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Why sin theta use in cross products and cos theta use in dot products?
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Why you use cosine theta with cross product?
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.
What is defined as force times distance?
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).
What are the applications of cross product and dot product?
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.
How do you prove two planes are coincident?
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
Why in dot product you use cos and in vector product sin?
We use the dot product cos and in vector we use the vector product sin because of the trigonometric triangle.
Why you use cosine theta with cross product?
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.
Prove that a constant vector always has a perpendicular derivative?
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
Find the Lagrangian of simple pendulum?
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)
Why we use sine angle in cross product while cos angle in dot product?
Because in dot product we take projection fashion and that is why we used cos and similar in cross product we used sin
What two condititons must be for work to be done?
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. ========================================
What is defined as force times distance?
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).
What are the applications of cross product and dot product?
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.
When a force F is being applied at an angle to the direction of the displacement d the work done on the object is calculated as?
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)
How do you prove two planes are coincident?
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
When cross and dot product equal what is angle between A and B?
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°
Why in dot product you use cos and in vector product sin?
We use the dot product cos and in vector we use the vector product sin because of the trigonometric triangle.
What are the applications of cross product and dot product in physics?
cross: torque dot: work
