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Why you use sin in cross product?
The sine function is used in the cross product because the magnitude of the cross product of two vectors is determined by the area of the parallelogram formed by those vectors. This area is calculated as the product of the magnitudes of the vectors and the sine of the angle between them. Specifically, the formula for the cross product (\mathbf{A} \times \mathbf{B}) includes (|\mathbf{A}||\mathbf{B}|\sin(\theta)), where (\theta) is the angle between the vectors, capturing the component of one vector that is perpendicular to the other. Thus, the sine function accounts for the directional aspect of the vectors in determining the resultant vector's magnitude and orientation.
When does the magnitude of dot product and cross product of vectors is equal?
If x is the angle between the two vectors then the magnitudes are equal if cos(x) = sin(x). That is, when x = pi/4 radians.
Can the vector product of two vectors be negative?
no .....the scalar product of two vectors never be negative Yes it can If A is a vector, and B = -A, then A.B = -A2 which is negative. Always negative when the angle is between the vectors is obtuse.
What is the difference between a 'dot product' and a 'cross product'?
Dot Product:Given two vectors, a and b, their dot product, represented as a ● b, is equal to their magnitudes multiplied by the cosine of the angle between them, θ, and is a scalar value.a ● b = ║a║║b║cos(θ)Cross Product:Given two vectors, a and b, their cross product, which is a vector, is represented as a X b and is equal to their magnitudes multiplied by the sine of the angle between them, θ, and then multiplied by a unit vector, n, which points perpendicularly away, via the right-hand rule, from the plane that a and bdefine.a X b= ║a║║b║sin(θ)n
Dot product of two vectors is equal to cross product what will be angle between them?
(A1) The dot product of two vectors is a scalar and the cross product is a vector? ================================== (A2) The cross product of two vectors, A and B, would be [a*b*sin(alpha)]C, where a = |A|; b = |B|; c = |C|; and C is vector that is orthogonal to A and B and oriented according to the right-hand rule (see the related link). The dot product of the two vectors, A and B, would be [a*b*cos(alpha)]. For [a*b*sin(alpha)]C to equal to [a*b*cos(alpha)], we have to have a trivial solution -- alpha = 0 and either a or b be zero, so that both expressions are zeroes but equal. ================================== Of course one is the number zero( scalar), and one is the zero vector. It is a small difference but worth mentioning. That is is to say if a or b is the zero vector, then a dot b must equal zero as a scalar. And similarly the cross product of any vector and the zero vector is the zero vector. (A3) The magnitude of the dot product is equal to the magnitude of the cross product when the angle between the vectors is 45 degrees.
