Problem 87 - A Complete Resource Book in Mathematics for JEE Main

Question

For any vector \(\vec{a}\), the value of \((\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2}\) is equal to (A) \(3 \vec{a}^{2}\) (B) \(\vec{a}^{2}\) (C) \(2 \vec{a}^{2}\) (D) \(4 \vec{a}^{2}\)

Step-by-Step Solution

Answer

The expression simplifies to \(2 \vec{a}^{2}\), hence the correct option is (C).

Step 1: Understanding the Problem

We need to evaluate the expression \((\vec{a} \times \hat{i})^2 + (\vec{a} \times \hat{j})^2 + (\vec{a} \times \hat{k})^2\), where \(\vec{a}\) is any vector, and \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors along the x, y, and z axes. The task is to express this in terms of \(\vec{a}^2\) and choose the correct option from the given choices.

Step 2: Vector Representation

Let the vector \(\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\), where \(a_x, a_y, a_z\) are the components of \(\vec{a}\) along the x, y, and z axes respectively.

Step 3: Calculate \(\vec{a} \times \hat{i}\)

The cross product \(\vec{a} \times \hat{i} = (0 \cdot (a_z \cdot \hat{k}) - a_y \cdot (1 \cdot \hat{k}), a_x \cdot (0) - a_z \cdot (1), a_y \cdot (1) - a_x \cdot (0)) \). This simplifies to \([-a_y \hat{k} + a_z \hat{j}]\).

Step 4: Calculate \((\vec{a} \times \hat{i})^2\)

The square of a vector \([-a_y \hat{k} + a_z \hat{j}]\) is its dot product with itself. So, \((\vec{a} \times \hat{i})^2 = (-a_y)^2 + (a_z)^2 = a_y^2 + a_z^2\).

Step 5: Calculate \(\vec{a} \times \hat{j}\)

The cross product \(\vec{a} \times \hat{j} = a_z \hat{i} - a_x \hat{k}\).

Step 6: Calculate \((\vec{a} \times \hat{j})^2\)

The square of a vector \(a_z \hat{i} - a_x \hat{k}\) is \((a_z)^2 + (-a_x)^2 = a_z^2 + a_x^2\).

Step 7: Calculate \(\vec{a} \times \hat{k}\)

The cross product \(\vec{a} \times \hat{k} = a_x \hat{j} - a_y \hat{i}\).

Step 8: Calculate \((\vec{a} \times \hat{k})^2\)

The square of the vector \(a_x \hat{j} - a_y \hat{i}\) is \((a_x)^2 + (-a_y)^2 = a_x^2 + a_y^2\).

Step 9: Sum the Results

Add the results from Steps 4, 6, and 8: \((a_y^2 + a_z^2) + (a_z^2 + a_x^2) + (a_x^2 + a_y^2) = 2(a_x^2 + a_y^2 + a_z^2) = 2 \vec{a}^2\).

Step 10: Conclusion

The expression \((\vec{a} \times \hat{i})^2 + (\vec{a} \times \hat{j})^2 + (\vec{a} \times \hat{k})^2\) simplifies to \(2 \vec{a}^2\). Therefore, the correct choice is (C) \(2 \vec{a}^2\).

Key Concepts

Cross Product

The cross product, often denoted by \( \vec{a} \times \vec{b} \), represents a vector that is perpendicular to both vectors \( \vec{a} \) and \( \vec{b} \). To determine the cross product, we can use the determinant of a 3x3 matrix involving the unit vectors \( \hat{i}, \hat{j}, \hat{k} \):

• The top row holds these unit vectors.
• The second row contains the components of vector \( \vec{a} \).
• The third row has the components of vector \( \vec{b} \).

Evaluating the determinant gives you the cross product. In simple terms, if \( \vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \) and \( \vec{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k} \), then the cross product \( \vec{a} \times \vec{b} \) will yield another vector. This new vector has components calculated as:

• \( (a_yb_z - a_zb_y) \hat{i} \)
• \( -(a_xb_z - a_zb_x) \hat{j} \)
• \( (a_xb_y - a_yb_x) \hat{k} \)

Understanding cross products helps in finding areas of parallelograms defined by two vectors and in understanding rotational vectors.

Vector Magnitude

The magnitude of a vector, sometimes referred to as the length or norm of the vector, is a measure of how long the vector is. For a vector \( \vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \), its magnitude \( |\vec{a}| \) is found using the formula:\[|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}\]The magnitude of a vector allows us to equate it in scale without regard to direction. It is always a non-negative scalar. In many applications, knowing the vector magnitude is essential, such as when normalizing a vector to determine its unit form.

Unit Vectors

Unit vectors are vectors with a magnitude of one. They are used to specify direction. In three-dimensional space, the most commonly used unit vectors are \( \hat{i}, \hat{j}, \hat{k} \), representing the x, y, and z axes, respectively.

• \( \hat{i} = (1, 0, 0) \)
• \( \hat{j} = (0, 1, 0) \)
• \( \hat{k} = (0, 0, 1) \)

To convert any vector \( \vec{a} \) into a unit vector \( \hat{a} \), divide the vector by its magnitude:\[\hat{a} = \frac{\vec{a}}{|\vec{a}|}\]Unit vectors are crucial in vector calculations, simplifying the mathematics and helping with direction-focused studies like physics and engineering.

Vector Components

Vector components are the projections of a vector along the axes of a coordinate system. They define the vector in terms of its directional influences along each axis. For a vector \( \vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \), \( a_x, a_y, \) and \( a_z \) are its components.

• These components clarify how the vector moves along the x, y, and z directions.
• Understanding components simplifies vector addition, subtraction, and scalar multiplication.

For example, decomposing a force into vector components allows easier analysis of how this force operates within a mechanical system. Simply put, vector components allow us to work with vectors by reducing three-dimensional problems to simpler one-dimensional issues.