In Exercises 61–64, find the magnitude ||v||, to the nearest hund...

Question

In Exercises 61–64, find the magnitude ||v||, to the nearest hundredth, and the direction angle θ, to the nearest tenth of a degree, for each given vector v.

v = -10i + 15j

Accepted Answer

Find the magnitude and direction angle of the vector A given below express your answer to one decimal place A equals negative 14 I plus 21 J. And we have four possible answers all with different magnitudes and direction angles. Now to solve this, we first need the standard form of the vector which is our vector V is equals to X I plus Y J. We can find the magnitude denoted by the vector V surrounded by double bars as the square root of X squared plus Y square. In our case, the magnitude of vector A will be the square root of negative 14 squared plus 21 squared. We can now simplify, we should get the square root of 6 37 which simplifies and approximates to about 25.2. Now that we have our magnitude, we can find our direction angle. Our direction angle is given by the arc tangent or the inverse tangent of our Y component divided by our X component. We know if our Y component is 21 our X component is negative 14, which if we throw this into the calculator, we should have come out with an angle of negative 56.3 degrees, we have our magnitude in another direction angle which means the answer to our problem is answer D 25. and negative 56.3 degrees. OK? I hope to help you solve the problem. Thank you for watching. Goodbye.

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In Exercises 61–64, find the magnitude ||v||, to the nearest hundredth, and the direction angle θ, to the nearest tenth of a degree, for each given vector v. v = -10i + 15j