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Chapter 1: Problem 84

Find the magnitude and direction of \(-7 \vec{B}+3 \vec{A}\), where \(\vec{A}=(23.0,59.0), \vec{B}=(90.0,-150.0)\)

Short Answer

Expert verified
Answer: The magnitude is approximately 1349.2 units, and the direction is approximately 114.2°.

Step by step solution

01

Multiply the vectors by the given scalars

First, we need to multiply each vector by their respective scalar. In this case, \(\vec{A}\) will be multiplied by 3, and \(\vec{B}\) will be multiplied by -7. This gives us: \(3\vec{A} = 3(23.0,59.0) = (3(23.0), 3(59.0)) = (69.0,177.0)\) \(-7\vec{B} = -7(90.0,-150.0) = (-7(90.0), -7(-150.0)) = (-630.𝟶,1050.0)\)
02

Add the multiplied vectors

Now, we need to add the multiplied vectors to find the resultant vector. This can be done component-wise: \(\vec R = (-630.0, 1050.0) + (69.0, 177.0) = (-630.0+69.0, 1050.0+177.0) = (-561.0, 1227.0)\)
03

Calculate the magnitude of the resultant vector

The magnitude of a vector is given by the formula: \(||\vec R|| = \sqrt{R_x^2 + R_y^2}\). Applying this formula to our resultant vector gives: \(||\vec R|| = \sqrt{(-561.0)^2 + (1227.0)^2} = \sqrt{314721 + 1505729} = \sqrt{1820450} \approx 1349.2\)
04

Calculate the direction of the resultant vector

To find the direction of the vector, we use the tangent function with the inverse tangent (also called arctangent) in degrees: \(\theta = \tan^{-1}{\frac{R_y}{R_x}}\) In our case, this gives us: \(\theta = \tan^{-1}{\frac{1227.0}{-561.0}} \approx -65.8^{\circ}\) However, since the vector is in the second quadrant, we need to add 180° to the angle: \(\theta = -65.8^{\circ} + 180^{\circ} = 114.2^{\circ}\) So the direction of the resultant vector is approximately \(114.2^{\circ}\). The magnitude and direction of \(-7 \vec{B}+3 \vec{A}\) are approximately 1349.2 units and \(114.2^{\circ}\), respectively.

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