Areas Of Parallelograms And Triangles Using Cross Products

This page will be about finding areas of parallelograms and triangles using vectors, norms and the cross product.


Table Of Contents


Geometric Intuition

Suppose we have two vectors in three space as shown in the picture below (I use vectors \textbf{u} and \textbf{v} while the picture uses \textbf{v} and \textbf{w}.

In the vector setting, distances are taken by using norms. In this case, we would have the norms ||\textbf{v}|| and ||\textbf{w}|| for vectors \textbf{v} and \textbf{w} respectively.

On one side of the parallelogram (the left side in this case), we draw an altitude line to the base which forms a triangle. In this triangle, the norm ||\textbf{v}|| is the hypotenuse and the angle \theta (with 0 \leq \theta \leq 180 degrees) is the angle of the parallelogram and the triangle.

We seek to find the side length opposite to the angle \theta. Using trigonometry (SOH CAH TOA) we have:

    \[\sin(\theta) = \dfrac{Altitude}{||\textbf{v}||}\]

and

    \[Altitude = ||\textbf{v}|| \sin(\theta)\]


Deriving The Area Of A Parallelogram Using Cross Products

Recall that one of the properties of cross products and dot products is Lagrange’s identity.

    \[||\textbf{u} \times \textbf{v}||^{2} = ||\textbf{u}||^2 ||\textbf{v}||^2 - (\textbf{u} \cdot \textbf{v})^2 \text{ (Lagrange's Identity) }\]

In addition, given an angle \theta with 0 \leq \theta \leq 180 degrees we have the dot product equation

    \[(\textbf{u} \cdot \textbf{v}) = ||\textbf{u}||||\textbf{v}|| \cos(\theta)\]

In Lagrange’s identity, we substitute \textbf{u} \cdot \textbf{v} with the above equation, factor and the equation \sin^{2}(\theta) + \cos^{2}(\theta) = 1 (or 1 - \cos^{2}(\theta) = \sin^{2}(\theta)) to get:

Source: http://quicklatex.com/cache3/80/ql_3cd4acd1e64c95db24247596da7a8280_l3.png

 

From ||\textbf{u} \times \textbf{v}||^{2} = ||\textbf{u}||^2 ||\textbf{v}||^2 \sin^{2}(\theta), we can take the square root of both sides to obtain

    \[||\textbf{u} \times \textbf{v}|| = ||\textbf{u}|| ||\textbf{v}|| \sin(\theta) \text{ .}\]


The Area Of The Parallelogram

The area of a parallelogram is the product of the base length (||\textbf{u}||) times the altitude. This formula can be expressed as:

    \[Area_{Paraellogram} = (Base) (Altitude) = ||\textbf{u}|| ||\textbf{v}|| \sin(\theta)\]

The above formula can also be expressed as Area_{Paraellogram} = (Base) (Altitude) = ||\textbf{u} \times \textbf{v}|| (because of the derivation earlier).


The Area Of A Triangle

It is known that the area of a triangle is half the area of a paraellogram. Using cross products and norms, the formula for the area of a triangle is:

    \[Area_{Triangle} = \dfrac{1}{2} (Base)(Altitude) = \dfrac{1}{2} ||\textbf{u}|| ||\textbf{v}|| \sin(\theta)\]


Examples

Example One

Given the vectors \textbf{u} = (1, -2, 5) and \textbf{v} = (2, 0, -1), Find the area of the parallelogram enclosed by these two vectors.

Answer

Given the two vectors \textbf{u} and \textbf{v}, we find the cross product \textbf{u} \times \textbf{v} first. The norm of this cross product will be calculated to obtain the area of the parallelogram enclosed by the two vectors.

One can show that the cross product \textbf{u} \times \textbf{v} is (2, 11, 4).

Taking the norm of this product yields:

    \[||\textbf{u} \times \textbf{v}|| = \sqrt{(2)^2 + (11)^2 + (4)^2} = \sqrt{4 + 121+ 16} = \sqrt{141} \approx 11.87 \text{ square units }\]


Example Two

Consider the points A (0, 1, 0), B (2,-2, 5) and C (3, -1, 5). Find the area of the triangle determined by the three points.

Answer

The strategy is to create two vectors from the three points, find the cross product of the two vectors and then take the half the norm of the cross product.

One vector is \overrightarrow{AB} = (2 - 0, -2 - 1, 5 - 0) = (2, -3, 5). The second vector would be \overrightarrow{AC} = (3 - 0, -1 - 1, 5 - 0) = (3, -2, 5)

One can show that the cross product of the vectors \overrightarrow{AB} and \overrightarrow{AC} is (-5, 5, 5).

The norm of this cross product computes to:

    \[||(-5, 5, 5)|| = \sqrt{(-5)^{2} + (5)^{2} + (5)^{2}} = \sqrt{75}\]

The area of a triangle is half the norm of the cross product. It computes to:

    \[Area_{Triangle} = \dfrac{\sqrt{75}}{2} \approx 4.33\]


Practice Problems 

Here are few practice problems. Solutions are in the next section.

1) Given the vectors \textbf{u} = (0, -1, 3) and \textbf{v} = (5, 2, -2), Find the area of the parallelogram enclosed by these two vectors.

2) Given the vectors \textbf{u} = (1, -2, 7) and \textbf{v} = (9, 0, -1), Find the area of the triangle enclosed by these two vectors.

3) Find the area of the triangle determined by the points A (-2, 2, 1), B (3,0, 1) and C (0, -1, 0).


Solutions

1) \textbf{u} \times \textbf{v} = (-4, 15, 5). The area of the parallelogram computes to \sqrt{266} \approx 16.31 square units.

2) \textbf{u} \times \textbf{v} = (2, 64, 18). The area of the triangle computes to \dfrac{\sqrt{4424}}{2} \approx 33.26 square units.

3) \overrightarrow{AB} = (5, -2, 0), \overrightarrow{AC} = (2, -3, 1); \overrightarrow{AB} \times \overrightarrow{AC} = (2, 5, -11); Area_{Triangle} = \dfrac{\sqrt{150}}{2} \approx 6.12 square units.


References

Elementary Linear Algebra (Tenth Edition) by Howard Anton

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