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Vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) have scalar product \(-\)6.00, and their vector product has magnitude \(+\)9.00. What is the angle between these two vectors?

Short Answer

Expert verified
The angle between the vectors is approximately 123.69 degrees.

Step by step solution

01

Understanding Given Data

We are provided with two pieces of information about vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \). First, their dot product is \( \overrightarrow{A} \cdot \overrightarrow{B} = -6.00 \). Secondly, their cross product magnitude is \( |\overrightarrow{A} \times \overrightarrow{B}| = 9.00 \). We seek the angle \( \theta \) between them.
02

Relation of Dot Product with Cosine

The dot product of two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \) is given by the formula: \[ \overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos(\theta) \] Using this, \( \cos(\theta) = \frac{-6.00}{|\overrightarrow{A}| |\overrightarrow{B}|} \).
03

Relation of Cross Product with Sine

The magnitude of the cross product is given by: \[ |\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \] From this, we find \( \sin(\theta) = \frac{9.00}{|\overrightarrow{A}| |\overrightarrow{B}|} \).
04

Solving for Angle \( \theta \)

To find \( \theta \), we need the quotient \( \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) \). Thus, \[ \tan(\theta) = \frac{9.00}{-6.00} = -\frac{3}{2} \] Therefore, \( \theta = \tan^{-1}\left(-\frac{3}{2}\right) \).
05

Finding Specific Angle

Using a calculator, solve \( \theta = \tan^{-1}(-\frac{3}{2}) \). We find the angle \( \theta \) to be approximately \( -56.31 \text{ degrees} \). However, the standard positive angle that is within \( 0 \) to \( 180 \text{ degrees} \) is \( 180 - 56.31 \approx 123.69 \text{ degrees} \), since the tangent function is periodic.

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Key Concepts

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

Scalar Product
The scalar product, often known as the dot product, is one of the fundamental operations you can perform on two vectors. It combines two vectors to produce a single scalar (number), which indicates some level of projection of one vector onto another.

For vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), their dot product is calculated as:
  • \( \overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos(\theta) \)
Here, \( \theta \) is the angle between the vectors, and \(|\overrightarrow{A}|\) and \(|\overrightarrow{B}|\) are their magnitudes.

This product is commutative, meaning \( \overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A} \). It's particularly useful in applications such as determining if vectors are perpendicular, as the dot product will be zero when they are at a right angle.
Vector Product
The vector product, or cross product, provides a powerful way to combine two vectors in three-dimensional space. Unlike the dot product, which results in a scalar, the cross product results in another vector.

For vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), the vector product is identified by a new vector whose magnitude is given by:
  • \( |\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \)
The direction of this new vector is perpendicular to both \( \overrightarrow{A} \) and \( \overrightarrow{B} \), following the right-hand rule.

This product is non-commutative, meaning \( \overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{B} \times \overrightarrow{A}) \). It's often used in physics and engineering to determine torque or rotational force.
Dot Product
The dot product is a type of scalar product, but it bears its own special emphasis, especially in geometry and physics. It's defined as:
  • \( \overrightarrow{A} \cdot \overrightarrow{B} = ax_b + by_b + cz_b \)
Where \(a, b, c\) are the respective components of \( \overrightarrow{A}\), and \(x, y, z\) are those of \( \overrightarrow{B} \).

One application of the dot product is finding the angle between two vectors. If the product yields zero, the vectors are perpendicular. For non-zero results, it can provide insight into how much two vectors "align" with each other.

This product is crucial in simplifying calculations in machine learning, guides collision detection in video games, and assists in projections in 3D graphics.
Cross Product
The cross product, being a specific form of vector product, serves a unique purpose in vector mathematics. It's represented as a vector whose magnitude and direction are derived from its operands.

The magnitude is calculated as:
  • \( |\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \)
This results in a new vector, essentially pointing out of the plane formed by \( \overrightarrow{A} \) and \( \overrightarrow{B} \).

In practice, this product is useful for determining the area of parallelograms defined by vectors, calculating moments and forces in physics, and aiding in operations like finding normals to surfaces in computer graphics. Its application is indispensable for tasks involving rotation and orientation.
Angle Between Vectors
Determining the angle between two vectors is crucial in various applications ranging from physics to computer graphics.

This angle, denoted as \(\theta\), can be computed using the relationships established by both the dot and cross product:
  • \( \cos(\theta) = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \)
  • \( \sin(\theta) = \frac{|\overrightarrow{A} \times \overrightarrow{B}|}{|\overrightarrow{A}||\overrightarrow{B}|} \)
To find the angle, you calculate the inverse trigonometric functions based on these relationships.

If tasked with finding the angle: work out both sine and cosine using the formulas above and then derive \( \theta \) using \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). This method gives a clear, process-oriented way to find the specific angle between two vectors in any scenario.

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Most popular questions from this chapter

Later in our study of physics we will encounter quantities represented by (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\). (a) Prove that for any three vectors \(\overrightarrow{A}\), \(\overrightarrow{B}\), and \(\overrightarrow{C}\), \(\overrightarrow{A}\) \(\cdot\) (\(\overrightarrow{B}\) \\(\times\\) \(\overrightarrow{C}\)) = (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\). (b) Calculate (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\) for vector \(\overrightarrow{A}\) with magnitude \(A\) = 5.00 and angle \(\theta$$_A\) = 26.0\(^{\circ}\) (measured from the +\(x\)-axis toward the \(+$$y\)-axis), vector \(\overrightarrow{B}\) with \(B\) = 4.00 and \(\theta$$_B\) = 63.0\(^{\circ}\), and vector \(\overrightarrow{C}\) with magnitude 6.00 and in the \(+$$z\)-direction. Vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) are in the \(x$$y\)-plane.

As a test of orienteering skills, your physics class holds a contest in a large, open field. Each contestant is told to travel 20.8 m due north from the starting point, then 38.0 m due east, and finally 18.0 m in the direction 33.0\(^{\circ}\) west of south. After the specified displacements, a contestant will find a silver dollar hidden under a rock. The winner is the person who takes the shortest time to reach the location of the silver dollar. Remembering what you learned in class, you run on a straight line from the starting point to the hidden coin. How far and in what direction do you run?

A square field measuring 100.0 m by 100.0 m has an area of 1.00 hectare. An acre has an area of 43,600 ft\(^2\). If a lot has an area of 12.0 acres, what is its area in hectares?

While driving in an exotic foreign land, you see a speed limit sign that reads 180,000 furlongs per fortnight. How many miles per hour is this? (One furlong is \\(\frac{1}{8}\\) mile, and a fortnight is 14 days. A furlong originally referred to the length of a plowed furrow.)

In the methane molecule, CH\(_4\), each hydrogen atom is at a corner of a regular tetrahedron with the carbon atom at the center. In coordinates for which one of the C\(-\)H bonds is in the direction of \(\hat{\imath}\) + \(\hat{\jmath}\) + \(\hat{k}\), an adjacent C\(-\)H bond is in the \(\hat{\imath}\) \(-\) \(\hat{\jmath}\) \(-\) \(\hat{k}\) direction. Calculate the angle between these two bonds.

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