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Chapter 2: Problem 25

For the following exercises, find vector \(\mathbf{v}\) with the given magnitude and in the same direction as vector \(\mathbf{u}\). \(\|\mathbf{v}\|=7, \mathbf{u}=\langle 3,4\rangle\)

Short Answer

Expert verified
\(\mathbf{v} = \left\langle \frac{21}{5}, \frac{28}{5} \right\rangle\)."

Step by step solution

01

Understand the Problem

We need to find a vector \(\mathbf{v}\) that has a magnitude of 7 and points in the same direction as \(\mathbf{u} = \langle 3, 4 \rangle\).
02

Find the Magnitude of the Original Vector

Calculate the magnitude of vector \(\mathbf{u}\) using the formula for the magnitude of a vector: \[\|\mathbf{u}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.\]
03

Find the Unit Vector in Direction of \(\mathbf{u}\)

To find a unit vector in the direction of \(\mathbf{u}\), divide \(\mathbf{u}\) by its magnitude:\[\mathbf{u}_{\text{unit}} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle.\]
04

Scale the Unit Vector by Given Magnitude

Multiply the unit vector by the magnitude 7 to find \(\mathbf{v}\): \[\mathbf{v} = 7 \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = \left\langle \frac{21}{5}, \frac{28}{5} \right\rangle.\]

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Key Concepts

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

Vector Magnitude
In vector calculus, the magnitude of a vector is essentially the length or size of the vector. If you imagine a vector as an arrow, the magnitude corresponds to its length. It's an important concept when working with vectors, as it helps us understand the scale of a vector.
To calculate the magnitude of a vector, we use the formula: \[ \|\mathbf{u}\| = \sqrt{x^2 + y^2} \]where \(x\) and \(y\) are the components of the vector.
In our problem, we have the vector \(\mathbf{u} = \langle 3, 4 \rangle\). We use the formula to find its magnitude:
  • Square each component: \(3^2 = 9\) and \(4^2 = 16\).
  • Sum these squares: \(9 + 16 = 25\).
  • Take the square root: \(\sqrt{25} = 5\).
Hence, the magnitude of vector \(\mathbf{u}\) is 5.
Unit Vector
A unit vector is a vector that has a magnitude of exactly 1. The idea behind a unit vector is to express direction without concerning the magnitude. It points in the same direction as the original vector but is standardized with a length of one.
To find the unit vector for any vector \(\mathbf{u} = \langle x, y \rangle\), we divide each component by the magnitude of the vector:\[ \mathbf{u}_{\text{unit}} = \left\langle \frac{x}{\|\mathbf{u}\|}, \frac{y}{\|\mathbf{u}\|} \right\rangle \]
In our case with \(\mathbf{u} = \langle 3, 4 \rangle\) and its magnitude \(5\):
  • \(\mathbf{u}_{\text{unit}} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle\).
This unit vector points in the same direction as \(\mathbf{u}\) but has been scaled down to a magnitude of 1.
Vector Direction
The direction of a vector is the orientation it points towards, which can be found from its components. When we talk about the direction, we're focused on the angle that the vector forms with a reference axis, typically the x-axis.
It's imperative to note that multiplying a vector by a positive scalar preserves its direction while modifying its magnitude. This characteristic is crucial when creating a new vector with a desired magnitude but the same direction.
In the exercise, we maintain direction with a unit vector before scaling it. This two-step process allows us to find a new vector, maintaining a normalized direction before applying the specific magnitude.
Scaling Vectors
Scaling a vector means changing its magnitude while maintaining its direction. We perform this by multiplying the vector by a scalar, a single number.
When we have a unit vector and wish to scale it, multiplying it by a scalar will change its length. It's important to note:
  • If the scalar is greater than one, the vector becomes longer.
  • If the scalar is less than one, the vector becomes shorter.
In this exercise, we found a unit vector \(\mathbf{u}_{\text{unit}} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle\) and scaled it by 7 to get vector \(\mathbf{v}\), which is:
\[ \mathbf{v} = 7 \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = \left\langle \frac{21}{5}, \frac{28}{5} \right\rangle \]This new vector \(\mathbf{v}\) has a magnitude of 7 and maintains the direction of \(\mathbf{u}\).

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