/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Show that the area of the triang... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that the area of the triangle contained between the vectors a and \(\mathbf{b}\) is one half of the magnitude of \(\mathbf{a} \times \mathbf{b}\).

Short Answer

Expert verified
The area of the triangle is \(\frac{1}{2}|\mathbf{a} \times \mathbf{b}|\), which confirms it is half the cross product's magnitude.

Step by step solution

01

Understand the vector cross product

The vector cross product \( \mathbf{a} \times \mathbf{b} \) results in a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), and its magnitude is equal to \( |\mathbf{a}| \times |\mathbf{b}| \times \sin(\theta) \), where \( \theta \) is the angle between vectors \( \mathbf{a} \) and \( \mathbf{b} \).
02

Recall the formula for the area of a triangle

The area of a triangle using vectors can be calculated by \( \frac{1}{2} \times \text{base} \times \text{height} \). When the vectors form two sides of a triangle, the base can be considered as the length of one vector, and the height corresponds to the perpendicular distance from the tip of the other vector to the line extended from the base vector.
03

Relate the cross product to the triangle area

The magnitude of the cross product \( | \mathbf{a} \times \mathbf{b} | \) gives the area of the parallelogram formed by vectors \( \mathbf{a} \) and \( \mathbf{b} \). The area of the triangle is half of this, as a triangle is half of a parallelogram.
04

Express area of triangle in terms of cross product

The area of the triangle formed by vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( \frac{1}{2} \times | \mathbf{a} \times \mathbf{b} | \), since it's half the area of the parallelogram. Thus, the area of the triangle is indeed half of the magnitude of the cross product.

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Key Concepts

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

Understanding the Cross Product
When dealing with vectors, understanding the cross product is crucial. It refers to an operation between two vectors in three-dimensional space that results in another vector, which is perpendicular to the original two vectors. This new vector's direction is determined by the right-hand rule, where your thumb points in the direction of the cross product when you point your fingers in the direction of vector \( \mathbf{a} \) and curl them towards vector \( \mathbf{b} \). The magnitude (or length) of the cross product can be found by the formula: \[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| \times |\mathbf{b}| \times \sin(\theta) \] where \( \theta \) is the angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \). This angle is located within the plane formed by the two vectors.
  • The cross product is all about directionality and magnitude.
  • If the vectors are parallel, their cross product will be a zero vector.
This concept is vital for calculating areas using vectors, as we'll explore further.
Triangle Area Using Vectors
The area of a triangle in the context of vectors can be found using geometric properties. Normally, you understand a triangle as a shape with a base and a height, but when dealing with vectors, these components take on new meanings. In this scenario, we consider two vectors that span two sides of a triangle. To find the area, utilize one half times the base times the height formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] When you view vectors as sides of the triangle, one vector acts as the base, while the perpendicular component to this base provided by the cross product determines the height. The cross product helps ensure we account for the accurate perpendicular distance, sometimes just referred to as the height, from the tip of one vector to the line along the other vector.
  • This method defines the triangle area through vector operations rather than direct measurement.
  • It's particularly useful in physics and engineering where spatial configurations are described using vectors.
Parallelogram Area
Vectors can not only help calculate triangle areas but also parallelogram areas. When two vectors are used to define a parallelogram, the magnitude of their cross product directly computes the area of the parallelogram. Think of the vectors as adjacent sides of a parallelogram; visually, you can see how they form such a shape in space. The formula for the area of a parallelogram using vectors is straightforward: \[ \text{Area of Parallelogram} = |\mathbf{a} \times \mathbf{b}| \] Here's a quick breakdown of the relationship between parallelogram and triangle areas:
  • The parallelogram area is fully covered by the cross product magnitude.
  • Since a triangle is half of a parallelogram, simply divide the parallelogram's area by two to find the triangle's area.
When working with any type of polygon derived from vectors, the cross product offers a consistent method to find areas accurately using just vector directions and magnitudes. This feature is extremely beneficial in multiple scientific disciplines where geometry and space are key.

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Most popular questions from this chapter

A solid cylinder of mass \(20 \mathrm{~kg}\) rotates about its axis with angular speed \(100 \mathrm{rad} \mathrm{s}^{-1}\). The radius of the cylinder is \(0.25 \mathrm{~m}\). What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be \(2 M R^{2} / 5\), where \(M\) is the mass of the sphere and \(R\) is the radius of the sphere. (b) Given the moment of inertia of a disc of mass \(M\) and radius \(R\) about any of its diameters to be \(M R^{2} / 4\), find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

From a uniform disk of radius \(R\), a circular hole of radius \(R / 2\) is cut out. The centre of the hole is at \(R / 2\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Prove the result that the velocity \(v\) of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height \(h\) is given by \(v^{2}=\frac{2 g h}{\left(1+k^{2} / R^{2}\right)}\) using dynamical consideration (i.e. by consideration of forces and torques). Note \(k\) is the radius of gyration of the body about its symmetry axis, and \(\mathrm{R}\) is the radius of the body. The body starts from rest at the top of the plane.

A bullet of mass \(10 \mathrm{~g}\) and speed \(500 \mathrm{~m} / \mathrm{s}\) is fired into a door and gets embedded exactly at the centre of the door. The door is \(1.0 \mathrm{~m}\) wide and weighs \(12 \mathrm{~kg}\). It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is \(M L^{2} / 3 .\) )

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