Chapter 1: Problem 52
For the following exercises, find vector \(\mathbf{u}\) with a magnitude that is given and satisfies the given conditions.\(\mathbf{v}=\langle 2,4,1\rangle,\|\mathbf{u}\|=15, \mathbf{u}\) and \(\mathbf{v}\) have the same direction
Short Answer
Expert verified
The vector \(\mathbf{u}\) is \(\left\langle \frac{10\sqrt{21}}{7}, \frac{20\sqrt{21}}{7}, \frac{5\sqrt{21}}{7} \right\rangle\).
Step by step solution
01
Understand the Problem
We need to find a vector \(\mathbf{u}\) that has the same direction as \(\mathbf{v} = \langle 2, 4, 1 \rangle\), but with a different magnitude, which is \(15\).
02
Calculate the Magnitude of \(\mathbf{v}\)
First, we find the magnitude of the vector \(\mathbf{v}\). The formula for the magnitude \(\|\mathbf{v}\|\) of vector \(\mathbf{v} = \langle a, b, c \rangle\) is: \[ \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \]. Substitute \(a = 2\), \(b = 4\), \(c = 1\): \[ \|\mathbf{v}\| = \sqrt{2^2 + 4^2 + 1^2} = \sqrt{4 + 16 + 1} = \sqrt{21} \].
03
Find the Unit Vector in the Direction of \(\mathbf{v}\)
The unit vector \(\mathbf{v}_u\) in the same direction as \(\mathbf{v}\) is given by dividing \(\mathbf{v}\) by its magnitude: \[ \mathbf{v}_u = \frac{1}{\sqrt{21}} \langle 2, 4, 1 \rangle = \left\langle \frac{2}{\sqrt{21}}, \frac{4}{\sqrt{21}}, \frac{1}{\sqrt{21}} \right\rangle \].
04
Scale the Unit Vector to the Desired Magnitude
To get a vector with magnitude 15 in the same direction, multiply the unit vector \(\mathbf{v}_u\) by 15: \[ \mathbf{u} = 15 \times \mathbf{v}_u = 15 \times \left\langle \frac{2}{\sqrt{21}}, \frac{4}{\sqrt{21}}, \frac{1}{\sqrt{21}} \right\rangle = \left\langle \frac{30}{\sqrt{21}}, \frac{60}{\sqrt{21}}, \frac{15}{\sqrt{21}} \right\rangle \].
05
Simplify the Vector \(\mathbf{u}\)
Simplify each component by rationalizing the denominator: \[ \mathbf{u} = \left\langle \frac{30\sqrt{21}}{21}, \frac{60\sqrt{21}}{21}, \frac{15\sqrt{21}}{21} \right\rangle \]. This simplifies to: \[ \mathbf{u} = \left\langle \frac{10\sqrt{21}}{7}, \frac{20\sqrt{21}}{7}, \frac{5\sqrt{21}}{7} \right\rangle \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
Calculating the magnitude of a vector is an essential concept when working with vectors. The magnitude of a vector provides the length or size of the vector, much like how distance gives the length between two points. It's like measuring how far a vector "stretches" in space.
The formula to calculate the magnitude \( \|\mathbf{v}\| \) for a vector \( \mathbf{v} = \langle a, b, c \rangle \) is:
This tells us how long the vector is in abstract space.
The formula to calculate the magnitude \( \|\mathbf{v}\| \) for a vector \( \mathbf{v} = \langle a, b, c \rangle \) is:
- \[ \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \]
- Square each component of the vector: \( a^2, b^2, \, and \, c^2 \).
- Add up these squares: \( a^2 + b^2 + c^2 \).
- Finally, take the square root to find the magnitude.
This tells us how long the vector is in abstract space.
Unit Vector
A unit vector is a vector with a magnitude of exactly 1—a special vector used to indicate direction. It retains the direction of the original vector but is "shrunken" down to a length of one.
To find the unit vector \( \mathbf{v}_u \) of any vector \( \mathbf{v} \), divide each component of the vector by its magnitude:
To find the unit vector \( \mathbf{v}_u \) of any vector \( \mathbf{v} \), divide each component of the vector by its magnitude:
- \[ \mathbf{v}_u = \frac{1}{\|\mathbf{v}\|} \langle a, b, c \rangle \]
- The unit vector maintains the direction of \( \mathbf{v} \).
- It ensures the length of the resultant vector is always 1.
Directional Vector Scaling
Once a unit vector has been established, it can be scaled to any magnitude desired, thus creating a new vector with the same direction as the original but a different "stretch." This process is known as directional vector scaling.
The formula for scaling a unit vector \( \mathbf{v}_u \) to a desired magnitude \( m \) is straightforward:
The formula for scaling a unit vector \( \mathbf{v}_u \) to a desired magnitude \( m \) is straightforward:
- \[ \mathbf{u} = m \times \mathbf{v}_u \]
- Direction remains unchanged, ensuring consistency in a specific orientation.
- The magnitude or length changes, adjusting for specific problems or context-based needs.
- Multiply each component by 15 to get \[ \langle \frac{30}{\sqrt{21}}, \frac{60}{\sqrt{21}}, \frac{15}{\sqrt{21}} \rangle \].