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If the vectors a and b form two of the sides of a triangle, prove that \(\frac{1}{2}|\mathbf{a} \times \mathbf{b}|\) is equal to the area of the triangle.

Short Answer

Expert verified
The area of the triangle is \( \frac{1}{2} |\mathbf{a} \times \mathbf{b}| \).

Step by step solution

01

Understand the Vector Cross Product

The cross product \( \mathbf{a} \times \mathbf{b} \) of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) in space is a vector that is perpendicular to both, and its magnitude is equal to the area of the parallelogram formed by \( \mathbf{a} \) and \( \mathbf{b} \).
02

Formula for the Area of a Parallelogram

The formula for the area of a parallelogram is given by the magnitude of the cross product of its adjacent sides, expressed as \( |\mathbf{a} \times \mathbf{b}| \).
03

Relate the Parallelogram to a Triangle

When \( \mathbf{a} \) and \( \mathbf{b} \) form two sides of a triangle, the triangle is exactly half of the parallelogram formed by \( \mathbf{a} \) and \( \mathbf{b} \).
04

Calculate the Area of the Triangle

Since the triangle is half of the parallelogram, the area of the triangle is \( \frac{1}{2} \times \text{area of parallelogram} = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| \).
05

Conclusion

By understanding and using the properties of vector cross product and parallelogram geometry, we've shown that the area of the triangle is \( \frac{1}{2} |\mathbf{a} \times \mathbf{b}| \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area of a Triangle
To calculate the area of a triangle formed by two vectors, we use the cross product. Imagine you have two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), forming two sides of a triangle. The cross product \( \mathbf{a} \times \mathbf{b} \) gives us a new vector whose magnitude represents the area of a parallelogram formed by \( \mathbf{a} \) and \( \mathbf{b} \).
Since a triangle is half of a parallelogram, the area of the triangle is half of the magnitude of the cross product. That's why the area is expressed as \( \frac{1}{2} |\mathbf{a} \times \mathbf{b}| \).
Thus, when trying to find the area of a triangle using vectors, think of it as finding half the area of the parallelogram they could form if extended.
Parallelogram Geometry
Parallelogram geometry is key to understanding the connection between vectors and the area of shapes they can form. A parallelogram has opposite sides that are equal in length and parallel, with but not necessarily congruent angles.
When two vectors are placed tail-to-tail, they create a shape that visually resembles a parallelogram. By using the cross product \( \mathbf{a} \times \mathbf{b} \), you determine a vector perpendicular to the plane containing \( \mathbf{a} \) and \( \mathbf{b} \), and its magnitude evaluates the area of that parallelogram.
Understanding the geometry of parallelograms helps in visualizing why the triangle formed by the same vectors is exactly half of this shape and thus its area is half the area given by the vector magnitude of the cross product.
Vector Magnitude
The magnitude of a vector is a measure of its length. For any vector \( \mathbf{v} \), the magnitude is denoted as \( |\mathbf{v}| \). It can be thought of as the size or strength of the vector in a geometric sense.
When computing the cross product of vectors \( \mathbf{a} \) and \( \mathbf{b} \), \( |\mathbf{a} \times \mathbf{b}| \) represents the area of the parallelogram they form. This is because the magnitude takes into account both the length of the vectors and the angle between them, which corresponds to the height of the parallelogram if the base is one of the vectors.
  • Therefore, understanding and calculating the magnitude of a vector is crucial when assessing the areas that vectors can encapsulate.
Vector Perpendicularity
In vector mathematics, perpendicular vectors play a pivotal role in numerous calculations, including the cross product. When you compute the cross product of two vectors, the result is a third vector that is perpendicular to the plane formed by the first two vectors.
This perpendicularity is essential when considering the geometry of shapes such as parallelograms and triangles. The vector resulting from \( \mathbf{a} \times \mathbf{b} \) not only helps in determining the area but also gives a direction relative to the original vectors' plane.
  • The perpendicular direction determined by the cross product is often utilized in physics and engineering to find normal forces, torques, and other vector-based quantities.
Understanding this perpendicular relationship helps reinforce how vector cross products can translate into practical geometric and physical applications.

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Most popular questions from this chapter

Many applications of conservation of momentum involve conservation of energy as well, and we haven't yet begun our discussion of energy. Nevertheless, you know enough about energy from your introductory physics course to handle some problems of this type. Here is one elegant example: An elastic collision between two bodies is defined as a collision in which the total kinetic energy of the two bodies after the collision is the same as that before. (A familiar example is the collision between two billiard balls, which generally lose extremely little of their total kinetic energy.) Consider an elastic collision between two equal mass bodies, one of which is initially at rest. Let their velocities be \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}=0\) before the collision, and \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) after. Write down the vector equation representing conservation of momentum and the scalar equation which expresses that the collision is elastic. Use these to prove that the angle between \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) is \(90^{\circ} .\) This result was important in the history of atomic and nuclear physics: That two bodies emerged from a collision traveling on perpendicular paths was strongly suggestive that they had equal mass and had undergone an elastic collision.

A shell traveling with speed \(v_{\mathrm{o}}\) exactly horizontally and due north explodes into two equal-mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed \(v_{\mathrm{o}} .\) What is the velocity of the other fragment?

Find the position of the center of mass of three particles lying in the \(x y\) plane at \(\mathbf{r}_{1}=(1,1,0)\) \(\mathbf{r}_{2}=(1,-1,0),\) and \(\mathbf{r}_{3}=(0,0,0),\) if \(m_{1}=m_{2}\) and \(m_{3}=10 m_{1} .\) Illustrate your answer with a sketch and comment.

The first couple of minutes of the launch of a space shuttle can be described very roughly as follows: The initial mass is \(2 \times 10^{6} \mathrm{kg}\), the final mass (after 2 minutes) is about \(1 \times 10^{6} \mathrm{kg}\), the average exhaust speed \(v_{\mathrm{ex}}\) is about \(3000 \mathrm{m} / \mathrm{s},\) and the initial velocity is, of course, zero. If all this were taking place in outer space, with negligible gravity, what would be the shuttle's speed at the end of this stage? What is the thrust during the same period and how does it compare with the initial total weight of the shuttle (on earth)?

In the early stages of the Saturn V rocket's launch, mass was ejected at about \(15,000 \mathrm{kg} / \mathrm{s}\), with a speed \(v_{\mathrm{ex}} \approx 2500 \mathrm{m} / \mathrm{s}\) relative to the rocket. What was the thrust on the rocket? Convert this to tons (1 ton \(\approx 9000\) newtons) and compare with the rocket's initial weight (about 3000 tons).

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