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Chapter 12: Problem 23

Find a unit vector that has the same direction as the given vector. $$ \langle 6,-2\rangle $$

Short Answer

Expert verified
The unit vector is \( \langle \frac{3\sqrt{10}}{10}, -\frac{\sqrt{10}}{10} \rangle \).

Step by step solution

01

Find the Magnitude of the Vector

To convert a vector into a unit vector, we first find its magnitude. The magnitude of a vector \( \langle a,b \rangle \) is calculated using the formula:\[\| \langle a,b \rangle \| = \sqrt{a^2 + b^2}\]Substituting in the given values of \( a = 6 \) and \( b = -2 \), the magnitude is:\[\| \langle 6,-2 \rangle \| = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}\]
02

Divide Each Component by the Magnitude

A unit vector \( \vec{u} \) in the same direction as \( \langle a,b \rangle \) is computed by dividing each component of the vector by its magnitude.\[\vec{u} = \left\langle \frac{a}{\|\langle a,b \rangle \|}, \frac{b}{\|\langle a,b \rangle \|} \right\rangle\]Substituting the values calculated and given, we find:\[\vec{u} = \left\langle \frac{6}{2\sqrt{10}}, \frac{-2}{2\sqrt{10}} \right\rangle = \left\langle \frac{3}{\sqrt{10}}, -\frac{1}{\sqrt{10}} \right\rangle\]To rationalize the denominator, multiply both numerator and denominator by \( \sqrt{10} \) for each component:\[\vec{u} = \left\langle \frac{3\sqrt{10}}{10}, -\frac{\sqrt{10}}{10} \right\rangle\]

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Key Concepts

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

Vector Magnitude
The magnitude of a vector is a measure of its length. It is calculated using a straightforward formula, similar to finding the hypotenuse of a right triangle. For any vector \( \langle a, b \rangle \), the magnitude is found by plugging the vector's components into the equation: \[\| \langle a, b \rangle \| = \sqrt{a^2 + b^2}\]In our example, the vector is \( \langle 6, -2 \rangle \). To determine its magnitude:
  • Square each component: \( 6^2 = 36 \) and \( (-2)^2 = 4 \).
  • Add these squares together: \( 36 + 4 = 40 \).
  • Take the square root: \( \sqrt{40} = 2\sqrt{10} \).
Understanding the magnitude helps us grasp how to manipulate vectors, leading us to find unit vectors, among other things.
Vector Direction
A unit vector maintains the original vector's direction but has a magnitude of exactly 1. It acts like a directional arrow. To create a unit vector from a non-unit vector, we divide each vector component by its magnitude. The original vector is \( \langle 6, -2 \rangle \). With its magnitude computed as \( 2\sqrt{10} \), the formula for our unit vector becomes:
  • Divide the horizontal component: \( \frac{6}{2\sqrt{10}} = \frac{3}{\sqrt{10}} \).
  • Divide the vertical component: \( \frac{-2}{2\sqrt{10}} = -\frac{1}{\sqrt{10}} \).
This process ensures that although the scale of the vector is adjusted, its direction remains unchanged. It鈥檚 essential for applications, like physics or engineering, where direction is key.
Rationalizing Denominator
Rationalizing the denominator is an essential math skill, especially with fractions in vector components. It particularly helps in making expressions cleaner and easier to interpret. When we have a fraction with a square root in the denominator, we multiply both the numerator and the denominator by the square root to remove it.For the unit vector, derived as \( \langle \frac{3}{\sqrt{10}}, -\frac{1}{\sqrt{10}} \rangle \), we rationalize as follows:
  • Multiply the numerators and denominators by \( \sqrt{10} \):
  • \( \frac{3\sqrt{10}}{10} \rightarrow \frac{3}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} \).
  • Similarly, \(-\frac{\sqrt{10}}{10} \rightarrow -\frac{1}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} \).
This results in \( \langle \frac{3\sqrt{10}}{10}, -\frac{\sqrt{10}}{10} \rangle \), eliminating any root in the denominator and simplifying further calculations and comprehension.

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