In unit-vector notation, what is the sum of vector a =(4m)i + (3m)j and vector b= (-13m)i +(7m)j?
a)I got (27m)i +(10m)j but the book said it should have been (-9m)i +(10m)j
b)what is the magnitude and direction of vector a + vector b (relative to i)?
to find the magnitude i did the sqrt((-9^2 + 10^2) and then I got 13.453
I wasn't sure how to find the direction.
I am wondering how you got 27i? The i component of a was 4, and the i component of b was -13. Adding them gets the book answer.
Magnitude of the sum: (-13)^2 + (10)^2= 269, and the square of that is ....
Direction: arctan (10/-13)
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To find the sum of two vectors in unit-vector notation, you simply add the corresponding components. Let's go through the calculations step by step for the given vectors:
Vector a = (4m)i + (3m)j
Vector b = (-13m)i + (7m)j
To find the sum, you add the i-components separately and the j-components separately:
Sum = (4m - 13m)i + (3m + 7m)j
= (-9m)i + (10m)j
So, the correct answer is indeed (-9m)i + (10m)j. It seems there might have been an error in your calculations.
Now, let's move on to the second part of your question about the magnitude and direction of the sum of vector a and vector b.
Magnitude: To find the magnitude of a vector, you need to use the Pythagorean theorem. The magnitude of the sum is given by the square root of the sum of the squares of its components. In this case, we have:
Magnitude = sqrt((-9m)^2 + (10m)^2)
= sqrt(81m^2 + 100m^2)
= sqrt(181m^2)
So, the magnitude of the sum is approximately sqrt(181)m.
Direction: To find the direction of the sum relative to the i-axis, you can use the arctan function. In this case, we take the ratio of the j-component to the i-component of the vector and apply the arctan function:
Direction = arctan(10m / -9m)
Evaluating the arctan, you can find the angle associated with this direction.
I hope this helps clarify the calculations and answer your questions. Let me know if you need any further assistance!