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Compute \(\mathbf{u} \cdot \mathbf{v}\) if \(\mathbf{u}\) is a unit vector, \(|\mathbf{v}|=2,\) and the angle between them is \(3 \pi / 4\).

Short Answer

Expert verified
Answer: The dot product of the two vectors is -√2.

Step by step solution

01

Recall the formula for the dot product using magnitudes and angles

The dot product of two vectors can be calculated using their magnitudes and the angle between them with the formula: \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| \cdot |\mathbf{v}| \cdot \cos (\theta)\), where \(\theta\) is the angle between the vectors (in radians).
02

Use the given information

We are given the following information: - \(\mathbf{u}\) is a unit vector, which means its magnitude is \(|\mathbf{u}|=1\). - The magnitude of \(\mathbf{v}\) is \(|\mathbf{v}|=2\). - The angle between the vectors is \(\theta=3\pi/4\).
03

Compute the dot product

Substitute the given values into the formula from Step 1: \(\mathbf{u} \cdot \mathbf{v} = (1) \cdot (2) \cdot \cos\left(\dfrac{3\pi}{4}\right) = 2 \cdot \cos\left(\dfrac{3\pi}{4}\right)\). Now compute the cosine of the given angle: \(\cos\left(\dfrac{3\pi}{4}\right) = -\sqrt{2}/2\). Finally, compute the dot product: \(\mathbf{u} \cdot \mathbf{v} = 2 \cdot \left(-\dfrac{\sqrt{2}}{2}\right) = -\sqrt{2}\). Hence the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) is \(\mathbf{u} \cdot \mathbf{v} = -\sqrt{2}\).

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Key Concepts

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

Unit Vector
A unit vector is a vector with a magnitude of exactly 1. It serves as a building block for constructing other vectors in any direction. Although its direction can vary, its length always remains the same—1 unit. This property makes unit vectors very useful for defining direction without affecting magnitude in vector operations.
  • Denoted often with a hat symbol, for example, \(\mathbf{u}\hat{}\).
  • Main property: \(|\mathbf{u}| = 1\).
  • Used to simplify calculations and standardize magnitude.
Recognizing a unit vector is crucial when working with dot products since it simplifies the calculation by setting one of the magnitudes to 1. In problems involving angles, as in our example, knowing that \(\mathbf{u}\) is a unit vector ensures that you're multiplying by 1, making calculations easier.
Angle Between Vectors
The angle between two vectors is a measure of their direction relative to each other. It ranges from 0 radians (or degrees) when vectors point in exactly the same direction, up to \(\pi\) radians (180 degrees) when they point in opposite directions.
  • An angle of \(0\) indicates parallel vectors.
  • An angle of \(\pi\) indicates antiparallel vectors.
In the dot product formula \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| \cdot |\mathbf{v}| \cdot \cos (\theta)\), the cosine of the angle tells us about the directional relationship:
  • \(\cos(\theta)\) is positive when vectors point roughly in the same direction; negative when they point oppositely.
For the given problem, the angle \(\theta\) is \(\frac{3\pi}{4}\), which is more than \(\pi/2\), leading to \(\cos(\theta)\) being negative, indicating opposite directional relations.
Magnitude of Vectors
The magnitude of a vector refers to its length or size. It's a scalar value that quantifies how much of the vector there is; think of it as the distance from the head to the tail of the arrow symbol that represents a vector.
  • Notation: usually denoted \(|\mathbf{v}|\).
  • Cannot be negative; it is always zero or a positive number.
Computing magnitude is straightforward with simple operations. For a vector \( \mathbf{v} = (x, y) \), the magnitude is calculated using the Pythagorean theorem: \(|\mathbf{v}| = \sqrt{x^2 + y^2}\) for 2D vectors, or \(|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}\) for 3D.
In the provided example, \(|\mathbf{v}|\) is given as 2. This information is crucial for determining the dot product, because the dot product scales with the magnitude of each vector, weighted by the directional relationship expressed through \(\cos(\theta)\).

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Most popular questions from this chapter

A golfer launches a tee shot down a horizontal fairway and it follows a path given by \(\mathbf{r}(t)=\left\langle a t,(75-0.1 a) t,-5 t^{2}+80 t\right\rangle,\) where \(t \geq 0\) measures time in seconds and \(\mathbf{r}\) has units of feet. The \(y\) -axis points straight down the fairway and the z-axis points vertically upward. The parameter \(a\) is the slice factor that determines how much the shot deviates from a straight path down the fairway. a. With no slice \((a=0),\) sketch and describe the shot. How far does the ball travel horizontally (the distance between the point the ball leaves the ground and the point where it first strikes the ground)? b. With a slice \((a=0.2),\) sketch and describe the shot. How far does the ball travel horizontally? c. How far does the ball travel horizontally with \(a=2.5 ?\)

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Write the vector \langle 2,-6\rangle in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(c(\mathbf{u} \cdot \mathbf{v})=(c \mathbf{u}) \cdot \mathbf{v}=\mathbf{u} \cdot(c \mathbf{v})\)

Prove that for integers \(m\) and \(n\), the curve $$\mathbf{r}(t)=\langle a \sin m t \cos n t, b \sin m t \sin n t, c \cos m t\rangle$$ lies on the surface of a sphere provided \(a^{2}+b^{2}=c^{2}\).

Determine the equation of the line that is perpendicular to the lines \(\mathbf{r}(t)=\langle-2+3 t, 2 t, 3 t\rangle\) and \(\mathbf{R}(s)=\langle-6+s,-8+2 s,-12+3 s\rangle\) and passes through the point of intersection of the lines \(\mathbf{r}\) and \(\mathbf{R}\).

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