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Q: How do you find the y-component of a vector if you are given the magnitude?

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Divide the vector by it's length (magnitude).

Given a vector, speed is the magnitude of the velocity vector, |v|. Consider vector V= IVx + JVy + KVz the magnitude is |V| = ( Vx2 + Vy2 + Vz2)1/2

We get the Unit Vector

If they are parallel, you can add them algebraically to get a resultant vector. Then you can resolve the resultant vector to obtain the vector components.

The magnitude of a vector is a geometrical value for hypotenuse.. The magnitude is found by taking the square root of the i and j components.

That's it! You know everything there is to know about it. It's not as if you have to wander through a crowd of vectors and find one that matches the description. "Find the vector" means figure out its magnitude and direction. If the problem already gave you the magnitude and direction, then it's unlikely that it's asking you to 'find' that same vector.

Use trigonometry.

One component = (magnitude) times (cosine of the angle).Other component = (magnitude) times (sine of the angle).In order to decide which is which, we have to know the angle with respect to what.

The magnitude of a vector can be found by taking the square root of each of the vector components squared. For example, if you had the vector 3i+4j, to find the magnitude, you take sqrt ( 3²+4² ) To get: sqrt ( 9+16 ) sqrt ( 25 ) = 5 Works the same in 3D or more, just put all the vector components in.

If you assume the vector is only in two dimensions, you can find the missing y-component with Pythagoras' Theorem: y = square root of (magnitude2 - x2).

You can't derive the direction only from the magnitude. A vector with the same magnitude can have different directions. You need some additional information to make conclusions about the direction.You can't derive the direction only from the magnitude. A vector with the same magnitude can have different directions. You need some additional information to make conclusions about the direction.You can't derive the direction only from the magnitude. A vector with the same magnitude can have different directions. You need some additional information to make conclusions about the direction.You can't derive the direction only from the magnitude. A vector with the same magnitude can have different directions. You need some additional information to make conclusions about the direction.

The Resultant Vector minus the other vector

The magnitude is the length of the vector (using any scaling factor that may have been employed).

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Both component (sin and cos) of the vector are found then the answers are squared, added and rooted to find the magnitude.

resultant

using the "dot product" formula, you can find the angle. where |a| denotes the length (magnitude) of a. More generally, if b is another vector : where |a| and |b| denote the length of a and b and θis the angle between them. Thus, given two vectors, the angle between them can be found by rearranging the above formula: : :

The magnitude of (i + 2j) is sqrt(5). The magnitude of your new vector is 2. If both vectors are in the same direction, then each component of one vector is in the same ratio to the corresponding component of the other one. The components of the known vector are 1 and 2, and its magnitude is sqrt(5). The magnitude of the new one is 2/sqrt(5) times the magnitude of the old one. So its x-component is 2/sqrt(5) times i, and its y-component is 2/sqrt(5) times 2j. The new vector is [ (2/sqrt(5))i + (4/sqrt(5))j ]. Since the components of both vectors are proportional, they're in the same direction.

The answer below assumes you are required to find the components of the vector. A vector with unity magnitude means that the magnitude of the vector equals to 1. Therefore its a simple case of calculating the values of sin(45) for the vertical components and cos(45) for the horizontal components. Both of these values equal to 1/sqrt(2) {one over square-root two}

vector PQ where P(-4, -3) and Q(-2, 2) equivalent vector P'Q' where P'(0, 0) and Q'(2, 5) the magnitude doesn't change so we can compute |P'Q'| = √(22 + 52) = √29

You need to take the magnitude of the cross-product of two position vectors. For example, if you had points A, B, C, and D, you could take the cross product of AB and BC, and then take the magnitude of the resultant vector.

There are two likely calculus applications of this problem. Both differential calculus and basic vector operations can be used to solve for power in a scenario, depending on how a problem is defined. Power is the dot-product of a force vector and a velocity vector and... Power is a change in energy over time, or in differential terms: dE/dt If you were given a function that defined a system's energy with respect to time, you could derive it to find a function for that system's power output. If you were given a force vector and a velocity vector and asked to find the total power applied to the system, you could take the dot product of the two vectors to find this. Or, if you are not taking a calculus approach to it: Average power is simply energy divided by time The magnitude of power given a force and velocity can be found with the formula: P=F*v*cos(theta) Where F is the magnitude of the force v is the magnitude of the velocity theta is the angle between the two quantities.

It depends on what other information you are given. If you are given the x-component and the angle, then you can still use trig by doing the following. Tan(angle)=y/x If you do not have the x component, then you will need to use other kinematic equations to find the answer. The different ways to find this information are too numerous to list out, but I will give an example of a common one. If you are given the horizontal distance and time the object flies, use d=vt to find the x component of velocity, then use the first strategy.

If the components are in the i and j directions, for example, then if the vector is mi + nj then the coefficients m and n can be used to find the magnitude and direction.The magnitude is the hypotenuse of a right triangle with legs m and n, so it is sqrt(mÂ² + nÂ²).

Suppose the magnitude of the vector is V and its direction makes an angle A with the x-axis, then the x component is V*Cos(A) and the y component is V*Sin(A)