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Normally you use sine theta with the cross product and cos theta with the vector product, so that the cross product of parallel vectors is zero while the dot product of vectors at right angles is zero.
Work is defined as the dot product of force times distance, or W = F * d = Fd cos (theta) where theta is the angle in between the force and distance vectors (if you are doing two dimensions). In three dimensions, use the standard definition for the dot product (using the component form of the vectors).
Cross product tests for parallelism and Dot product tests for perpendicularity. Cross and Dot products are used in applications involving angles between vectors. For example given two vectors A and B; The parallel product is AxB= |AB|sin(AB). If AXB=|AB|sin(AB)=0 then Angle (AB) is an even multiple of 90 degrees. This is considered a parallel condition. Cross product tests for parallelism. The perpendicular product is A.B= -|AB|cos(AB) If A.B = -|AB|cos(AB) = 0 then Angle (AB) is an odd multiple of 90 degrees. This is considered a perpendicular condition. Dot product tests for perpendicular.
By "coincident", do you mean that they lie on the same plane (i.e. they are instances of the same plane)? Assuming that's what you're after: To begin with, you should have a plane equation defining each plane in the following form: ax + by + cz + d = 0 This form can be re-written in vector form as: (a,b,c) dot (x,y,z) + d = 0 Written in this form, (a,b,c) is a vector that defines the normal of the plane (a vector perpendicular to the plane), Which determines the orientation of the plane around the origin. The scalar value d defines a distance that the plane is offset from the origin, times the length of vector (a,b,c).In case you aren't familiar with the dot product, it is a way of multiplying two vectors resulting in a scalar value that represents the angle and the magnitude of the two vectors. Specifically: v1 dot v2 = |v1| |v2| cos(theta)where: * theta is the angle between the vectors * |v| is the length of a vector To compare the two planes, the first thing that we will want to do is "normalize" the plane equations, that is, scale them so that the length of (a,b,c) is 1.0. To do this, divide each constant in the equation by the length of the vector (a,b,c): (ax + by + cz + d) / |(a, b, c)| = 0 The normalized version of the plane equation defines the same plane as the original plane equation. The difference is that now we can readily use the dot product to determine the angle between the two plane's normals, as "v1 dot v2 = cos(theta)" when |v1| and |v2| are both equal to 1.0. The cosine of zero degrees is 1.0, so if we calculate the dot product of the normal vectors of the two planes we can determine if the normals, and thus the planes, are parallel: cos(theta) = (a1,b1,c1) dot (a2,b2,c2)IF cos(theta) < 1.0the planes are not parallel, and thus cannot be coincidental.IF cos(theta) = 1.0the planes are parallel, and may be coincidentalIF cos(theta) = -1.0the planes are parallel, with normals facing opposite directionsthe planes may be coincidental If the planes are not parallel, you don't have to test any further, they are not coincidental. If they are parallel, then you can determine if they are coincidental by comparing the normalized d from each equation to see if they are equal, flipping the sign if the normals are facing opposite directions: IF d1 = cos(theta) d2the planes are coincident
We use the dot product cos and in vector we use the vector product sin because of the trigonometric triangle.
Normally you use sine theta with the cross product and cos theta with the vector product, so that the cross product of parallel vectors is zero while the dot product of vectors at right angles is zero.
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
The Lagrangian of a simple pendulum can be given as: [ L = T - V ] where the kinetic energy (T) is given by (\frac{1}{2} m l^2 \dot{\theta}^2) and the potential energy (V) is given by (mgl(1 - \cos(\theta))), where (m) is the mass of the pendulum bob, (l) is the length of the pendulum, (g) is the acceleration due to gravity, (\theta) is the angle of the pendulum with the vertical, and (\dot{\theta}) is the angular velocity.
Because in dot product we take projection fashion and that is why we used cos and similar in cross product we used sin
work = the dot product of the force (F) and displacement vectors (D) = f * d * cos (theta), where 'f' and 'd' the magnitude of F and D, respectively; 'theta' is the angle between the two vectors. If theta = 90o, cos(theta) = 0. No work is done. That is, F is orthogonal to D. If d = 0, no work is done. That is, if the object is returning to the starting point, D = 0. ========================================
The angle between two vectors a and b can be found using the dot product formula: a · b = |a| |b| cos(theta), where theta is the angle between the two vectors. Rearranging the formula, we can solve for theta: theta = arccos((a · b) / (|a| |b|)).
The angle between two vectors can be found using the dot product formula: A · B = |A| |B| cos(theta). In this case, the dot product of the two given unit vectors is (1)(0) + (1)(1) + (0)(1) = 1. The magnitudes of the vectors are √2 and √2, therefore cos(theta) = 1 / (2)(2) = 1/4, giving theta = arccos(1/4) ≈ 75.5 degrees.
Work is defined as the dot product of force times distance, or W = F * d = Fd cos (theta) where theta is the angle in between the force and distance vectors (if you are doing two dimensions). In three dimensions, use the standard definition for the dot product (using the component form of the vectors).
The work done on the object is calculated as the force F multiplied by the component of the displacement d in the direction of the force, which is given by the dot product of F and d. The formula for work done in this case is W = F * d * cos(theta), where theta is the angle between the force and displacement vectors.
Cross product tests for parallelism and Dot product tests for perpendicularity. Cross and Dot products are used in applications involving angles between vectors. For example given two vectors A and B; The parallel product is AxB= |AB|sin(AB). If AXB=|AB|sin(AB)=0 then Angle (AB) is an even multiple of 90 degrees. This is considered a parallel condition. Cross product tests for parallelism. The perpendicular product is A.B= -|AB|cos(AB) If A.B = -|AB|cos(AB) = 0 then Angle (AB) is an odd multiple of 90 degrees. This is considered a perpendicular condition. Dot product tests for perpendicular.
By "coincident", do you mean that they lie on the same plane (i.e. they are instances of the same plane)? Assuming that's what you're after: To begin with, you should have a plane equation defining each plane in the following form: ax + by + cz + d = 0 This form can be re-written in vector form as: (a,b,c) dot (x,y,z) + d = 0 Written in this form, (a,b,c) is a vector that defines the normal of the plane (a vector perpendicular to the plane), Which determines the orientation of the plane around the origin. The scalar value d defines a distance that the plane is offset from the origin, times the length of vector (a,b,c).In case you aren't familiar with the dot product, it is a way of multiplying two vectors resulting in a scalar value that represents the angle and the magnitude of the two vectors. Specifically: v1 dot v2 = |v1| |v2| cos(theta)where: * theta is the angle between the vectors * |v| is the length of a vector To compare the two planes, the first thing that we will want to do is "normalize" the plane equations, that is, scale them so that the length of (a,b,c) is 1.0. To do this, divide each constant in the equation by the length of the vector (a,b,c): (ax + by + cz + d) / |(a, b, c)| = 0 The normalized version of the plane equation defines the same plane as the original plane equation. The difference is that now we can readily use the dot product to determine the angle between the two plane's normals, as "v1 dot v2 = cos(theta)" when |v1| and |v2| are both equal to 1.0. The cosine of zero degrees is 1.0, so if we calculate the dot product of the normal vectors of the two planes we can determine if the normals, and thus the planes, are parallel: cos(theta) = (a1,b1,c1) dot (a2,b2,c2)IF cos(theta) < 1.0the planes are not parallel, and thus cannot be coincidental.IF cos(theta) = 1.0the planes are parallel, and may be coincidentalIF cos(theta) = -1.0the planes are parallel, with normals facing opposite directionsthe planes may be coincidental If the planes are not parallel, you don't have to test any further, they are not coincidental. If they are parallel, then you can determine if they are coincidental by comparing the normalized d from each equation to see if they are equal, flipping the sign if the normals are facing opposite directions: IF d1 = cos(theta) d2the planes are coincident
A · B = |A| |B| cos(Θ)A x B = |A| |B| sin(Θ)If [ A · B = A x B ] then cos(Θ) = sin(Θ).Θ = 45°