or duplicate along this line. This is useful for breaking down vector relationships and problems in physics and engineering.
The projection of vector \(\bar{v}\) onto \(\bar{u}\) is represented by \(\text{proj}_{\bar{u}} \bar{v} = \left( \frac{\bar{v} \cdot \bar{u}}{|\bar{u}|^2} \right) \bar{u}\). Here, the formula involves:

• The dot product, \(\bar{v} \cdot \bar{u}\), which captures the part of \(\bar{v}\) in the direction of \(\bar{u}\).
• The magnitude squared of \(\bar{u}\), \(|\bar{u}|^2\), to normalize the projection along \(\bar{u}\).

A key detail in vector projection is that the length of this projection is always less than or equal to the length of the vector being projected. In the problem, projections of \(\bar{v}\) and \(\bar{w}\) on \(\bar{u}\) are equal. This indicates an important relationship between these vectors.

Dot Product

The dot product is a way to multiply two vectors that results in a scalar, rather than another vector. It is found by multiplying the corresponding components of the vectors and then summing those products.
The formula for the dot product of two vectors \(\bar{a} = (a_1, a_2, a_3)\) and \(\bar{b} = (b_1, b_2, b_3)\) is:

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

This scalar value can also be calculated using the magnitudes of the vectors and the cosine of the angle between them: \(\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}|\cos(\theta)\).
In our exercise, the dot product helps us set up the relationship between the vectors, especially because we know \(\bar{v} \cdot \bar{w} = 0\), signifying that \(\bar{v}\) and \(\bar{w}\) are perpendicular.

Orthogonal Vectors

Two vectors are orthogonal when they are perpendicular to each other. This means the angle between them is exactly 90 degrees. The mathematical confirmation of orthogonality is simple: if the dot product between two vectors is zero, they are orthogonal.
For vectors \(\bar{v}\) and \(\bar{w}\) in the exercise, we have:

• \(\bar{v} \cdot \bar{w} = 0\), indicating that \(\bar{v}\) and \(\bar{w}\) meet the condition for orthogonality.

Orthogonality is a powerful concept in vector analysis as it implies no component of one vector exists along the direction of another. This property greatly simplifies many vector calculations and analyses.

Vector Magnitude

The magnitude or "length" of a vector \(\bar{v} = (v_1, v_2, v_3)\) is calculated using the formula:
\(|\bar{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\)
The magnitude is always a non-negative quantity that represents the vector's distance from the origin in a coordinate system.
In the provided problem statement:

• \(|\bar{u}| = 1\)
• \(|\bar{v}| = 2\)
• \(|\bar{w}| = 3\)

Knowing these magnitudes allows us to substitute the values directly into equations, simplifying calculations significantly. These magnitudes play a crucial role in solving the vector equation, where the compound vector \(|\bar{u} - \bar{v} + \bar{w}|\) must be evaluated.