Problem 97 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{u}=\hat{i}+\hat{j}, \vec{v}=\hat{i}-\hat{j}\) and \(\vec{w}=\hat{i}+2 \hat{j}+3 \hat{k}\). If \(\hat{n}\) is a unit vector such that \(\vec{u}, \hat{n}=0\) and \(\vec{v} \cdot \hat{n}=0\), then \(|\vec{w} \cdot \hat{n}|\) is equal to \(\quad[2003]\) (a) 3 (b) 0 (c) 1 (d) 2

Step-by-Step Solution

Answer

The answer is (a) 3.

Step 1: Understanding the Problem

We are given three vectors, \(\vec{u}, \vec{v},\) and \(\vec{w}\), as well as a unit vector \(\hat{n}\) that is orthogonal to both \(\vec{u}\) and \(\vec{v}\). We need to find the absolute value of \(\vec{w} \cdot \hat{n}\).

Step 2: Express the Vectors

Write the given vectors in component form:\[ \vec{u} = \hat{i} + \hat{j} = (1, 1, 0) \]\[ \vec{v} = \hat{i} - \hat{j} = (1, -1, 0) \]\[ \vec{w} = \hat{i} + 2\hat{j} + 3\hat{k} = (1, 2, 3) \]

Step 3: Find Orthogonal Conditions

Since \(\hat{n}\) is orthogonal to \(\vec{u}\) and \(\vec{v}\), we write the equations:\[ \vec{u} \cdot \hat{n} = 1 \cdot n_1 + 1 \cdot n_2 + 0 \cdot n_3 = 0 \rightarrow n_1 + n_2 = 0 \]\[ \vec{v} \cdot \hat{n} = 1 \cdot n_1 - 1 \cdot n_2 + 0 \cdot n_3 = 0 \rightarrow n_1 - n_2 = 0 \]

Step 4: Solve Orthogonal Conditions

From the equations \(n_1 + n_2 = 0\) and \(n_1 - n_2 = 0\), we solve for \(n_1\) and \(n_2\):Adding both equations gives \(2n_1 = 0\), so \(n_1 = 0\).Substituting \(n_1 = 0\) into \(n_1 + n_2 = 0\) gives \(n_2 = 0\).

Step 5: Determine Third Component of Unit Vector

Since \(\hat{n}\) is a unit vector, its magnitude should be 1:\[ \sqrt{n_1^2 + n_2^2 + n_3^2} = 1 \].We have \(n_1 = 0\) and \(n_2 = 0\), thus \(\sqrt{n_3^2} = 1\) implies \(|n_3| = 1\).Therefore, \(n_3 = \pm 1\), and we choose either \( (0, 0, 1) \) or \( (0, 0, -1) \).

Step 6: Compute Dot Product

Calculate \(\vec{w} \cdot \hat{n}\):If \(\hat{n} = (0, 0, 1)\), \[ \vec{w} \cdot \hat{n} = 1\cdot0 + 2\cdot0 + 3\cdot1 = 3 \].If \(\hat{n} = (0, 0, -1)\), \[ \vec{w} \cdot \hat{n} = 1\cdot0 + 2\cdot0 + 3\cdot(-1) = -3 \].

Step 7: Consider Absolute Value of Dot Product

Finally, we find the absolute value:\[ |\vec{w} \cdot \hat{n}| = |3| = 3 \]. So, \(|\vec{w} \cdot \hat{n}| = 3\).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is an important operation in vector algebra. It combines two vectors to give a scalar (a single number), reflecting how much one vector extends in the direction of another.

The formula for the dot product of two vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \) is:

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

This formula is the sum of the products of their corresponding components.

The dot product is zero if and only if the vectors are orthogonal (perpendicular). This aspect is crucial for solving problems involving orthogonality, as it provides the condition needed to identify orthogonal vectors.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at right angles. If two vectors are orthogonal, their dot product is zero, reflecting the absence of any overlap in their directions.

For vectors \( \vec{u} \) and \( \vec{v} \), they are orthogonal if \( \vec{u} \cdot \vec{v} = 0 \).

Here's why orthogonal vectors are useful:

• The orthogonality condition simplifies many vector problems, allowing you to determine unknown components easily.
• Orthogonal vectors are foundational in creating coordinate systems that simplify complex geometry into manageable components.

Understanding this concept becomes particularly practical when working with unit vectors or finding projections of one vector onto another.

Unit Vector

Unit vectors have a magnitude (or length) of 1, functioning as direction indicators.

Unit vectors are normally used to describe directions and are crucial in decomposing vectors into their basic components.

To verify if a vector \( \hat{n} = (n_1, n_2, n_3) \) is a unit vector, calculate:

• \( \sqrt{n_1^2 + n_2^2 + n_3^2} = 1 \)

In our problem, the unit vector \( \hat{n} \) was chosen to be orthogonal to two given vectors \( \vec{u} \) and \( \vec{v} \), which simplified the condition to find its components. Given that it needed to be orthogonal, we established it had to lie along the axis perpendicular to both \( \vec{u} \) and \( \vec{v} \). This emphasized that its non-zero component was simply \( n_3 = \pm 1 \).

Magnitude of a Vector

The magnitude of a vector measures its length, and gives insight into the size of the vector without regard to its direction.

For a vector \( \vec{a} = (a_1, a_2, a_3) \), its magnitude is calculated as:

• \( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

Having knowledge about magnitude helps in understanding physical quantities represented as vectors, like velocity or force. For a unit vector, it's especially straightforward because its magnitude is always 1, confirming the vector doesn't stretch beyond its current direction.

In the exercise, this property was used to verify the unit vector \( \hat{n} \), ensuring its total length was appropriate even after determining its components.