Example Question - vectors
Here are examples of questions we've helped users solve.
Constructing Points in Vector Geometry
<p>Soit \( \vec{w} \) et \( \vec{v} \) deux vecteurs donnés sur un plan. Pour placer les points E, F, et G, on suit les étapes suivantes :</p> <p>1. Choisir un point arbitraire A comme point de départ pour la construction.</p> <p>2. Construire le point E tel que \( \overrightarrow{AE} = \vec{w} + \vec{v} \) en ajoutant les vecteurs \( \vec{w} \) et \( \vec{v} \) en partant du point A.</p> <p>3. Construire le point F tel que \( \overrightarrow{EF} = \vec{u} + \vec{v} \), où \( \vec{u} \) doit être défini ou donné dans le problème. On ajoute ces vecteurs en partant du point E que nous avons déjà placé.</p> <p>4. Construire le point G en se servant de la relation \( \overrightarrow{FG} = \vec{u} + \vec{v} \) en ajoutant les vecteurs \( \vec{u} \) et \( \vec{v} \) en partant du point F que nous avons placé à l'étape précédente.</p> <p>Note: Comme les vecteurs \( \vec{u} \) et \( \vec{A} \) ne sont pas définis dans le problème, on suppose qu'ils sont donnés ou que leurs valeurs seront choisies convenablement pour la construction.</p>
Determining the Resultant Force and its Direction
<p>Given:</p> <p>\(\vec{F_1} = 800 \text{ N} \ (\text{upwards along the y-axis})\)</p> <p>\(\vec{F_2} = 600 \text{ N} \ (\text{30 degrees below the -x-axis})\)</p> <p>To solve for the resultant force \(\vec{R}\), break \(\vec{F_2}\) into its x and y components.</p> <p>\(F_{2x} = 600 \cos(30^\circ)\)</p> <p>\(F_{2y} = -600 \sin(30^\circ)\) (The y-component is negative because it is downward.)</p> <p>Calculate \(F_{2x}\) and \(F_{2y}\):</p> <p>\(F_{2x} = 600 \times \frac{\sqrt{3}}{2} = 300\sqrt{3}\) N</p> <p>\(F_{2y} = -600 \times \frac{1}{2} = -300\) N</p> <p>The resultant force \(\vec{R}\) in the x and y components is:</p> <p>\(R_x = F_{2x}\) because there is no x-component for \(\vec{F_1}\).</p> <p>\(R_y = F_{1} + F_{2y}\)</p> <p>Calculate \(R_x\) and \(R_y\):</p> <p>\(R_x = 300\sqrt{3}\) N</p> <p>\(R_y = 800 - 300 = 500\) N</p> <p>Now calculate the magnitude of the resultant vector \(|\vec{R}|\):</p> <p>\(|\vec{R}| = \sqrt{R_x^2 + R_y^2}\)</p> <p>\(|\vec{R}| = \sqrt{(300\sqrt{3})^2 + (500)^2}\)</p> <p>\(|\vec{R}| = \sqrt{(900 \times 3) + (250000)}\)</p> <p>\(|\vec{R}| = \sqrt{2700 + 250000}\)</p> <p>\(|\vec{R}| = \sqrt{252700}\)</p> <p>\(|\vec{R}| \approx 502.7\) N</p> <p>To find the direction θ measured counterclockwise from the positive x-axis:</p> <p>\(\tan(\theta) = \frac{R_y}{R_x}\)</p> <p>\(\theta = \arctan\left(\frac{500}{300\sqrt{3}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{500}{300 \times \frac{\sqrt{3}}{2}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{500}{150\sqrt{3}}\right)\)</p> <p>\(\theta \approx \arctan\left(\frac{500}{259.81}\right)\)</p> <p>\(\theta \approx \arctan(1.925)\)</p> <p>\(\theta \approx 62.5^\circ\) (above the negative x-axis)</p> <p>Finally, from the positive x-axis, add 180 degrees because it is on the left half of the coordinate system:</p> <p>\(\theta_{\text{from positive x-axis}} = 180^\circ + \theta\)</p> <p>\(\theta_{\text{from positive x-axis}} = 180^\circ + 62.5^\circ\)</p> <p>\(\theta_{\text{from positive x-axis}} = 242.5^\circ\)</p>
Determining the Resultant Force and its Direction
<p>To find the magnitude of the resultant force, we can break the forces into their x and y components. For the 600 N force:</p> <p>\[ F_{x1} = 600 \cos(30^\circ) \]</p> <p>\[ F_{y1} = 600 \sin(30^\circ) \]</p> <p>For the 800 N force, which is purely in the y-direction:</p> <p>\[ F_{x2} = 0 \]</p> <p>\[ F_{y2} = 800 \]</p> <p>Now, add the x-components and y-components of the forces to find the resultant:</p> <p>\[ R_x = F_{x1} + F_{x2} \]</p> <p>\[ R_y = F_{y1} + F_{y2} \]</p> <p>Substitute the values to find \( R_x \) and \( R_y \):</p> <p>\[ R_x = 600 \cos(30^\circ) \]</p> <p>\[ R_y = 600 \sin(30^\circ) + 800 \]</p> <p>Use the Pythagorean theorem to find the magnitude of the resultant force (R):</p> <p>\[ R = \sqrt{R_x^2 + R_y^2} \]</p> <p>Calculate the angle (\( \theta \)) from the positive x-axis:</p> <p>\[ \theta = \arctan\left(\frac{R_y}{R_x}\right) \]</p> <p>Solve for \( R \) and \( \theta \) to get the magnitude and direction of the resultant force.</p>
Mathematical Problem with Vectors in French
The image shows a handwritten math problem in French, which includes two parts. The first part is about computing the vectorial product of two vectors, and the second part is about determining the expression of a vector. Let's solve both parts step by step. 1. Compute the vector product (vecteur) of: \(\overrightarrow{u}(x_u, y_u, z_u) = (x - y, y^2 - x^2, 3z)\) \(\overrightarrow{v}(x_v, y_v, z_v) = (x, y, z)\) The cross product of two vectors in three-dimensional space is given by the determinant of the following matrix: \[\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x_u & y_u & z_u \\ x_v & y_v & z_v \\ \end{vmatrix}\] Substituting the given vector components: \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x - y & y^2 - x^2 & 3z \\ x & y & z \\ \end{vmatrix} \] = \(\hat{i}((y^2 - x^2)z - y(3z))\) - \(\hat{j}((x - y)z - (3z)x)\) + \(\hat{k}((x - y)y - (y^2 - x^2)x)\) Now let's compute each component individually: For \(\hat{i}\): \[ (y^2 - x^2)z - y(3z) = y^2z - x^2z - 3yz \] For \(\hat{j}\) (note the change of sign because it's the second component): \[ -((x - y)z - (3z)x) = -xz + yz - 3zx = 2zx - yz \] For \(\hat{k}\): \[ (x - y)y - (y^2 - x^2)x = xy - y^2 - y^2x + x^3 \] Finally, put these components together to get the cross product: \(\overrightarrow{u} \times \overrightarrow{v} = (y^2z - x^2z - 3yz)\hat{i} + (2zx - yz)\hat{j} + (xy - y^2 - y^2x + x^3)\hat{k}\) 2. Determine the expression of the vector \( \overrightarrow{w} \) as a linear combination of the base vectors \( \hat{i}, \hat{j}, \hat{k} \) knowing that \( \overrightarrow{w} \cdot \overrightarrow{u} = 3 \) and \( \overrightarrow{w} \cdot \overrightarrow{v} = 1 \) The scalar product (dot product) of two vectors is defined as \( \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \), where \( a_x, a_y, a_z \) and \( b_x, b_y, b_z \) are the components of vectors \( \vec{a} \) and \( \vec{b} \), respectively. Let's assume \( \overrightarrow{w} = (w_x, w_y, w_z) \). We have two equations from the problem statement: \( \overrightarrow{w} \cdot \overrightarrow{u} = w_x(x - y) + w_y(y^2 - x^2) + w_z(3z) = 3 \) \( \overrightarrow{w} \cdot \overrightarrow{v} = w_xx + w_yy + w_zz = 1 \) Unfortunately, with only two equations, we cannot uniquely determine three unknowns (\( w_x, w_y, w_z \)). Additional information would be needed to find a unique solution for \( \vec{w} \), but we can write \( \vec{w} \) as the following, leaving the coefficients \( w_x, w_y, w_z \) as variables: \( \vec{w} = w_x \hat{i} + w_y \hat{j} + w_z \hat{k} \) Without more information, we can't determine the exact values for \( w_x, w_y, w_z \) that satisfy both given conditions.
Generating the yz-Plane with Vectors
The question is asking to show that the yz-plane, given by the set W = {(0, b, c) | b, c ∈ ℝ}, can be generated by the indicated sets of vectors. The yz-plane in ℝ³ is the set of all points where the x-coordinate is 0. The general point in this plane can be written as (0, y, z), where y and z can take any real values. (i) To show that W is generated by the vectors (0, 1, 1) and (0, 2, -1), we need to express any vector (0, b, c) in W as a linear combination of (0, 1, 1) and (0, 2, -1). Let's try to express any point (0, b, c) as a linear combination of (0, 1, 1) and (0, 2, -1): (0, b, c) = α(0, 1, 1) + β(0, 2, -1) Expanding this, we have: (0, b, c) = (0, α + 2β, α - β) We want to solve for α and β such that the second and third components of the vectors match. This gives us two equations: α + 2β = b α - β = c We can solve these equations simultaneously to find α and β in terms of b and c. Adding the two equations, we get: 2α + β = b + c Subtracting the second equation from the first one, we get: 3β = b - c Thus, β = (b - c) / 3. Substitute β back into one of the original equations to get α. For example, using the first equation: α = b - 2β α = b - 2(b - c) / 3 α = (3b - 2b + 2c) / 3 α = (b + 2c) / 3 So any vector (0, b, c) can be represented as a linear combination of (0, 1, 1) and (0, 2, -1), with the coefficients α = (b + 2c) / 3 and β = (b - c) / 3. (ii) Similarly, to show that W is generated by the vectors (0, 1, 2), (0, 2, 3) and (0, 3, 1), we need to express any vector (0, b, c) in W as a linear combination of these vectors. (0, b, c) = α(0, 1, 2) + β(0, 2, 3) + γ(0, 3, 1) Expanding this, we have: (0, b, c) = (0, α + 2β + 3γ, 2α + 3β + γ) Again, set up a system of equations to solve for α, β, and γ: α + 2β + 3γ = b 2α + 3β + γ = c This system has two equations with three unknowns, making it underdetermined. However, since the third component isn't given (as there is no x-component in the vectors generating W), we can freely choose γ (for example, γ = 0), and the other two can be determined from the equations above. This way, every vector in the yz-plane can be expressed as a combination of these three vectors. For any given b and c, we can always find α and β (and choose γ) such that (0, b, c) is a linear combination of (0, 1, 2), (0, 2, 3), and (0, 3, 1), confirming that these vectors generate the yz-plane.
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