Chapter 11: Problem 63
Let \(\vec{v}=\left\langle v_{1}, v_{2}\right\rangle\) be any non-zero vector. Show that \(\frac{1}{\|\vec{v}\|} \vec{v}\) has length 1 .
Short Answer
Expert verified
The vector \( \frac{1}{\|\vec{v}\|} \vec{v} \) has a length of 1, proving it is a unit vector.
Step by step solution
01
Understand the Vector and its Magnitude
A vector \( \vec{v} = \langle v_1, v_2 \rangle \) is given. The magnitude of this vector is calculated using the formula \( \|\vec{v}\| = \sqrt{v_1^2 + v_2^2} \). This tells us how long the vector is when mapped from its origin to its endpoint.
02
Normalize the Vector
To normalize the vector \( \vec{v} \), we divide it by its magnitude. Thus, the normalized vector is \( \frac{1}{\|\vec{v}\|} \vec{v} = \left\langle \frac{v_1}{\|\vec{v}\|}, \frac{v_2}{\|\vec{v}\|} \right\rangle \). This essentially scales the vector to have a magnitude of 1.
03
Calculate the Length of the Normalized Vector
The length of the normalized vector \( \frac{1}{\|\vec{v}\|} \vec{v} = \left\langle \frac{v_1}{\|\vec{v}\|}, \frac{v_2}{\|\vec{v}\|} \right\rangle \) is found using its magnitude: \( \left\| \frac{1}{\|\vec{v}\|} \vec{v} \right\| = \sqrt{\left(\frac{v_1}{\|\vec{v}\|}\right)^2 + \left(\frac{v_2}{\|\vec{v}\|}\right)^2} \).
04
Simplify the Expression
Simplify the magnitude expression: \[ \left\| \frac{1}{\|\vec{v}\|} \vec{v} \right\| = \sqrt{\frac{v_1^2}{\|\vec{v}\|^2} + \frac{v_2^2}{\|\vec{v}\|^2}} = \sqrt{\frac{v_1^2 + v_2^2}{\|\vec{v}\|^2}} \]. Since \( \|\vec{v}\| = \sqrt{v_1^2 + v_2^2} \), substitute back to get \( \|\vec{v}\|^2 = v_1^2 + v_2^2 \).
05
Verify the Length is One
The expression simplifies to \( \sqrt{\frac{v_1^2 + v_2^2}{v_1^2 + v_2^2}} = \sqrt{1} = 1 \). Thus, the length of the vector \( \frac{1}{\|\vec{v}\|} \vec{v} \) is indeed 1, proving it is a unit vector.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude of a Vector
The magnitude of a vector is an essential concept in vector algebra that gives you a sense of how long a vector is. If you think of a vector as an arrow, the magnitude tells you the length of that arrow. Consider the vector \( \vec{v} = \langle v_1, v_2 \rangle \). To find its magnitude, use the formula:\[\|\vec{v}\| = \sqrt{v_1^2 + v_2^2}\]
- The squared terms \( v_1^2 \) and \( v_2^2 \) represent the x and y components of the vector, respectively.
- Adding these squared components and then taking the square root gives the overall length from the vector’s start to its endpoint.
Unit Vector
A unit vector is a vector that has a magnitude of exactly one. It is a useful tool because it indicates direction only and doesn't carry any information about the vector's length. Any vector can be transformed into a unit vector by dividing it by its magnitude.The formula to achieve this transformation for a vector \( \vec{v} \) is:\[\frac{\vec{v}}{\|\vec{v}\|} = \left\langle \frac{v_1}{\|\vec{v}\|}, \frac{v_2}{\|\vec{v}\|} \right\rangle\]
- This operation resizes the vector to have a length of 1 while maintaining its original direction.
- Unit vectors are particularly useful in determining directions in physics and engineering.
Normalizing a Vector
Normalizing a vector is the process of converting any non-zero vector into a unit vector. This process is crucial for simplifying complex vector calculations and ensuring clarity in directional components.To normalize a vector \( \vec{v} = \langle v_1, v_2 \rangle \):1. First, find the magnitude, \( \|\vec{v}\| \), as described previously.2. Then, apply the normalization formula:\[\frac{1}{\|\vec{v}\|} \vec{v} = \left\langle \frac{v_1}{\|\vec{v}\|}, \frac{v_2}{\|\vec{v}\|} \right\rangle\]
- This scales the original vector so that its magnitude becomes 1.
- The resulting vector still points in the same direction as \( \vec{v} \).