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

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 \(\bar{u}-\bar{v}+\bar{w} \mid\) equals (A) 2 (B) \(\sqrt{7}\) (C) \(\sqrt{14}\) (D) 14

Step-by-Step Solution

Answer

The answer is (C) \(\sqrt{14}\).

Step 1: Understanding Projection

The projection of a vector \(\bar{v}\) onto \(\bar{u}\), denoted as \(\text{proj}_{\bar{u}} \bar{v}\), is given by the formula \(\text{proj}_{\bar{u}} \bar{v} = \frac{\bar{v} \cdot \bar{u}}{\bar{u} \cdot \bar{u}} \bar{u}\). Given \(|\bar{u}| = 1\), this simplifies to \(\bar{v} \cdot \bar{u}\). Hence, \(\text{proj}_{\bar{u}} \bar{v} = \bar{v} \cdot \bar{u}\).

Step 2: Setting up Equal Projections

We know that \(\text{proj}_{\bar{u}} \bar{v} = \text{proj}_{\bar{u}} \bar{w}\). This implies that \(\bar{v} \cdot \bar{u} = \bar{w} \cdot \bar{u}\).

Step 3: Calculating Magnitudes

The vectors \(\bar{v}\) and \(\bar{w}\) are perpendicular, so \(\bar{v} \cdot \bar{w} = 0\). The magnitudes are \(|\bar{v}|=2\) and \(|\bar{w}|=3\). We use these values later in calculations.

Step 4: Finding \(\bar{u} - \bar{v} + \bar{w}\)

Calculate \(|\bar{u} - \bar{v} + \bar{w}|\) by using the formula for the magnitude of a vector, \(|\bar{a} - \bar{b} + \bar{c}| = \sqrt{(\bar{a} - \bar{b} + \bar{c}) \cdot (\bar{a} - \bar{b} + \bar{c})}\).

Step 5: Substitute Known Values

Substitute the known values: Use \(|\bar{u}|=1\), \(|\bar{v}|=2\), \(|\bar{w}|=3\), and \(\bar{v} \cdot \bar{w} = 0\). We find \(\bar{u} \cdot \bar{v} = \bar{u} \cdot \bar{w}\). By letting \(\bar{u} \cdot \bar{v} = a\), we have \(\bar{u} \cdot \bar{w} = a\).

Step 6: Simplify the Expression

Expanding the magnitude calculation: \(|\bar{u} - \bar{v} + \bar{w}|^2 = (\bar{u} \cdot \bar{u} - 2 \bar{u} \cdot \bar{v} + \bar{v} \cdot \bar{v} + 2 \bar{u} \cdot \bar{w} - 2 \bar{v} \cdot \bar{w} + \bar{w} \cdot \bar{w})\) which with substitutions becomes \(1 - 2a + 4 + 2a + 0 + 9 = 14\). Thus, the magnitude \(\sqrt{14}\).

Step 7: Determine the Correct Choice

From the previous calculation, \(|\bar{u} - \bar{v} + \bar{w}| = \sqrt{14}\). Thus, the correct answer is (C) \(\sqrt{14}\).

Key Concepts

Vector Magnitudes

The magnitude of a vector, often represented as \(|\bar{v}|\), is a measure of its length. It's like the size of the vector. To compute the magnitude, you apply the formula for the Euclidean norm, which is \(|\bar{v}| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2}\), where each \(v_i\) is a component of the vector.
Here in the exercise, we have three vectors: \(\bar{u}\), \(\bar{v}\), and \(\bar{w}\), with their magnitudes given as \(1\), \(2\), and \(3\) respectively. These magnitudes help us understand the relative sizes of these vectors.

• \(|\bar{u}| = 1\): This indicates that \(\bar{u}\) is a unit vector, which is the simplest form having a magnitude of one.
• \(|\bar{v}| = 2\): This means \(\bar{v}\) is twice as long as \(\bar{u}\).
• \(|\bar{w}| = 3\): Similarly, \(\bar{w}\) is three times longer than the unit vector \(\bar{u}\).

Understanding these magnitudes is crucial, as they are directly involved in calculations such as projections and finding the resultant magnitude of vector combinations.

Perpendicular Vectors

Perpendicular vectors, also known as orthogonal vectors, have a special property in vector mathematics: their dot product is zero. The dot product of two vectors, say \(\bar{v}\) and \(\bar{w}\), is calculated as \(\bar{v} \cdot \bar{w} = v_1w_1 + v_2w_2 + ... + v_nw_n\).
When vectors are perpendicular, \(\bar{v} \cdot \bar{w} = 0\). This property indicates that they form a 90-degree angle with each other. In our exercise, it's given that \(\bar{v}\) and \(\bar{w}\) are perpendicular.
This leads to:

• Calculation simplification: Knowing this simplifies many calculations, as any component involving the dot product \(\bar{v} \cdot \bar{w}\) can be ignored, assuming a value of zero.
• Geometric visualization: It helps to visualize how these vectors might look geometrically, i.e., forming a right angle in space.

Understanding the perpendicularity of vectors is important in problems involving projections and when determining vector relationships, like in this example.

Vector Addition

Vector addition is the process of combining two or more vectors to create a new vector, known as the resultant vector. This process is straightforward: you add the corresponding components of the vectors. For instance, for vectors \(\bar{a} = (a_1, a_2)\) and \(\bar{b} = (b_1, b_2)\), the resultant \(\bar{c} = \bar{a} + \bar{b}\) is \(\bar{c} = (a_1+b_1, a_2+b_2)\).
In our exercise, we consider the expression \(\bar{u} - \bar{v} + \bar{w}\). Understanding vector addition helps us:

• Determine resulting vectors: Use vector addition rules to find the new vector formed by adding or subtracting initial vectors.
• Magnitude calculation: The magnitude of this new vector is found by applying the Euclidean norm to its components – using the formula \(|\bar{c}| = \sqrt{c_1^2 + c_2^2}\).
• Simplification: Each operation affects the vector components, so it's essential for simplification and ensuring accurate results.

Ultimately, vector addition helps in finding combinations like \(\bar{u} - \bar{v} + \bar{w}\) in our problem, leading us to the final result's magnitude, which is \(\sqrt{14}\).