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Why in dot product you use cos and in vector product sin?
Kunaljashnani15338
We use the dot product cos and in vector we use the vector product sin because of the trigonometric triangle.
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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.
What is scalar and vector product?
The scalar product of two vectors, A and B, is a number, which is a * b * cos(alpha), where a = |A|; b = |B|; and alpha = the angle between A and B. The vector product of two vectors, A and B, is a vector, which is a * b * sin(alpha) *C, where C is unit vector orthogonal to both A and B and follows the right-hand rule (see the related link). ============================ The scalar AND vector product are the result of the multiplication of two vectors: AB = -A.B + AxB = -|AB|cos(AB) + |AB|sin(AB)UC where UC is the unit vector perpendicular to both A and B.
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
How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
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.
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
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.
What is scalar and vector product?
The scalar product of two vectors, A and B, is a number, which is a * b * cos(alpha), where a = |A|; b = |B|; and alpha = the angle between A and B. The vector product of two vectors, A and B, is a vector, which is a * b * sin(alpha) *C, where C is unit vector orthogonal to both A and B and follows the right-hand rule (see the related link). ============================ The scalar AND vector product are the result of the multiplication of two vectors: AB = -A.B + AxB = -|AB|cos(AB) + |AB|sin(AB)UC where UC is the unit vector perpendicular to both A and B.
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°
What is the difference between the ''dot product'' and the ''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 b define.a X b = ║a║║b║sin(θ)n
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
How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
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.
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.
Why only cos theta is used in vector functions?
It's not. Cos(Θ) only gives you the x-component of a vector. In order to find its y-component, you also need to use sin(Θ).
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
