Find the magnitude and direction angle for each vector. Round ...

Question

Find the magnitude and direction angle for each vector. Round angle measures to the nearest tenth, as necessary. See Example 1.

〈8√2, -8√2〉

Accepted Answer

Welcome back, everyone. For the given vector calculate its magnitude and direction angle. Express your answer to the nearest sent if required. We're given a vector in its component form. It is 2823. 2823, and there are four answer choices. A says 45.0 and 68.6 degrees. B 45.0 and 68.6 degrees. C 68.6 and 45.0 degrees, and D 68.6 and 45.0 degrees. So first of all, we're given our vector and specifically there are two components. Let's call this vector U. Its X component UX is 2823. And its Y component is -2823. First of all, let's identify the magnitude of you. And we have to recall the formula which allows us to calculate. The magnitude of a vector and we essentially have to take the X component we have to square it. We have to add the square of the y component, and then when we add those squares, well, we are taking square root of the result. So square root of U X2 plus U Y2. And now we simply want to substitute those values. We are taking 2823 squad. And we're adding the square of -2823. Now, let's simplify. Let's remember that whenever we square a negative number, it becomes a positive number. So we get 28 squad. Squared of 3 squared gives us 3. And then we're adding The same number, 28 squared multiplied by 3. And this gives us We, because we can factor it out. Multiplied by 28 squared. Multiplied by 2, right? And from here, well, essentially we can just combine 3 and 2 because we have a square, right? So we can take square root of a perfect square. We're going to get square root of 3 multiplied by 2 gives us 6, and 6 is multiplied by 28 squad. Now let's simplify square root of 28 squared is just 28, and then we're left with squared of 6. So we have the magnitude and now. Let's evaluate that magnitude. We're going to round to the nearest sun. It is going to be approximately equal to 68.6. Now let's identify the angle. Whenever we are considering angles, well, it is really important to understand where the vector is. We have our X. Y plane and we can see that X is positive. Y is negative. This means that we are in quadron number 4 and the angle is going to be negative, right? It is below the x axis. We want to evaluate that angle theta. For quadron 4 we can use the formula. Theta is equal to. In verse of 10. Of Y divided by X, in this case, U Y, the Y component divided by U X component. Now let's substitute the values. We're going to take the inverse of 10. Of our UI is negative. 2823 and our U-axis 2823. So we're basically taking the inverse of 10. Off-1 Which is 45 degrees and we want to use. The nearest scent, even though it's exact, we can add 0.0. And that's it, we have the required angle, we have the magnitude. We can conclude that the correct answer to this problem corresponds to the answer choice C 68.6 and 45.0 degrees. Thank you for watching.

Find the magnitude and direction angle for each vector. Round angle measures to the nearest tenth, as necessary. See Example 1.
〈8√2, -8√2〉

Key Concepts

Magnitude of a Vector

Direction Angle of a Vector

Quadrants in the Coordinate Plane

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