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Let \(\mathbf{u}=\langle 3,-4\rangle, \mathbf{v}=\langle 1,1\rangle,\) and \(\mathbf{w}=\langle-1,0\rangle .\) Carry out the following computations. Which has the greater magnitude, \(\mathbf{u}-\mathbf{v}\) or \(\mathbf{w}-\mathbf{u} ?\)

Short Answer

Expert verified
Given vectors \(\mathbf{u}=\langle 3, -4\rangle\), \(\mathbf{v}=\langle 1, 1\rangle\), and \(\mathbf{w}=\langle -1, 0\rangle\). Answer: The vector \(\mathbf{w}-\mathbf{u}\) has a greater magnitude than the vector \(\mathbf{u}-\mathbf{v}\).

Step by step solution

01

Subtract the Vectors

We first need to calculate the vectors \(\mathbf{u}-\mathbf{v}\) and \(\mathbf{w}-\mathbf{u}\). To do this, subtract the corresponding components of the vectors. For \(\mathbf{u}-\mathbf{v}\): \[(3 - 1, -4 - 1) = (2, -5)\] For \(\mathbf{w}-\mathbf{u}\): \[(-1 - 3, 0 - (-4)) = (-4, 4)\] So, \(\mathbf{u}-\mathbf{v}=\langle 2, -5\rangle\) and \(\mathbf{w}-\mathbf{u}=\langle-4, 4\rangle\).
02

Calculate the Magnitudes

Next, we need to calculate the magnitudes of these two vectors. The magnitude of a vector \(\langle x, y \rangle\) is given by the formula \(\sqrt{x^2 + y^2}\). For the vector \(\mathbf{u}-\mathbf{v}=\langle 2, -5\rangle\): \[||\mathbf{u}-\mathbf{v}|| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}\] For the vector \(\mathbf{w}-\mathbf{u}=\langle -4, 4\rangle\): \[||\mathbf{w}-\mathbf{u}|| = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32}\]
03

Compare the Magnitudes

Now, we compare the magnitudes of the vectors \(\mathbf{u}-\mathbf{v}\) and \(\mathbf{w}-\mathbf{u}\). Since \(\sqrt{29} \approx 5.39\) and \(\sqrt{32} \approx 5.66\), we see that: \[||\mathbf{w}-\mathbf{u}|| > ||\mathbf{u}-\mathbf{v}||\] Thus, the vector \(\mathbf{w}-\mathbf{u}\) has a greater magnitude than the vector \(\mathbf{u}-\mathbf{v}\).

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Key Concepts

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

Vector Subtraction
Vector subtraction is a fundamental concept in vector algebra that helps us determine the difference between two vectors by subtracting their corresponding components.
It's similar to subtracting numbers, but here we deal with pairs of numbers called components.
  • Each vector is represented as an ordered pair or triplet, depending on the dimensions.
  • To subtract vectors, subtract the corresponding components.
Given vectors \(\mathbf{u} = \langle 3, -4 \rangle\) and \(\mathbf{v} = \langle 1, 1 \rangle\), subtracting \(\mathbf{v}\) from \(\mathbf{u}\) means:
\[\mathbf{u} - \mathbf{v} = \langle 3 - 1, -4 - 1 \rangle = \langle 2, -5 \rangle\]
Now, for \(\mathbf{w} = \langle -1, 0 \rangle\) and subtracting \(\mathbf{u}\), the subtraction \( \mathbf{w} - \mathbf{u} \) is:
\[\mathbf{w} - \mathbf{u} = \langle -1 - 3, 0 - (-4) \rangle = \langle -4, 4 \rangle\] This calculation forms the basis for further analysis, such as finding magnitudes or distances between points.
Euclidean Distance
The Euclidean distance is a measure that shows the straight-line distance between two points in Euclidean space.
Often used in geometry, it's calculated using the Pythagorean theorem.
When discussing vectors, the distance can translate to a vector's length or magnitude:
  • In 2D space, if a vector \( \mathbf{v} = \langle x, y \rangle \), its Euclidean distance or magnitude from the origin is \( \sqrt{x^2 + y^2} \).
  • For vectors \( \mathbf{u} - \mathbf{v} \) and \( \mathbf{w} - \mathbf{u} \), we find this distance by calculating the magnitude.
For instance, for \( \mathbf{u} - \mathbf{v} = \langle 2, -5 \rangle \):
\[||\mathbf{u} - \mathbf{v}|| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}\]
Similarly, for \( \mathbf{w} - \mathbf{u} = \langle -4, 4 \rangle \):
\[||\mathbf{w} - \mathbf{u}|| = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32}\] These calculations help find which vector is longer and therefore has a greater distance from its origin.
Vector Components
Vector components are the individual values that make up a vector.
They determine the direction and magnitude of the vector when combined.
In a 2D vector, each component corresponds to horizontal and vertical distances on a plane:
  • A vector in two dimensions is often written as \( \langle x, y \rangle \), where \( x \) is the horizontal component, and \( y \) is the vertical component.
  • The components indicate how far and in which direction the vector extends from its origin.
Take the vector \( \langle 2, -5 \rangle \):
  • Here, 2 is the horizontal component, moving right (positive direction) on the x-axis.
  • -5 is the vertical component, moving down (negative direction) on the y-axis.
Similarly, for the vector \( \langle -4, 4 \rangle \):
  • -4 extends left on the x-axis, indicating the negative direction.
  • 4 extends upwards on the y-axis, indicating the positive direction.
Understanding these components is crucial for performing vector operations like addition, subtraction, and calculating their magnitude.

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Most popular questions from this chapter

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3\rangle, \mathbf{v}=\langle 1,1\rangle\)

Assume that \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in \(\mathrm{R}^{3}\) that form the sides of a triangle (see figure). Use the following steps to prove that the medians intersect at a point that divides each median in a 2: 1 ratio. The proof does not use a coordinate system. a. Show that \(\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{0}\) b. Let \(\mathbf{M}_{1}\) be the median vector from the midpoint of \(\mathbf{u}\) to the opposite vertex. Define \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) similarly. Using the geometry of vector addition show that \(\mathbf{M}_{1}=\mathbf{u} / 2+\mathbf{v} .\) Find analogous expressions for \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) c. Let \(a, b,\) and \(c\) be the vectors from \(O\) to the points one-third of the way along \(\mathbf{M}_{1}, \mathbf{M}_{2},\) and \(\mathbf{M}_{3},\) respectively. Show that \(\mathbf{a}=\mathbf{b}=\mathbf{c}=(\mathbf{u}-\mathbf{w}) / 3\) d. Conclude that the medians intersect at a point that divides each median in a 2: 1 ratio.

An object moves clockwise around a circle centered at the origin with radius \(5 \mathrm{m}\) beginning at the point (0,5) a. Find a position function \(\mathbf{r}\) that describes the motion if the object moves with a constant speed, completing 1 lap every 12 s. b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(e^{-t}\)

Relationship between \(\mathbf{T}, \mathbf{N},\) and a Show that if an object accelerates in the sense that \(d^{2} s / d t^{2}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). If an object decelerates in the sense that \(d^{2} s / d t^{2}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N}\)

Parabolic trajectory In Example 7 it was shown that for the parabolic trajectory \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle, \mathbf{a}=\langle 0,2\rangle\) and \(\mathbf{a}=\frac{2}{\sqrt{1+4 t^{2}}}(\mathbf{N}+2 t \mathbf{T}) .\) Show that the second expression for \(\mathbf{a}\) reduces to the first expression.

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