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Magazine Chapter 8: Problem 37
Find a unit vector that has the same direction as the given vector. $$\mathbf{r}=\langle- 2,-8\rangle$$
Short Answer
Expert verified
The unit vector is \(\left\langle \frac{-\sqrt{17}}{17}, \frac{-4\sqrt{17}}{17} \right\rangle\).
Step by step solution
01
Understand the Given Vector
The given vector is \(\mathbf{r} = \langle -2, -8 \rangle\). A unit vector has a magnitude of 1 and points in the same direction as the original vector.
02
Find the Magnitude of the Vector
The magnitude of a vector \(\mathbf{r} = \langle x, y \rangle\) is given by \[ \|\mathbf{r}\| = \sqrt{x^2 + y^2} \]Substitute \(x = -2\) and \(y = -8\) to find the magnitude:\[ \|\mathbf{r}\| = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17} \]
03
Divide Each Component by the Magnitude
To find the unit vector, divide each component of \(\mathbf{r}\) by its magnitude \(\|\mathbf{r}\| = 2\sqrt{17} \):\[ \mathbf{u} = \left\langle \frac{-2}{2\sqrt{17}}, \frac{-8}{2\sqrt{17}} \right\rangle = \left\langle \frac{-1}{\sqrt{17}}, \frac{-4}{\sqrt{17}} \right\rangle \]
04
Rationalize the Denominators (Optional)
To make the unit vector more presentable, rationalize the denominators:\[ \mathbf{u} = \left\langle \frac{-1}{\sqrt{17}}, \frac{-4}{\sqrt{17}} \right\rangle = \left\langle \frac{-1 \cdot \sqrt{17}}{\sqrt{17} \cdot \sqrt{17}}, \frac{-4 \cdot \sqrt{17}}{\sqrt{17} \cdot \sqrt{17}} \right\rangle = \left\langle \frac{-\sqrt{17}}{17}, \frac{-4\sqrt{17}}{17} \right\rangle \]
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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 quantifies its length or size. For a vector \(\backslashmathbf{r} = \backslashlangle x, y \backslashrangle\), the magnitude is found using the Pythagorean theorem and is given by the formula: \[ \backslash|\backslashmathbf{r}\backslash| = \backslashsqrt{x^2 + y^2} \] In our exercise, the vector is \(\backslashmathbf{r} = \backslashlangle -2, -8 \backslashrangle\), and so:
\[ \backslash|\backslashmathbf{r}\backslash| = \backslashsqrt{(-2)^2 + (-8)^2} = \backslashsqrt{4 + 64} = \backslashsqrt{68} = 2 \backslashsqrt{17} \] This gives you the magnitude of the original vector, which is 2鈭�17.
Magnitude is crucial because it helps in understanding the vector's size regardless of its direction. This is the first step in finding a unit vector, which we will discuss next.
\[ \backslash|\backslashmathbf{r}\backslash| = \backslashsqrt{(-2)^2 + (-8)^2} = \backslashsqrt{4 + 64} = \backslashsqrt{68} = 2 \backslashsqrt{17} \] This gives you the magnitude of the original vector, which is 2鈭�17.
Magnitude is crucial because it helps in understanding the vector's size regardless of its direction. This is the first step in finding a unit vector, which we will discuss next.
vector normalization
Vector normalization is the process of scaling a vector so that it maintains its direction but has a magnitude of 1. This is known as finding the unit vector.
To normalize a vector, you divide each of its components by its magnitude. For a vector \(\backslashmathbf{r} = \backslashlangle x, y \backslashrangle\), its unit vector is calculated as: \[ \backslashmathbf{u} = \backslashleft\backslashlangle \backslashfrac{x}{\backslash|\backslashmathbf{r}\backslash|}, \backslashfrac{y}{\backslash|\backslashmathbf{r}\backslash|} \backslashright\backslashrangle \] Using our example vector \(\backslashmathbf{r} = \backslashlangle -2, -8 \backslashrangle\), with a magnitude \(\backslash|\backslashmathbf{r}\backslash| = 2 \backslashsqrt{17}\):
\[ \backslashmathbf{u} = \backslashleft\backslashlangle \backslashfrac{-2}{2\backslashsqrt{17}}, \backslashfrac{-8}{2\backslashsqrt{17}} \backslashright\backslashrangle = \backslashleft\backslashlangle \backslashfrac{-1}{\backslashsqrt{17}}, \backslashfrac{-4}{\backslashsqrt{17}} \backslashright\backslashrangle \] The resulting vector \(\backslashmathbf{u}\) has the same direction as the original vector \(\backslashmathbf{r}\) but with a magnitude of 1. Normalizing vectors is a common practice in many applications such as physics and computer graphics.
To normalize a vector, you divide each of its components by its magnitude. For a vector \(\backslashmathbf{r} = \backslashlangle x, y \backslashrangle\), its unit vector is calculated as: \[ \backslashmathbf{u} = \backslashleft\backslashlangle \backslashfrac{x}{\backslash|\backslashmathbf{r}\backslash|}, \backslashfrac{y}{\backslash|\backslashmathbf{r}\backslash|} \backslashright\backslashrangle \] Using our example vector \(\backslashmathbf{r} = \backslashlangle -2, -8 \backslashrangle\), with a magnitude \(\backslash|\backslashmathbf{r}\backslash| = 2 \backslashsqrt{17}\):
\[ \backslashmathbf{u} = \backslashleft\backslashlangle \backslashfrac{-2}{2\backslashsqrt{17}}, \backslashfrac{-8}{2\backslashsqrt{17}} \backslashright\backslashrangle = \backslashleft\backslashlangle \backslashfrac{-1}{\backslashsqrt{17}}, \backslashfrac{-4}{\backslashsqrt{17}} \backslashright\backslashrangle \] The resulting vector \(\backslashmathbf{u}\) has the same direction as the original vector \(\backslashmathbf{r}\) but with a magnitude of 1. Normalizing vectors is a common practice in many applications such as physics and computer graphics.
rationalizing denominators
Rationalizing denominators is a method used to eliminate square roots or irrational numbers from the denominators of fractions. This makes expressions easier to handle and understand.
Given our example unit vector: \[ \backslashmathbf{u} = \backslashleft\backslashlangle \backslashfrac{-1}{\backslashsqrt{17}}, \backslashfrac{-4}{\backslashsqrt{17}} \backslashright\backslashrangle \], we rationalize its components by multiplying the numerator and the denominator by \(\backslashsqrt{17}\):
\[ \backslashfrac{-1}{\backslashsqrt{17}} \backslashrightarrow \backslashfrac{-1 \backslashcdot \backslashsqrt{17}}{\backslashsqrt{17} \backslashcdot \backslashsqrt{17}} = \backslashfrac{-\backslashsqrt{17}}{17} \]
\[ \backslashfrac{-4}{\backslashsqrt{17}} \backslashrightarrow \backslashfrac{-4 \backslashcdot \backslashsqrt{17}}{\backslashsqrt{17} \backslashcdot \backslashsqrt{17}} = \backslashfrac{-4 \backslashsqrt{17}}{17} \]
After rationalizing, our unit vector becomes:
\[ \backslashmathbf{u} = \backslashleft\backslashlangle \backslashfrac{-\backslashsqrt{17}}{17}, \backslashfrac{-4 \backslashsqrt{17}}{17} \backslashright\backslashrangle \] This process ensures the denominators are free of square roots, making the components clearer.
Given our example unit vector: \[ \backslashmathbf{u} = \backslashleft\backslashlangle \backslashfrac{-1}{\backslashsqrt{17}}, \backslashfrac{-4}{\backslashsqrt{17}} \backslashright\backslashrangle \], we rationalize its components by multiplying the numerator and the denominator by \(\backslashsqrt{17}\):
\[ \backslashfrac{-1}{\backslashsqrt{17}} \backslashrightarrow \backslashfrac{-1 \backslashcdot \backslashsqrt{17}}{\backslashsqrt{17} \backslashcdot \backslashsqrt{17}} = \backslashfrac{-\backslashsqrt{17}}{17} \]
\[ \backslashfrac{-4}{\backslashsqrt{17}} \backslashrightarrow \backslashfrac{-4 \backslashcdot \backslashsqrt{17}}{\backslashsqrt{17} \backslashcdot \backslashsqrt{17}} = \backslashfrac{-4 \backslashsqrt{17}}{17} \]
After rationalizing, our unit vector becomes:
\[ \backslashmathbf{u} = \backslashleft\backslashlangle \backslashfrac{-\backslashsqrt{17}}{17}, \backslashfrac{-4 \backslashsqrt{17}}{17} \backslashright\backslashrangle \] This process ensures the denominators are free of square roots, making the components clearer.
