Question Video: Finding the Components of a Vector Given Its Magnitude and Argument | Nagwa Question Video: Finding the Components of a Vector Given Its Magnitude and Argument | Nagwa

Question Video: Finding the Components of a Vector Given Its Magnitude and Argument Physics • First Year of Secondary School

The diagram shows a vector, 𝐀, that has a magnitude of 29. The angle between the vector and the 𝑥-axis is 60°. Give this vector in component form. Round all of the numbers in your answer to the nearest whole number.

03:21

Video Transcript

The diagram shows a vector 𝐀 that has a magnitude of 29. The angle between the vector and the 𝑥-axis is 60 degrees. Give this vector in component form. Round all of the numbers in your answer to the nearest whole number.

In this question, we’re being asked to give the vector 𝐀 in component form. This means that we need to find the horizontal and vertical components of 𝐀, which we typically call 𝐴 subscript 𝑥 and 𝐴 subscript 𝑦. Then, we can write the vector 𝐀 in the form 𝐀 is equal to 𝐴 subscript 𝑥 times 𝐢 hat plus 𝐴 subscript 𝑦 times 𝐣 hat. Here, 𝐢 hat and 𝐣 hat are the unit vectors in the 𝑥- and 𝑦-directions.

Let’s first find the horizontal component of the vector 𝐴 subscript 𝑥. To find this horizontal component, recall that we can use the formula 𝐴 subscript 𝑥 is equal to 𝐴 times cos 𝜃, where 𝐴 is the magnitude of the vector and 𝜃 is the argument of the vector. We are told in the question that the magnitude of the vector 𝐀 is 29. We are also given an angle of 60 degrees between the vector and the negative 𝑥-axis. But we need to be careful here. This angle is not the argument of the vector. The argument of a vector is defined as the angle between the vector and the positive 𝑥-axis, measured counterclockwise from the positive 𝑥-axis.

So, to find the argument of this vector, we need to add 180 degrees to this angle of 60 degrees. Adding 60 degrees plus 180 degrees, we find an argument 𝜃 of 240 degrees. Substituting these values for the amplitude and argument into our equation, we find that 𝐴 subscript 𝑥 is equal to 29 multiplied by the cos of 240 degrees. Completing this calculation, we find that the horizontal component 𝐴 subscript 𝑥 is equal to negative 14.5.

We want this answer to the nearest whole number. So we can round down negative 14.5 to negative 15. We have then that the value of the horizontal component of this vector is equal to negative 15. Let’s make a note of this value and clear some space on screen.

We now need to find the vertical component of the vector 𝐴 subscript 𝑦. To find this vertical component, recall that we can use the formula 𝐴 subscript 𝑦 equals 𝐴 times sin 𝜃, where as before 𝐴 is the magnitude and 𝜃 is the argument of the vector. The values of the magnitude and argument of the vector are unchanged. So we can substitute those values into this equation. And we find that the vertical component 𝐴 subscript 𝑦 is equal to 29 multiplied by the sin of 240 degrees.

Completing this calculation, we find that the vertical component 𝐴 subscript 𝑦 is equal to negative 25.11. Again, we want this to the nearest whole number, so we can round up negative 25.11 to negative 25. This gives us the value of the vertical component of the vector as negative 25.

We have now found both components of this vector. And so the vector 𝐀 can be written in component form as 𝐀 equals negative 15 times 𝐢 hat minus 25 times 𝐣 hat. This is our final answer to this question.

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