Area of Triangle and Parallelogram

Area of Triangle and Parallelogram

Welcome, JEE aspirants! In this lesson, we'll explore how the cross product, which we learned in the previous lesson, becomes a powerful tool for calculating areas of parallelograms and triangles in 2D and 3D space. This isn't just abstract theory; these concepts are frequently tested in JEE Main, often combined with other vector algebra and 3D geometry principles. Mastering this will give you a significant edge!

Conceptual Explanation

Imagine two vectors, $\vec{a}$ and $\vec{b}$, originating from the same point. They define a parallelogram. The magnitude of the cross product, $|\vec{a} \times \vec{b}|$, gives us the area of this parallelogram. Why? Recall that $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin{\theta}$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. $|\vec{b}| \sin{\theta}$ is precisely the height of the parallelogram if we consider $\vec{a}$ as the base. Therefore, $|\vec{a}| |\vec{b}| \sin{\theta}$ is base times height, which is the area of the parallelogram.

Now, a triangle formed by $\vec{a}$ and $\vec{b}$ as adjacent sides is exactly half of the parallelogram. Hence, the area of the triangle is $(1/2)|\vec{a} \times \vec{b}|$. This simple geometric relationship makes the cross product incredibly useful.

What if you're given the position vectors of the vertices of a triangle? No problem! Just create the vectors representing the sides using those position vectors, and then apply the same formula.

Finally, the cross product also gives us a vector perpendicular to both $\vec{a}$ and $\vec{b}$. By normalizing this vector (dividing by its magnitude), we get a unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. This is crucial in many 3D geometry problems.

Important Formulas

1. Area of parallelogram with adjacent sides $\vec{a}$, $\vec{b}$

As explained earlier, this formula directly relates the magnitude of the cross product to the area. Remember, $\vec{a}$ and $\vec{b}$ must be adjacent sides, meaning they share a common vertex.

2. Area of triangle with adjacent sides $\vec{a}$, $\vec{b}$

This is a direct consequence of the parallelogram area. The triangle is simply half the parallelogram formed by the same adjacent sides.

3. Area of triangle ABC with position vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ for vertices A, B, C respectively

Here, $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are the vectors representing the sides AB and AC of the triangle. We obtain these vectors by subtracting the position vector of the initial point from the position vector of the terminal point. For example, $\overrightarrow{AB} = \vec{b} - \vec{a}$.

4. Unit vector perpendicular to both $\vec{a}$ and $\vec{b}$

The cross product $\vec{a} \times \vec{b}$ results in a vector perpendicular to both $\vec{a}$ and $\vec{b}$. To make it a unit vector, we divide by its magnitude. Remember that there are two unit vectors perpendicular to $\vec{a}$ and $\vec{b}$ – this one and its negative. The "right-hand rule" will help you determine the direction of $\vec{a} \times \vec{b}$.

Example: Find the area of the triangle with vertices A(1, 1, 1), B(1, 2, 3), and C(2, 3, 1). First, find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. $\overrightarrow{AB} = \vec{b} - \vec{a} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + 1\hat{j} + 2\hat{k}$ $\overrightarrow{AC} = \vec{c} - \vec{a} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = 1\hat{i} + 2\hat{j} + 0\hat{k}$ Now, calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$: $$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = (0 - 4)\hat{i} - (0 - 2)\hat{j} + (0 - 1)\hat{k} = -4\hat{i} + 2\hat{j} - \hat{k}$$ Next, find the magnitude: $|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-4)^2 + (2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$ Finally, the area of the triangle is: $$Area = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{\sqrt{21}}{2}$$

Example: Find a unit vector perpendicular to the vectors $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{j} + \hat{k}$. First, find the cross product $\vec{a} \times \vec{b}$: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = (1 - 0)\hat{i} - (1 - 0)\hat{j} + (1 - 0)\hat{k} = \hat{i} - \hat{j} + \hat{k}$$ Now, find the magnitude: $|\vec{a} \times \vec{b}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{3}$ Finally, the unit vector is: $$\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{1}{\sqrt{3}}\hat{i} - \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$$

By understanding the relationship between the cross product and area, and by practicing these formulas, you'll be well-prepared to tackle any JEE Main problem on this topic. Good luck!