Question

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

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

Step-by-step solution

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: \(3\vec{A} = 3(23.0,59.0) = (3(23.0), 3(59.0)) = (69.0,177.0)\) \(-7\vec{B} = -7(90.0,-150.0) = (-7(90.0), -7(-150.0)) = (-630.𝟶,1050.0)\)

Add the multiplied vectors

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

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: \(||\vec R|| = \sqrt{(-561.0)^2 + (1227.0)^2} = \sqrt{314721 + 1505729} = \sqrt{1820450} \approx 1349.2\)

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.

Key concepts

Vector Magnitude

The magnitude of a vector can be thought of as its size or length, providing us with a scalar quantity that tells us how long the vector is. It helps quantify the extent of a vector's reach in terms of distance. To calculate the magnitude, we use the Pythagorean theorem, which makes sense because by plotting the vector, its components become the sides of a right triangle.

Here’s a simple way to find the magnitude of a vector \(\vec{V} = (x, y)\):

• Use the formula: \(||\vec{V}|| = \sqrt{x^2 + y^2}\), where \(x\) and \(y\) are the components of the vector.
• Square the components individually, add them together, and then take the square root of the result.

So for a vector with components \((-561.0, 1227.0)\), we find the magnitude to be \(||\vec{R}|| = \sqrt{314721 + 1505729} = \sqrt{1820450} \approx 1349.2\) units. This value tells us how long the vector \(-7 \vec{B}+3 \vec{A}\) stretches in space.

Vector Direction

The direction of a vector indicates where the vector is pointing, usually measured as an angle from a reference direction. Understanding direction is crucial as it completes the description of a vector, alongside magnitude. To find this direction, we often use trigonometric functions.

For any vector with components \(x\) and \(y\), like \((x, y)\), the direction angle \(\theta\), measured from the positive \(x\)-axis, can be found using the inverse tangent function:

• Use \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\) to determine the angle.
• Adjust the angle according to which quadrant the resultant vector lies in. For example, if you end up in the second quadrant, you need to add 180°.

In the outlined exercise, our calculations led us to an angle \(\theta = -65.8^{\circ}\) initially, which necessitated an adjustment since the vector was in the second quadrant, giving the final direction as \(114.2^{\circ}\). This final direction tells us exactly where the vector \(-7 \vec{B}+3 \vec{A}\) points relative to a reference direction.

Scalar Multiplication

Scalar multiplication in the context of vectors involves multiplying a vector by a scalar (a single number). This operation affects the magnitude of the vector but not its direction, unless the scalar is negative. When a vector \(\vec{V} = (x, y)\) is multiplied by a scalar \(k\), the result is a vector \(k\vec{V} = (kx, ky)\). This operation scales the vector's components by the scalar value:

• If \(k > 1\), the vector is stretched.
• If \(0 < k < 1\), the vector is shrunk.
• If \(k < 0\), the vector is reversed in direction.

In our problem:

• We multiplied \(\vec{A} = (23.0,59.0)\) by 3, getting \(3\vec{A} = (69.0,177.0)\).
• We multiplied \(\vec{B} = (90.0,-150.0)\) by -7, resulting in \(-7\vec{B} = (-630.0,1050.0)\).

Thus, scalar multiplication allowed us to adjust the vectors for subsequent calculations of their resultant, which respects the original direction of \(\vec{A}\) while reversing \(\vec{B}\). This concept is fundamental in vector algebra as it allows for flexibility in adjusting vector properties to suit the problem's needs.