Problem 81 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\vec{u}=\hat{j}+4 \hat{k}, \vec{v}=\hat{i}+3 \hat{k}\) and \(\vec{w}=\cos \theta \hat{i}+\sin \theta \hat{j}\) are vectors in 3 -dimensional space, then the maximum possible value of \(|\vec{u} \times \vec{v} \cdot \vec{w}|\) is \(\quad\) [Online May 12, 2012] (a) \(\sqrt{3}\) (b) 5 (c) \(\sqrt{14}\) (d) 7

Step-by-Step Solution

Answer

The maximum possible value is 5.

Step 1: Calculate the Cross Product \( \vec{u} \times \vec{v} \)

Find the cross product of vectors \( \vec{u} = \hat{j} + 4\hat{k} \) and \( \vec{v} = \hat{i} + 3\hat{k} \). Use the determinant formula for cross products: \[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & 4 \ 1 & 0 & 3 \end{vmatrix} \]\[ = \hat{i}(1 \cdot 3 - 4 \cdot 0) - \hat{j}(0 \cdot 3 - 4 \cdot 1) + \hat{k}(0 \cdot 0 - 1 \cdot 1) \]\[ = 3 \hat{i} + 4 \hat{j} - \hat{k} \]Thus, \( \vec{u} \times \vec{v} = 3\hat{i} + 4\hat{j} - \hat{k} \).

Step 2: Calculate the Dot Product \( (\vec{u} \times \vec{v}) \cdot \vec{w} \)

Let \( \vec{w} = \cos \theta \hat{i} + \sin \theta \hat{j} \). Compute the dot product of \( \vec{u} \times \vec{v} \) and \( \vec{w} \):\[ (3\hat{i} + 4\hat{j} - \hat{k}) \cdot (\cos \theta \hat{i} + \sin \theta \hat{j}) = 3\cos \theta + 4\sin \theta \]The term involving \( \hat{k} \) in \( \vec{u} \times \vec{v} \) does not contribute since \( \vec{w} \) has no \( \hat{k} \) component.

Step 3: Maximize the Expression \( |3\cos \theta + 4\sin \theta| \)

We need to maximize the magnitude of the expression \( |3\cos \theta + 4\sin \theta| \).This can be rewritten using the trigonometric identity:\[ a\cos \theta + b\sin \theta = \sqrt{a^2 + b^2} \cos(\theta - \phi) \]where \( \tan \phi = \frac{b}{a} \).Here, \( a = 3 \) and \( b = 4 \), so\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]Thus, the maximum possible value of \( |3\cos \theta + 4\sin \theta| \) is 5.

Key Concepts

Cross Product

The cross product is an essential operation in vector calculus. It results in a vector that is perpendicular to the plane formed by two given vectors in three-dimensional space. To calculate the cross product of two vectors, say \( \vec{u} \) and \( \vec{v} \), you use the determinant of a 3x3 matrix. This matrix includes unit vectors for the axes \( \hat{i}, \hat{j}, \hat{k} \), and the components of the respective vectors. For example, for \( \vec{u} = \hat{j} + 4\hat{k} \) and \( \vec{v} = \hat{i} + 3\hat{k} \), the cross product \( \vec{u} \times \vec{v} \) is calculated as follows:- Create a determinant where the first row consists of the unit vectors \( \hat{i}, \hat{j}, \hat{k} \).- The second row comprises the components of \( \vec{u} \), i.e., \( 0 \) for \( \hat{i} \), \( 1 \) for \( \hat{j} \), and \( 4 \) for \( \hat{k} \).- The third row contains components of \( \vec{v} \), which are \( 1 \) for \( \hat{i} \), \( 0 \) for \( \hat{j} \), and \( 3 \) for \( \hat{k} \).Upon solving the determinant, the resulting vector \( 3\hat{i} + 4\hat{j} - \hat{k} \) is perpendicular to both \( \vec{u} \) and \( \vec{v} \). This operation is frequently used in physics and engineering to find normals to surfaces.

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar value. It's a measure of the extent to which two vectors point in the same direction. To compute the dot product of two vectors, you multiply their respective components and sum these products.For our vectors, consider \( \vec{u} \times \vec{v} = 3\hat{i} + 4\hat{j} - \hat{k} \) and \( \vec{w} = \cos \theta \hat{i} + \sin \theta \hat{j} \). The dot product \( (\vec{u} \times \vec{v}) \cdot \vec{w} \) simplifies as follows:- Only components with matching directions are multiplied.- Multiply the \( \hat{i} \) components: \( 3 \times \cos \theta \).- Multiply the \( \hat{j} \) components: \( 4 \times \sin \theta \).Add the results to yield \( 3\cos \theta + 4\sin \theta \). Note that the \( \hat{k} \) component does not come into play since \( \vec{w} \) lacks a \( \hat{k} \) component. Dot products are crucial for finding projections and determining angles between vectors.

Trigonometric Identities

Trigonometric identities are pivotal tools in simplifying and solving expressions involving trigonometric functions. When maximizing or minimizing expressions like \( 3\cos \theta + 4\sin \theta \), particular trigonometric identities make the process straightforward. The expression \( a\cos \theta + b\sin \theta \) can be rewritten using the identity: - \[ a\cos \theta + b\sin \theta = \sqrt{a^2 + b^2} \cos(\theta - \phi) \]where \( \tan \phi = \frac{b}{a} \) and \( \phi \) is an angle that aligns the combined effect of both sine and cosine terms into a single cosine wave.Here, by setting \( a = 3 \) and \( b = 4 \), you find:- \[ \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \]This identity shows that the maximum magnitude of \( 3\cos \theta + 4\sin \theta \) is indeed 5, attained when \( \cos(\theta - \phi) = 1 \). Such identities are immensely helpful in physics and engineering to transform and evaluate periodic functions.