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Chapter 9: Problem 13

Find \((a) u \cdot v\) and \((b)\) the angle between \(u\) and \(v\) to the nearest degree. $$\mathbf{u}=\mathbf{i}+3 \mathbf{j}, \quad \mathbf{v}=4 \mathbf{i}-\mathbf{j}$$

Short Answer

Expert verified
(a) The dot product is 1. (b) The angle is approximately 86 degrees.

Step by step solution

01

Calculate the Dot Product

To find the dot product \( u \cdot v \), use the formula: \( u \cdot v = u_1v_1 + u_2v_2 \). For vectors \( u = i + 3j \) and \( v = 4i - j \), this becomes \( u \cdot v = 1\cdot4 + 3\cdot(-1) \). Calculate this: \( 4 - 3 = 1 \). Thus, \( u \cdot v = 1 \).
02

Find the Magnitude of u

Calculate the magnitude of vector \( u = i + 3j \). The magnitude formula is \( \|u\| = \sqrt{u_1^2 + u_2^2} \). Substitute the values: \( \|u\| = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \).
03

Find the Magnitude of v

Calculate the magnitude of vector \( v = 4i - j \). The magnitude formula is \( \|v\| = \sqrt{v_1^2 + v_2^2} \). Substitute the values: \( \|v\| = \sqrt{4^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17} \).
04

Calculate the Cosine of the Angle

The cosine of the angle \( \theta \) between \( u \) and \( v \) is given by the formula: \( \cos \theta = \frac{u \cdot v}{\|u\| \|v\|} \). Using previous results, \( \cos \theta = \frac{1}{\sqrt{10} \cdot \sqrt{17}} \). Simplify to get \( \cos \theta = \frac{1}{\sqrt{170}} \).
05

Calculate the Angle

To find the angle \( \theta \) in degrees, use the inverse cosine function: \( \theta = \cos^{-1}\left( \frac{1}{\sqrt{170}} \right) \). Calculate this using a calculator and round to the nearest degree. The result is approximately \( 86^{\circ} \).

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Key Concepts

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

Angle Between Vectors
When it's time to figure out the angle between two vectors, we're essentially measuring how far apart they are directionally. You might wonder why this is important. Well, it helps us understand whether the vectors point in similar or very different directions. Here is how we do it:
  • First, use the dot product. This product combines the components of the vectors without worrying about their lengths. In our exercise, we found the dot product of vectors \( u \) and \( v \) to be 1.
  • Then, find the magnitudes of each vector. This gives us their lengths, acting like a measure of their size.
  • With these two pieces of information, we use the formula \( \cos \theta = \frac{u \cdot v}{\|u\| \|v\|} \), which gives us the cosine of the angle \( \theta \) between the vectors. In simple terms, this equation measures how much one vector is aligned with another.
Plugging our values into the formula, the cosine was found to be approximately \( \frac{1}{\sqrt{170}} \), which corresponds to an angle of around \( 86^{\circ} \). Understanding the angle between vectors helps determine their relative orientation.
Magnitude of a Vector
The magnitude of a vector is like its length or size in space. It's important because it helps to grasp how large or small the vector is compared to others. Calculating it is straightforward.
  • The magnitude is determined by using the formula \( \|u\| = \sqrt{u_1^2 + u_2^2} \) for a vector \( u \) with components \( (u_1, u_2) \).
  • For example, for the vector \( u = i + 3j \), the magnitude was computed as \( \sqrt{10} \).
  • This calculation involves squaring each component of the vector, adding these squared values, and finally, taking the square root of the sum.
Knowing the magnitude gives you an insight into how much space the vector covers. It's like figuring out the length of an arrow laid out in a 2D or 3D space.
Inverse Cosine Function
The inverse cosine function, often noted as \( \cos^{-1} \), is crucial when finding angles between vectors. It allows you to retrieve the actual angle once you know the cosine of that angle.
  • In our context, once we used the dot product and magnitudes to find the cosine value, which was \( \frac{1}{\sqrt{170}} \), we needed to find the angle \( \theta \). That's where the inverse cosine function steps in.
  • By applying \( \theta = \cos^{-1}(x) \), you convert the cosine value \( x \) back into an angle. This angle is usually in radians first unless you specify degrees or convert it.
  • In practice, you'd often use a calculator or a software tool to perform this function because it's complex to do manually. In our exercise, the angle between vectors was computed to be approximately \( 86^{\circ} \).
Utilizing the inverse cosine helps give meaningful, easily understood angles from otherwise abstract cosine values.

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Most popular questions from this chapter

Let a \(=\langle 2,2,2\rangle\) \(\mathbf{b}=\langle- 2,-2,0\rangle,\) and \(\mathbf{r}=\langle x, y, z\rangle\) (a) Show that the vector equation \((\mathbf{r}-\mathbf{a}) \cdot(\mathbf{r}-\mathbf{b})=0\) represents a sphere, by expanding the dot product and simplifying the resulting algebraic equation. (b) Find the center and radius of the sphere. (c) Interpret the result of part (a) geometrically, using the fact that the dot product of two vectors is 0 only if the vectors are perpendicular. endpoints of the vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{r},\) noting that the end. points of a and b are the endpoints of a diameter and the endpoint of \(\mathbf{r} \text { is an arbitrary point on the sphere. }]\) (d) Using your observations from part (a), find a vector equation for the sphere in which the points \((0,1,3)\) and \((2,-1,4)\) form the endpoints of a diameter. Simplify the vector equation to obtain an algebraic equation for the sphere. What are its center and radius?

Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors i and j. $$|\mathbf{v}|=800, \quad \theta=125^{\circ}$$

Every line can be described by infinitely many different sets of parametric equations, since any point on the line and any vector parallel to the line can be used to construct the equations. But how can we tell whether two sets of parametric equations represent the same line? Consider the following two sets of parametric equations: Line \(1: \quad x=1-t, \quad y=3 t, \quad z=-6+5 t\) Line \(2: \quad x=-1+2 t, \quad y=6-6 t, \quad z=4-10 t\) (a) Find two points that lie on Line 1 by setting \(t=0\) and \(t=1\) in its parametric equations. Then show that these points also lie on Line 2 by finding two values of the parameter that give these points when substituted into the parametric equations for Line 2 . (b) Show that the following two lines are not the same by finding a point on Line 3 and then showing that it does not lie on Line 4 Line \(3: \quad x=4 t, \quad y=3-6 t, \quad z=-5+2 t\) Line \(4: \quad x=8-2 t, \quad y=-9+3 t, \quad z=6-t\)

Two vectors \(u\) and \(v\) are given. Find the angle (expressed in degrees) between \(\mathbf{u}\) and \(v\) $$\mathbf{u}=\langle 4,0,2\rangle, \quad \mathbf{v}=\langle 2,-1,0\rangle$$

Find the area of the parallelogram determined by the given vectors. $$\mathbf{u}=2 \mathbf{i}-\mathbf{j}+4 \mathbf{k}, \quad \mathbf{v}=\frac{1}{2} \mathbf{i}+2 \mathbf{j}-\frac{3}{2} \mathbf{k}$$
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