Problem 61 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\alpha=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) and \(\vec{\beta}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\). If \(\vec{\beta}=\vec{\beta}_{1}-\vec{\beta}_{2}\), where \(\vec{\beta}_{1}\) is parallel to \(\bar{\alpha}\) and \(\vec{\beta}_{2}\) is perpendicular to \(\bar{\alpha}\), then \(\vec{\beta}_{1} \times \vec{\beta}_{2}\) is equal to: \(\quad\) [April 09, 2019 (I)] (a) \(-3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) (b) \(3 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) (c) \(\frac{1}{2}(-3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})\) (d) \(\frac{1}{2}(3 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})\)

Step-by-Step Solution

Answer

Option (c): \( \frac{1}{2}(-3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \) is correct.

Step 1: Understand the Vectors

We have two vectors \( \vec{\alpha} = 3 \hat{\mathrm{i}} + \hat{\mathrm{j}} \) and \( \vec{\beta} = 2 \hat{\mathrm{i}} - \hat{\mathrm{j}} + 3 \hat{\mathrm{k}} \). The task is to express \( \vec{\beta} \) as \( \vec{\beta}_1 - \vec{\beta}_2 \) where \( \vec{\beta}_1 \) is parallel to \( \vec{\alpha} \) and \( \vec{\beta}_2 \) is perpendicular to \( \vec{\alpha} \).

Step 2: Projection of \( \vec{\beta} \) onto \( \vec{\alpha} \)

To find \( \vec{\beta}_1 \), we project \( \vec{\beta} \) onto \( \vec{\alpha} \) using the formula: \( \vec{\beta}_1 = \frac{\vec{\beta} \cdot \vec{\alpha}}{\vec{\alpha} \cdot \vec{\alpha}} \vec{\alpha} \). First, we compute \( \vec{\beta} \cdot \vec{\alpha} = 2 \times 3 + (-1) \times 1 + 3 \times 0 = 5 \) and \( \vec{\alpha} \cdot \vec{\alpha} = 3^2 + 1^2 = 10 \). Therefore, \( \vec{\beta}_1 = \left(\frac{5}{10}\right)(3 \hat{\mathrm{i}} + \hat{\mathrm{j}}) = \frac{1}{2}(3 \hat{\mathrm{i}} + \hat{\mathrm{j}}) = \frac{3}{2} \hat{\mathrm{i}} + \frac{1}{2} \hat{\mathrm{j}} \).

Step 3: Calculate \(\vec{\beta}_2\)

Since \( \vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2 \), we have \( \vec{\beta}_2 = \vec{\beta}_1 - \vec{\beta} \). Therefore, \( \vec{\beta}_2 = \frac{3}{2} \hat{\mathrm{i}} + \frac{1}{2} \hat{\mathrm{j}} - (2 \hat{\mathrm{i}} - \hat{\mathrm{j}} + 3 \hat{\mathrm{k}}) = (-\frac{1}{2} \hat{\mathrm{i}} + \frac{3}{2} \hat{\mathrm{j}} - 3 \hat{\mathrm{k}}) \).

Step 4: Calculate \(\vec{\beta}_1 \times \vec{\beta}_2\)

Use the cross-product formula to find \( \vec{\beta}_1 \times \vec{\beta}_2 \):\[ \vec{\beta}_1 \times \vec{\beta}_2 = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ \frac{3}{2} & \frac{1}{2} & 0 \ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix} \]Expanding the determinant:\[ = \hat{\mathrm{i}} \left( \frac{1}{2}(-3) - 0 \cdot \frac{3}{2} \right) - \hat{\mathrm{j}} \left( \frac{3}{2}(-3) - 0 \cdot -\frac{1}{2} \right) + \hat{\mathrm{k}} \left( \frac{3}{2} \cdot \frac{3}{2} - \frac{1}{2}\cdot-\frac{1}{2} \right) \]\[ = \hat{\mathrm{i}} (-\frac{3}{2}) - \hat{\mathrm{j}} (-\frac{9}{2}) + \hat{\mathrm{k}} \left( \frac{9}{4} + \frac{1}{4} \right) \]\[ = -\frac{3}{2} \hat{\mathrm{i}} + \frac{9}{2} \hat{\mathrm{j}} + 2.5 \hat{\mathrm{k}} \]Evaluating gives \( \vec{\beta}_1 \times \vec{\beta}_2 = -\frac{3}{2} \hat{\mathrm{i}} + \frac{9}{2} \hat{\mathrm{j}} + 2.5 \hat{\mathrm{k}} \).

Step 5: Compare with Options

The result of the cross-product \( -\frac{3}{2} \hat{\mathrm{i}} + \frac{9}{2} \hat{\mathrm{j}} + 2.5 \hat{\mathrm{k}} \) simplifies as \( \frac{1}{2}(-3 \hat{\mathrm{i}} + 9 \hat{\mathrm{j}} + 5 \hat{\mathrm{k}}) \). This matches option (c).

Key Concepts

Vector Projection

When working with vectors, projecting one vector onto another is a common task. To visualize this, imagine shining a flashlight from one vector onto another, casting a shadow on it. This shadow is the projection. The vector projection is used to find a component of one vector along another vector's direction. For instance, if you have two vectors, like \(\vec{\alpha}\) and \(\vec{\beta}\), the projection helps to see how much of \(\vec{\beta}\)'s direction aligns with \(\vec{\alpha}\).

The formula for projecting \(\vec{\beta}\) onto \(\vec{\alpha}\) is:

• \( \vec{\beta}_1 = \frac{\vec{\beta} \cdot \vec{\alpha}}{\vec{\alpha} \cdot \vec{\alpha}} \vec{\alpha} \)

Where \(\vec{\beta} \cdot \vec{\alpha}\) is the dot product of \(\vec{\beta}\) and \(\vec{\alpha}\). This formula shows how to calculate the vector that lies along \(\vec{\alpha}\) with the same length proportionate to \(\vec{\beta}\). Through this, we find components of vectors parallel and perpendicular to other vectors.

Vector Cross Product

The cross product of two vectors results in a vector that is perpendicular to both of the original vectors. Unlike the dot product which gives a scalar, the cross product gives a vector. The formula for the cross product \( \vec{a} \times \vec{b} \) involves determinants and is essential in finding a vector perpendicular to a plane defined by two vectors.

In the example given, we used the cross product to compute \( \vec{\beta}_1 \times \vec{\beta}_2 \). The calculation involves setting up a determinant with:

• The unit vectors \( \hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}} \)
• The components of \(\vec{\beta}_1\) and \(\vec{\beta}_2\) as rows beneath these unit vectors.

This determinant evaluates to a vector that is orthogonal to both input vectors.

The result indicates the relative orientation and length of the perpendicular vector generated, which is used in various applications such as calculating torque, and determining axis of rotations.

Parallel and Perpendicular Vectors

Understanding parallel and perpendicular vectors is key to vector algebra. Two vectors are parallel if they point in the same direction or exactly opposite. Mathematically, this occurs when one vector is a scalar multiple of the other. For example, \( \vec{\beta}_1 \) is parallel to \( \vec{\alpha} \) after we calculate its projection.

Perpendicular vectors, on the other hand, have a dot product of zero. They meet at right angles to each other, indicating no component of one exists in the direction of the other. When \( \vec{\beta}_2 \) is derived by having the projection reduced from \( \vec{\beta} \), it becomes perpendicular to \( \vec{\alpha} \).

This concept helps in breaking vectors into components, such as resolving forces in physics or achieving certain vector transformations in computer graphics. Recognizing parallel and perpendicular vector relationships can simplify problems greatly.

Determinants in Vector Calculations

Determinants appear frequently in vector calculations, particularly in cross products. They help in organizing and computing vector operations. A determinant provides a scalar value derived from a square matrix, which in three-dimensional space can simplify to a pattern used to find areas, volumes, and cross products.

For two vectors \( \vec{a} \) and \( \vec{b} \), the determinant set up in the cross product \( \vec{a} \times \vec{b} \) is a 3x3 matrix including:

• Top row: The unit vectors \( \hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}} \)
• Subsequent rows: The components of \(\vec{a}\) and \(\vec{b}\).

Solving this provides the needed vector perpendicular to both.

These calculations are vital in many fields such as physics for calculating rotational effects, and in computer graphics to determine normal vectors for surfaces. Determinants offer a structured approach to handle such three-dimensional problems easily.