Question

A Berechnen Sie jeweils den Einheitsvektor \(\overrightarrow{e_{a}}\) in Richtung von \(\vec{a}\) a) \(\vec{a}=(3,-1,2)\) b) \(\vec{a}=(2,-1,-2)\) B Berechnen Sie den Abstand \(\vec{d}\) der Punkte \(P_{1}\) und \(P_{2}\) a) \(P_{1}=(3,2,0)\) b) \(P_{1}=(-2,-1,3)\) $$ P_{2}=(-1,4,2) $$ $$ P_{2}=(4,-2,-1) $$ C Ein Flugzeug fliege auf Nordkurs. Seine Geschwindigkeit gegenüber der Luft beträgt $$ \overrightarrow{v_{1}}=\left(0 \frac{\mathrm{km}}{\mathrm{h}}, 300 \frac{\mathrm{km}}{\mathrm{h}}\right) $$ Geben Sie die Geschwindigkeit des Flugzeugs über Land für drei verschiedene Windgeschwindigkeiten an: a) \(\overrightarrow{v_{2}}=\left(0 \frac{k m}{h},-50 \frac{k m}{h}\right)\) (Gegenwind) b) \(\overrightarrow{v_{3}}=\left(50 \frac{\mathrm{km}}{\mathrm{h}}, 0 \frac{\mathrm{km}}{\mathrm{h}}\right)\) (Seitenwind) c) \(\overrightarrow{v_{4}}=\left(0 \frac{k m}{\hbar},+50 \frac{k m}{h}\right)\) (Rückenwind) ? Geben Sie den Betrag der Absolutgeschwindigkeit über dem Erdboden für alle drei Fälle an d) \(\left|\overrightarrow{v_{1}}+\overrightarrow{v_{2}}\right|\) e) \(\left|\overrightarrow{v_{1}}+\overrightarrow{v_{3}}\right|\) f) \(\left|\overrightarrow{v_{1}}+\overrightarrow{v_{4}}\right|\)

Short answer

Unit vectors: (3/√14, -1/√14, 2/√14) and (2/3, -1/3, -2/3). Distances: 2√6 and √53. Ground speeds: 250 km/h, 304.1 km/h, and 350 km/h.

Step-by-step solution

- Calculate the magnitude of vector \(\textbackslash vec{a}\)

\[ \|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

- Calculate the unit vector for \(\textbackslash vec{a}\)

\[ \overrightarrow{e_{a}} = \frac{\vec{a}}{\|\vec{a}\|} \]

Application for vector \(\textbackslash vec{a} = (3, -1, 2)\)

First, find the magnitude: \[\|\vec{a}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}\]Now, calculate the unit vector: \[\overrightarrow{e_{a}} = \frac{(3, -1, 2)}{\sqrt{14}} = \left(\frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}}\right)\]

Application for vector \(\textbackslash vec{a} = (2, -1, -2)\)

First, find the magnitude: \[\|\vec{a}\| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9}\]Now, calculate the unit vector: \[\overrightarrow{e_{a}} = \frac{(2, -1, -2)}{\sqrt{9}} = \left(\frac{2}{3}, \frac{-1}{3}, \frac{-2}{3}\right)\]

- Calculate distance \textbackslash vec{d}\) between points \(P_{1}\) and \(P_{2}\)

\[ \|\vec{d}\| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Application for points \(P_{1} = (3,2,0)\) and \(P_{2} = (-1,4,2)\)

Calculate the distance: \[\|\vec{d}\| = \sqrt{(-1 - 3)^2 + (4 - 2)^2 + (2 - 0)^2} = \sqrt{(-4)^2 + 2^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}\]

Application for points \(P_{1} = (-2,-1,3)\) and \(P_{2} = (4,-2,-1)\)

Calculate the distance: \[\|\vec{d}\| = \sqrt{(4 - (-2))^2 + (-2 - (-1))^2 + (-1 - 3)^2} = \sqrt{(4 + 2)^2 + (-2 + 1)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-1)^2 + (-4)^2} = \sqrt{36 + 1 + 16} = \sqrt{53}\]

Determine the resultant speed of the plane \(\overrightarrow{v_{\text{ground}}} = \overrightarrow{v_1} + \overrightarrow{v_2}\)

For the resultant speed, add vectors: \[\overrightarrow{v_{\text{ground}}} = \overrightarrow{v_1} + \overrightarrow{v_2}\]

Application for wind speed vector \(\overrightarrow{v_2} = (0, -50)\) (Headwind)

Calculate the resultant speed: \[\overrightarrow{v_{\text{ground}}} = (0, 300) + (0, -50) = (0, 250)\]Magnitude: \[|\overrightarrow{v_{\text{ground}}}| = \sqrt{0^2 + 250^2} = 250\, \text{km/h}\]

Application for wind speed vector \(\overrightarrow{v_3} = (50, 0)\) (Crosswind)

Calculate the resultant speed: \[\overrightarrow{v_{\text{ground}}} = (0, 300) + (50, 0) = (50, 300)\]Magnitude: \[|\overrightarrow{v_{\text{ground}}}| = \sqrt{50^2 + 300^2} = \sqrt{2500 + 90000} = \sqrt{92500} \approx 304.1\, \text{km/h}\]

Application for wind speed vector \(\overrightarrow{v_4} = (0, 50)\) (Tailwind)

Calculate the resultant speed: \[\overrightarrow{v_{\text{ground}}} = (0, 300) + (0, 50) = (0, 350)\]Magnitude: \[|\overrightarrow{v_{\text{ground}}}| = \sqrt{0^2 + 350^2} = 350\, \text{km/h}\]

Key concepts

Unit Vector

A Unit Vector is a vector with a magnitude of 1. Unit vectors are often used to represent directions. To find a unit vector in the same direction as a given vector \( \overrightarrow{a} \), you must divide the vector by its magnitude. Here's the formula:
\( \overrightarrow{e_{a}} = \frac{\overrightarrow{a}}{\| \overrightarrow{a} \|} \)

Let's apply this to an example. Given a vector \( \overrightarrow{a} = (3, -1, 2) \), we first calculate its magnitude:
\( \| \overrightarrow{a} \| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \).
Then, the unit vector is:
\( \overrightarrow{e_a} = \frac{(3, -1, 2)}{\sqrt{14}} = \left( \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \right) \).

Thus, you get a clear direction of the vector, normalized to a magnitude of 1.

Distance Between Points

The distance between two points in space can be found using the distance formula, which is derived from the Pythagorean theorem. For points \( P_1 = (x_1, y_1, z_1) \) and \( P_2 = (x_2, y_2, z_2) \), the distance \( \vec{d} \) is given by:
\( \| \vec{d} \| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \).

For example, to find the distance between \( P_1 = (3, 2, 0) \) and \( P_2 = (-1, 4, 2) \):
\( \| \vec{d} \| = \sqrt{(-1 - 3)^2 + (4 - 2)^2 + (2 - 0)^2} = \sqrt{(-4)^2 + 2^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \).

This calculation ensures you understand the spatial relationship between the two points.

Vector Magnitude

The magnitude of a vector \( \overrightarrow{a} \) represents its length. To compute it, use the formula:
\( \| \overrightarrow{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).

If we have a vector \( \overrightarrow{a} = (2, -1, -2) \), its magnitude is:
\( \| \overrightarrow{a} \| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).
Knowing the magnitude is crucial for various applications in physics, such as calculating unit vectors or understanding the strength of forces represented by vectors.

Resultant Velocity

Resultant Velocity is the vector sum of two or more individual velocity vectors. This concept is vital in determining the actual speed and direction of an object in motion.
Consider a plane with a velocity \( \overrightarrow{v_1} = (0, 300) \) km/h heading north.
Now assume a headwind with speed \( \overrightarrow{v_2} = (0, -50) \) km/h.
The resultant velocity over the ground \( \overrightarrow{v_{ground}} \) can be found by adding these vectors:
\( \overrightarrow{v_{ground}} = (0, 300) + (0, -50) = (0, 250) \).
The magnitude of resultant speed is:
\( | \overrightarrow{v_{ground}} | = \sqrt{0^2 + 250^2} = 250 \) km/h.
This calculation helps in various scenarios like navigation of aircraft where wind speeds can affect the actual travel speed and direction.

Coordinate Geometry in Physics

Coordinate Geometry in Physics helps to describe the position and movement of objects in space. Using coordinates (x, y, z), we can analyze points, vectors, and the relationships between them.
For instance, locating points in 3D space allows us to compute distances, as shown with the distance formula.
Moreover, vectors defined by coordinates enables us to calculate magnitudes, directions, and perform operations like addition to find resultant vectors.
This branch of geometry becomes particularly powerful when analyzing motions and forces in physics, bringing clarity to how different objects interact in terms of their positions and velocities.