Problem 39 - Master Resource Book in JEE Main Physics

Question

The adjacent sides of a parallelogram are represented by co-initial vectors \(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}\) and \(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}\) The area of the parallelogram is (a) 5 units along \(z\)-axis (b) 5 units in \(x-y\) plane (c) 3 units in \(x-z\) plane (d) 3 units in \(y-z\) plane

Step-by-Step Solution

Answer

(b) 5 units in x-y plane

Step 1: Understanding the Formula for Area of Parallelogram

The area of a parallelogram formed by two vectors \( \vec{a} \) and \( \vec{b} \) is given by the magnitude of their cross product: \(|\vec{a} \times \vec{b}|\).

Step 2: Identify the Vectors

We are given the vectors \( \vec{a} = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}} \) and \( \vec{b} = \hat{\mathbf{i}} + 4\hat{\mathbf{j}} \).

Step 3: Calculate the Cross Product

The cross product between two vectors in three dimensions \( \vec{a} = a_1\hat{\mathbf{i}} + a_2\hat{\mathbf{j}} + a_3\hat{\mathbf{k}} \) and \( \vec{b} = b_1\hat{\mathbf{i}} + b_2\hat{\mathbf{j}} + b_3\hat{\mathbf{k}} \) is given by the determinant:\[\vec{a} \times \vec{b} = \begin{vmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \2 & 3 & 0 \1 & 4 & 0\end{vmatrix} = (3\cdot0 - 0\cdot4)\hat{\mathbf{i}} - (2\cdot0 - 0\cdot1)\hat{\mathbf{j}} + (2\cdot4 - 3\cdot1)\hat{\mathbf{k}}\]This simplifies to \(\vec{a} \times \vec{b} = 5\hat{\mathbf{k}}\).

Step 4: Find the Magnitude of the Cross Product

The magnitude of \(\vec{a} \times \vec{b} = 5\hat{\mathbf{k}}\) is just \(|5| = 5\).

Step 5: Determine the Orientation of the Parallelogram

The vector \(5\hat{\mathbf{k}}\) indicates that the area vector is oriented along the \(z\)-axis. This means the parallelogram lies in the \(x-y\) plane.

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus that allows us to find a vector that is perpendicular to two given vectors in three-dimensional space. It is denoted by the symbol \( \times \), which reads as "cross." The resulting vector from the cross product has both a direction and a magnitude.

Here’s how the cross product works:

• If you have vectors \( \vec{a} = a_1\hat{\mathbf{i}} + a_2\hat{\mathbf{j}} + a_3\hat{\mathbf{k}} \) and \( \vec{b} = b_1\hat{\mathbf{i}} + b_2\hat{\mathbf{j}} + b_3\hat{\mathbf{k}} \), their cross product \( \vec{a} \times \vec{b} \) is calculated using the determinant of a 3x3 matrix.
• The standard formula for the cross product is given by:\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\hat{\mathbf{i}} - (a_1b_3 - a_3b_1)\hat{\mathbf{j}} + (a_1b_2 - a_2b_1)\hat{\mathbf{k}}.\]
• This results in a new vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \).

This operation is crucial for many applications, including finding the area of shapes like parallelograms and triangles when their sides are represented as vectors.

Area of Parallelogram

When studying vectors and shapes in geometry, finding the area of a parallelogram is a practical application of the cross product. The area is found using the magnitude of the cross product of the two vectors representing the adjacent sides of the parallelogram.

Here's a simple process to calculate it:

• Let's say we have two vectors \( \vec{a} \) and \( \vec{b} \). Their cross product, \( \vec{a} \times \vec{b} \), gives us a vector perpendicular to both.
• To find the area, we take the magnitude of this cross product vector, denoted by \(|\vec{a} \times \vec{b}|\).
• The formula for the magnitude |a vector| is given by \( \sqrt{(i ext{'s coefficient})^2 + (j ext{'s coefficient})^2 + (k ext{'s coefficient})^2} \).

In our given exercise, after finding the cross product as \( 5\hat{\mathbf{k}} \), the magnitude is \( |5| = 5 \). Therefore, the area of the parallelogram is 5 square units.

Coordinate Geometry

Coordinate geometry combines algebra and geometry using coordinates for precise calculations. It involves using a coordinate plane, typically the \( x \), \( y \), and \( z \)-axes, to define positions and solve geometric problems.

In the context of vectors:

• Vectors originate from the origin and are described in terms of their components along the \( x \), \( y \), and \( z \)-axes.
• For example, the vector \( 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}} \) in our exercise translates to a position in the plane by moving 2 units along the \( x \)-axis and 3 units along the \( y \)-axis.
• This allows for geometrical constructs like lines, planes, and shapes like parallelograms being analyzed using algebraic methods.

Coordinate geometry effectively allows the visualization and calculation of vector relationships and shapes in a structured and organized manner.