In vector mathematics, the cross product involves the sine of the angle because it measures the area of the parallelogram formed by the two vectors, which is maximized when the vectors are perpendicular (90 degrees) and zero when they are parallel (0 degrees). On the other hand, the dot product uses the cosine of the angle because it quantifies the extent to which one vector extends in the direction of another, achieving its maximum when the vectors are aligned (0 degrees) and zero when they are perpendicular (90 degrees). This geometric interpretation aligns with the respective relationships of sine and cosine to angles in right triangles.
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.
The dot product measures the extent to which two vectors align in the same direction, which is directly related to the cosine of the angle between them; thus, ( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) ). In contrast, the cross product gives a vector that is perpendicular to the plane formed by the two vectors, and its magnitude is proportional to the sine of the angle between them; hence, ( |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin(\theta) ). This distinction arises from the geometric interpretations of these operations in relation to the angle between the vectors.
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
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 generalized coordinate for the pendulum is the angle of the arm off vertical, theta. Theta is 0 when the pendulum arm is down and pi when the arm is up. M = mass of pendulum L = length of pendulum arm g = acceleration of gravity \theta = angle of pendulum arm off vertical \dot{\theta} = time derivative of \theta What are the kinetic and potential energies? Kinetic energy: T = (1/2)*M*(L*\dot{\theta})^2 Potential energy: V' = MLg(1-cos(\theta)) V = -MLg*cos(\theta) --note: we can shift the potential by any constant, so lets choose to drop the MLg The Lagrangian is L=T-V: L = (1/2)ML^2\dot{\theta}^2 + MLg*cos(\theta)
Because in dot product we take projection fashion and that is why we used cos and similar in cross product we used sin
The dot product measures the extent to which two vectors align in the same direction, which is directly related to the cosine of the angle between them; thus, ( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) ). In contrast, the cross product gives a vector that is perpendicular to the plane formed by the two vectors, and its magnitude is proportional to the sine of the angle between them; hence, ( |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin(\theta) ). This distinction arises from the geometric interpretations of these operations in relation to the angle between the vectors.
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).
There are two completely separate ways to look at this. The first is using simply the properties of vectors. Break the force vector into a parallel and perpendicular component (relative to the displacement). From here, we know that work done in perpendicular to the direction can not contribute. So now only the parallel segment can contribute to work, and with a little bit of geometry and trigonometry, it can be shown that W=(F)(d)cos(theta) Alternatively, the definition of work is W=F dot d. Knowing that the magnitude of the dot product of a and b equals (a)(b)cos(theta) where theta is the angle formed by the tails when the are placed with ends touching, immediately gives you the answer W=(F)(d)cos(theta)
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