'N' no.of vectors of equal magnitude act at a point making an angle of 2 pie /N with each other . What is the resultant?
Answers
Answer:
You've to use a parallelogram law of addition of vectors. For that, we've to lay 2 vectors at a common point & complete the parallelogram. The diagonal of the parallelogram starting from the same common point denotes addition of 2 vectors. You can find the length of this diagonal using some trigonometry. It's the magnitude of the resultant vector which equals :
V =√(V1)^2+(V2)^2+2V1V2*cosθ
Now V1 = V2 = A.
Magnitude of a resultant vector
= √2A^2+2A^2*cosθ.
= A√[2(1+cosθ)].
As both vectors are equal in magnitude, a quadrilateral formed is a rhombus. We know that a diagonal of a rhombus bisects its angle so the resultant vector makes an angle (theta/2) with both the vectors. It's the direction of a resultant vector.
2 vectors A and B of equal magnitude act at a point in different directions. If one of the vectors is halved, the angle which the resultant makes with the other is also halved. What is the angle between them?
Three vectors, each of magnitude 'a', are present at an angle of 60 degrees with each other. What is the sum of their resultant?
The resultant of two vectors having equal magnitude F makes an angle of 120 degrees with the vectors. What is the magnitude and direction of the resultant?
What is the magnitude and direction if two vectors have equal magnitude A and makes angle theta?
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Interesting question. An answer doesn’t just spring out of my mind like it usually does with these seemingly trivial problems, so let’s do the algebra. We have two vectors ⟨acosθ,asinθ⟩ and ⟨bcos(θ+α),bsin(θ+α)⟩ , using a conversion from polar coordinates, where a is the norm of the first vector, b is the norm of the second, θ is the angle the first vector is rotated from the x-axis, and α is the angle between the two vectors.
Since rotations do not change the length of a vector (easy to see in Euclidean space),
Productivity maximized.
Let each vector has a magnitude a which makes an angle θ between them hence the resultant ( R ) of these equal vectors
R=a2+a2+2a⋅acosθ−−−−−−−−−−−−−−−−√
=2a2+2a2cosθ−−−−−−−−−−−−√
=a2–√1+cosθ−−−−−−−√
=a2–√1+2cos2θ2−1−−−−−−−−−−−−√
=a2–√2cos2θ2−−−−−−√
=a2–√2–√cosθ2
=2acosθ2
& the resultant will be equally inclined at an angle θ2 with the equal vectors or the resultant acts along the angle bisector of both the equal vectors
I understand the question thusly: "Two vectors of magnitude A have an angle theta between them and are added to each other. What is the magnitude and direction of the resultant?" The three vectors form an isosceles triangle. Let the magnitude of the resultant be C. By the law of cosines, C= A(2(1-cos(theta)))^(1/2). The inclination of the resultant with respect to either vector of length A is (1/2)(pi-theta) radians because the three angles have to sum to pi radians. If you like degrees as a measure of angle, replace pi radians with 180 degrees and assume that theta is expressed in degrees.
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What is the magnitude of a resultant of 3 vectors, each having a magnitude A acting at a point, and any two consecutive vectors in the same plane having an angle of 60° between them?
Let’s add the vectors tip-to-tail to get the resultant:
Now we have a simple triangle to solve. Use cosine law to determine the magnitude of the resultant:
R2=A2+A2−2(A)(A)cos(180−θ)
but cos(180−θ)=−cosθ
∴ R2=A2+A2−2(A)(A)(−cosθ)
or
R2=A2+A2+2(A)(A)cosθ
R2=2A2+2(A2)cosθ
R2=2A2(1+cosθ)
R=A2(1+cosθ)−−−−−−−−−√
resultant will be along a line in between 2 vectors.
component of a A, along this line is A cos ( theta/2 ).
so the resultant is, 2 A cos ( theta/2 ).
You can use the resultant formula
Under root of Psquare + Q square +2PQ cos theta
P=Q in this case
2cos(ang\1/2),tan(ang4/2
2a
90
2