Example Question - vector addition
Here are examples of questions we've helped users solve.
Vector Addition Problem
<p>Let the two vectors be </p> <p> \[ \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 & y \\ x & -2 \end{bmatrix} = \begin{bmatrix} 1 \\ -4 \\ 9 \\ -2 \end{bmatrix} \] </p> <p>Set up the equation for each component:</p> <p>1. \(2 - 1 = 1\)</p> <p>2. \(3 + y = -4\)</p> <p>3. \(0 + x = -2\)</p> <p>From the first equation: \(1\) is valid.</p> <p>From the second equation: \(y = -4 - 3 = -7\).</p> <p>From the third equation: \(x = -2\).</p> <p>Thus, \(x = -2\) and \(y = -7\).</p>
Vector Displacement and Distance Calculation
<p>Ilustramos el problema con un diagrama de vectores:</p> <p>Punto inicial O. El coche se mueve 52 km al Este, llegando al punto A. Luego se mueve 27 km al Sur, llegando al punto B.</p> <p>\[\vec{OA} = 52 \text{ km Este}, \quad \vec{AB} = 27 \text{ km Sur}\]</p> <p>La distancia es la longitud total del camino recorrido, es decir:</p> <p>\[ \text{Distancia} = |\vec{OA}| + |\vec{AB}| = 52 \text{ km} + 27 \text{ km} = 79 \text{ km} \]</p> <p>El desplazamiento es el vector resultante desde el punto de inicio al punto final, es decir \(\vec{OB}\). Calculamos su magnitud utilizando el teorema de Pitágoras, donde el desplazamiento es la hipotenusa de un triángulo rectángulo con lados 52 km y 27 km:</p> <p>\[ |\vec{OB}| = \sqrt{52^2 + 27^2} \]</p> <p>\[ |\vec{OB}| = \sqrt{2704 + 729} \]</p> <p>\[ |\vec{OB}| = \sqrt{3433} \]</p> <p>\[ |\vec{OB}| \approx 58.59 \text{ km} \]</p> <p>Por tanto, la distancia recorrida es de 79 km y el desplazamiento es aproximadamente de 58.59 km.</p>
Calculating Distance and Displacement
<p>La distancia es la longitud total del camino recorrido por el auto, que es igual a la suma de los desplazamientos en cada dirección:</p> \[ \text{Distancia} = 52 \text{ km} + 27 \text{ km} = 79 \text{ km} \] <p>El desplazamiento es el vector que va desde el punto inicial al final.</p> <p>Para calcular la magnitud del desplazamiento, usamos el teorema de Pitágoras, considerando un desplazamiento de 52 km hacia el Este y 27 km hacia el Sur, que forman un triángulo rectángulo.</p> \[ \text{Desplazamiento} = \sqrt{(52 \text{ km})^2 + (27 \text{ km})^2} \] \[ \text{Desplazamiento} = \sqrt{2704 + 729} \] \[ \text{Desplazamiento} = \sqrt{3433} \] \[ \text{Desplazamiento} \approx 58.6 \text{ km} \]
Graph Interpretation and Vector Components
<p>Considérons les coordonnées des points A et B sur le graphique :</p> <p>Point A : (2, 3)</p> <p>Point B : (7, 1)</p> <p>Calculons les composantes du vecteur \(\overrightarrow{AB}\) :</p> <p>\(\Delta x = x_B - x_A = 7 - 2 = 5\)</p> <p>\(\Delta y = y_B - y_A = 1 - 3 = -2\)</p> <p>Les composantes du vecteur \(\overrightarrow{AB}\) sont donc (5, -2).</p> <p>Utilisons ces composantes pour trouver les vecteurs \(\overrightarrow{u}\) et \(\overrightarrow{v}\) :</p> <p>\(\overrightarrow{u}\) + \(\overrightarrow{v}\) = \(\overrightarrow{AB}\)</p> <p>\(\overrightarrow{u}\) + \(\overrightarrow{v}\) = (5, -2)</p> <p>Selon l'énoncé, \(\overrightarrow{u}\) = (2, a) et \(\overrightarrow{v}\) = (b, -3).</p> <p>Donc (2 + b, a - 3) = (5, -2).</p> <p>Égalons les composantes correspondantes :</p> <p>2 + b = 5 => b = 3</p> <p>a - 3 = -2 => a = 1</p> <p>Par conséquent, les valeurs de a et b sont respectivement 1 et 3.</p>
Vector Equality Implication
<p>Soit deux vecteurs \(\overrightarrow{AB}\) et \(\overrightarrow{BC}\), avec la condition \(\overrightarrow{AB} = \overrightarrow{BC}\).</p> <p>En utilisant la propriété de l'addition des vecteurs:</p> <p>\(\overrightarrow{BA} = -\overrightarrow{AB}\)</p> <p>\(\overrightarrow{BA} + \overrightarrow{BC} = -\overrightarrow{AB} + \overrightarrow{BC}\)</p> <p>Étant donné que \(\overrightarrow{AB} = \overrightarrow{BC}\), on peut substituer \(\overrightarrow{BC}\) par \(\overrightarrow{AB}\) dans l'équation:</p> <p>\(\overrightarrow{BA} + \overrightarrow{BC} = -\overrightarrow{AB} + \overrightarrow{AB}\)</p> <p>Cela nous donne:</p> <p>\(\overrightarrow{BA} + \overrightarrow{BC} = \overrightarrow{0}\)</p> <p>La réponse est donc \(\overrightarrow{BA} + \overrightarrow{BC} = \overrightarrow{0}\).</p>
Vectors Representation Exercise
<p>The image seems to display graphical representation of vector addition and subtraction. However, without the context of the problem such as specific questions or instructions related with the vectors, no unique solution can be provided.</p> <p>The first box possibly shows two vectors being added head-to-tail.</p> <p>The second box might be showing vector subtraction by adding the negative of a vector.</p> <p>The third box possibly represents two equal vectors pointing in opposite directions, suggesting that their sum is zero.</p> <p>In the fourth box, it could be an example of the Parallelogram rule for vector addition.</p>
Determination of Resultant Force Magnitude and Direction
For Prob. F2-1: <p>\(R_{x} = 6 \cos(45^\circ)N - 2N\)</p> <p>\(R_{x} = 4.2426N - 2N\)</p> <p>\(R_{x} = 2.2426N\)</p> <p>\(R_{y} = 6 \sin(45^\circ)N\)</p> <p>\(R_{y} = 4.2426N\)</p> <p>\(R = \sqrt{R_{x}^2 + R_{y}^2}\)</p> <p>\(R = \sqrt{2.2426^2 + 4.2426^2}N\)</p> <p>\(R = \sqrt{5.0291 + 18.0000}N\)</p> <p>\(R = \sqrt{23.0291}N\)</p> <p>\(R \approx 4.8001N\)</p> <p>\(\theta = \arctan\left(\frac{R_{y}}{R_{x}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{4.2426}{2.2426}\right)\)</p> <p>\(\theta \approx 62.1021^\circ\) clockwise from the +x-axis</p> For Prob. F2-2: Isolated from the image and omitted based on the provided instructions. For Prob. F2-3: <p>\(R_{x} = 800 \cos(30^\circ)N - 600 \cos(45^\circ)N\)</p> <p>\(R_{x} = 692.82N - 424.26N\)</p> <p>\(R_{x} = 268.56N\)</p> <p>\(R_{y} = 800 \sin(30^\circ)N + 600 \sin(45^\circ)N\)</p> <p>\(R_{y} = 400N + 424.26N\)</p> <p>\(R_{y} = 824.26N\)</p> <p>\(R = \sqrt{R_{x}^2 + R_{y}^2}\)</p> <p>\(R = \sqrt{268.56^2 + 824.26^2}N\)</p> <p>\(R = \sqrt{72105.59 + 679186.67}N\)</p> <p>\(R = \sqrt{751292.27}N\)</p> <p>\(R \approx 866.77N\)</p> <p>\(\theta = \arctan\left(\frac{R_{y}}{R_{x}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{824.26}{268.56}\right)\)</p> <p>\(\theta \approx 71.5651^\circ\) counterclockwise from the +x-axis</p>
Analysis of Resultant Forces in Various Configurations
<p>For F2-1:</p> <p>Using vector addition, the resultant force \( R \) is the vector sum of the two forces.</p> <p>\( R_x = 6\cos(45^\circ) + 2\cos(45^\circ) \)</p> <p>\( R_x = (6 + 2) \cdot \frac{\sqrt{2}}{2} \)</p> <p>\( R_x = 8 \cdot \frac{\sqrt{2}}{2} \)</p> <p>\( R_x = 4\sqrt{2} \text{ N} \)</p> <p>\( R_y = 6\sin(45^\circ) - 2\sin(45^\circ) \)</p> <p>\( R_y = (6 - 2) \cdot \frac{\sqrt{2}}{2} \)</p> <p>\( R_y = 4 \cdot \frac{\sqrt{2}}{2} \)</p> <p>\( R_y = 2\sqrt{2} \text{ N} \)</p> <p>The magnitude of the resultant force is:</p> <p>\( R = \sqrt{R_x^2 + R_y^2} \)</p> <p>\( R = \sqrt{(4\sqrt{2})^2 + (2\sqrt{2})^2} \)</p> <p>\( R = \sqrt{32 + 8} \)</p> <p>\( R = \sqrt{40} \)</p> <p>\( R = 2\sqrt{10} \text{ N} \)</p> <p>The direction is given by:</p> <p>\( \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \)</p> <p>\( \theta = \tan^{-1}\left(\frac{2\sqrt{2}}{4\sqrt{2}}\right) \)</p> <p>\( \theta = \tan^{-1}\left(\frac{1}{2}\right) \)</p> <p>The direction is \( \theta \) degrees above the +x-axis.</p> <p>For F2-2:</p> <p>This part of the question corresponds to a different problem about forces acting on a hook.</p> <p>For F2-3:</p> <p>This part of the question corresponds to yet another different problem about determining the magnitude of the resultant force and its direction.</p> <p>Note: Since the question is about F2-1, only the solution for F2-1 is provided, and not for F2-2 or F2-3.</p>
Analysis of Resultant Forces in Different Configurations
For problem F2-1: <p>The two forces can be added using vector addition. Let's break them down into their x and y components. \(F_{1x} = 6 \cos(45^\circ), F_{1y} = 6 \sin(45^\circ)\) \(F_{2x} = -2 \sin(45^\circ), F_{2y} = 2 \cos(45^\circ)\) \(R_x = F_{1x} + F_{2x}\) \(R_y = F_{1y} + F_{2y}\)</p> <p>The magnitude of the resultant force \(R\) can be found using Pythagoras' theorem: \(R = \sqrt{R_x^2 + R_y^2}\) The direction \(\theta\) relative to the x-axis is: \(\theta = \arctan\left(\frac{R_y}{R_x}\right)\)</p> For problem F2-2: <p>To find the resultant force acting on the hook, we add the two forces vectorially. \(F_{1x} = 200 \cos(30^\circ), F_{1y} = 200 \sin(30^\circ)\) \(F_{2x} = -500 \sin(60^\circ), F_{2y} = 500 \cos(60^\circ)\) \(R_x = F_{1x} + F_{2x}\) \(R_y = F_{1y} + F_{2y}\)</p> <p>Thus, the magnitude of the resultant force \(R\) is: \(R = \sqrt{R_x^2 + R_y^2}\)</p> For problem F2-3: <p>The resultant force is the vector sum of the three forces. \(F_{1x} = 800 \cos(30^\circ), F_{1y} = 800 \sin(30^\circ)\) \(F_{2x} = -600\) \(F_{3x} = -600 \cos(45^\circ), F_{3y} = -600 \sin(45^\circ)\) \(R_x = F_{1x} + F_{2x} + F_{3x}\) \(R_y = F_{1y} + F_{3y}\)</p> <p>The magnitude of the resultant force \(R\) is: \(R = \sqrt{R_x^2 + R_y^2}\) And the direction \(\theta\) measured counterclockwise from the positive x-axis is: \(\theta = \arctan\left(\frac{R_y}{R_x}\right)\)</p>
Determining the Resultant Force in Various Configurations
For Prob. F2-1: <p>\( R_x = 6\cos(45^\circ) + 2\sin(45^\circ) \)</p> <p>\( R_y = 6\sin(45^\circ) - 2\cos(45^\circ) \)</p> <p>\( R = \sqrt{R_x^2 + R_y^2} \)</p> <p>\( \theta = \arctan\left(\frac{R_y}{R_x}\right) \)</p> For Prob. F2-2: <p>\( R_x = 500\cos(30^\circ) - 200\sin(60^\circ) \)</p> <p>\( R_y = 500\sin(30^\circ) + 200\cos(60^\circ) \)</p> <p>\( R = \sqrt{R_x^2 + R_y^2} \)</p> For Prob. F2-3: <p>\( R_x = 800\cos(30^\circ) - 600\sin(45^\circ) \)</p> <p>\( R_y = 800\sin(30^\circ) + 600\cos(45^\circ) \)</p> <p>\( R = \sqrt{R_x^2 + R_y^2} \)</p> <p>\( \theta = \arctan\left(\frac{R_y}{R_x}\right) \)</p>
Simplifying Vector Equations
The question is asking to simplify vector equations. Let's solve the first part a): **a) \(\vec{AM} + \vec{AC} + \vec{MC}\)** To simplify this, we use the concept that a vector going from point A to point C (\(\vec{AC}\)) is the same as going from A to M (\(\vec{AM}\)) added to going from M to C (\(\vec{MC}\)). Symbolically: **\(\vec{AM} + \vec{MC} = \vec{AC}\)** Using this property, we can replace \(\vec{AM} + \vec{MC}\) in the initial equation with \(\vec{AC}\). So the equation simplifies to: **\(\vec{AC} + \vec{AC}\)** When we add a vector to itself, we are just doubling the length of the vector in the same direction. Therefore, the final simplified expression is: **2\(\vec{AC}\)**
Vector Operations and Magnitudes
Természetesen, segítek megoldani a feladatot. Az \( a \) és \( b \) vektorokat már megadták nekünk: \[ a = 5i - 12j \] \[ b = 4i - 3j \] **a) rész:** Az \( a + b \) vektor koordinátái: \[ a + b = (5i - 12j) + (4i - 3j) \] \[ a + b = (5 + 4)i + (-12 - 3)j \] \[ a + b = 9i - 15j \] A \( 3a + 2b \) vektor koordinátái: \[ 3a + 2b = 3(5i - 12j) + 2(4i - 3j) \] \[ 3a + 2b = (3 \cdot 5)i + (3 \cdot -12)j + (2 \cdot 4)i + (2 \cdot -3)j \] \[ 3a + 2b = 15i - 36j + 8i - 6j \] \[ 3a + 2b = (15 + 8)i + (-36 - 6)j \] \[ 3a + 2b = 23i - 42j \] **b) rész:** A vektorok hosszát vagy abszolút értéket a következő képlettel számoljuk: \[ |v| = \sqrt{x^2 + y^2} \] ahol \( v = xi + yj \). \( |a| \) értéke: \[ |a| = \sqrt{(5)^2 + (-12)^2} \] \[ |a| = \sqrt{25 + 144} \] \[ |a| = \sqrt{169} \] \[ |a| = 13 \] \( |b| \) értéke: \[ |b| = \sqrt{(4)^2 + (-3)^2} \] \[ |b| = \sqrt{16 + 9} \] \[ |b| = \sqrt{25} \] \[ |b| = 5 \] Végül a \( |a + b| \) értéke: \[ |a + b| = \sqrt{(9)^2 + (-15)^2} \] \[ |a + b| = \sqrt{81 + 225} \] \[ |a + b| = \sqrt{306} \] \[ |a + b| = \sqrt{9 \cdot 34} \] \[ |a + b| = 3\sqrt{34} \] (közelítőleg \( |a + b| \) értéke kerekítve lesz ami a gyök 34-nek felel meg, mivel a gyök alatti szám nem négyzetszám) Ez adja a részfeladatok megoldását.
Vector Addition for Airplane Velocity Calculation
To solve this problem, you need to use vector addition to calculate the resultant velocity of the airplane relative to the ground, taking into account the wind's effect. The airplane has an initial velocity of 70 m/s, and the wind has a velocity of 18 m/s at a 120° angle to the airplane's direction. First, let's define the vectors: - The airplane's velocity vector, \( \vec{V}_a \), is 70 m/s in the direction of the positive x-axis. - The wind's velocity vector, \( \vec{V}_w \), is 18 m/s at a 120° angle from the airplane's direction (which means it is 30° above the negative x-axis, or equivalently, 150° from the positive x-axis). Now, decompose \( \vec{V}_w \) into x and y components: - \( V_{wx} = \vec{V}_w \cos(150°) = 18 \cos(150°) = -18 \cos(30°) \) (since cos(150°) = -cos(30°)) - \( V_{wy} = \vec{V}_w \sin(150°) = 18 \sin(150°) = 18 \sin(30°) \) (since sin(150°) = sin(30°)) Calculate these components: - \( V_{wx} = -18 \times \cos(30°) = -18 \times \sqrt{3}/2 \) - \( V_{wy} = 18 \times \sin(30°) = 18 \times 1/2 \) Now, add the wind's components to the airplane's velocity to find the resultant velocity vector, \( \vec{V}_r \): - \( V_{rx} = V_{ax} + V_{wx} = 70 + (-18 \times \sqrt{3}/2) \) - \( V_{ry} = V_{ay} + V_{wy} = 0 + 18 \times 1/2 \) Calculate the magnitude of the resultant velocity: - \( |\vec{V}_r| = \sqrt{V_{rx}^2 + V_{ry}^2} \) Lastly, you'll want to calculate the distance traveled over 6 minutes. You can convert the time to seconds (6 minutes * 60 seconds/minute = 360 seconds) and then use \( d = |\vec{V}_r| \times t \) to find the distance \( d \). Plug in the values and calculate the numerical results.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com