Question

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

Short answer

Question: Determine the magnitude and direction of the vector \(-\vec{A}+\vec{B}\), given \(\vec{A}=(23.0, 59.0)\) and \(\vec{B}=(90.0, -150.0)\). Answer: The magnitude of the vector \(-\vec{A}+\vec{B}\) is approximately 145.18 units and its direction is approximately 320.92 degrees.

Step-by-step solution

Find the components of the resultant vector

First, we need to find the components of the vector \(-\vec{A}+\vec{b}\): $$ -\vec{A}+\vec{B} = (-\vec{A}_x,\,-\vec{A}_y) + (\vec{B}_x,\,\vec{B}_y) $$ where \(\vec{A}=(\vec{A}_x, \vec{A}_y)\) and \(\vec{B}=(\vec{B}_x,\,\vec{B}_y)\) Given: $$ \vec{A} = (23.0, 59.0)\quad and \quad \vec{B} = (90.0, -150.0) $$ then the resultant vector is: $$ -\vec{A}+\vec{B} = (-23.0 - 90.0, -59.0 + 150.0) = (-113.0, 91.0) $$

Find the magnitude of the resultant vector

Now that we have the components of the resulting vector, we can find its magnitude using the Pythagorean theorem: $$ |\vec{R}| = \sqrt{(-113.0)^2 + (91.0)^2} \approx 145.18 $$

Find the direction of the resultant vector

Finally, we need to find the direction of the resultant vector. We can use the inverse tangent function (arctangent) to find the angle with respect to the positive x-axis. The formula for the direction \(\theta\) from the components of the vector is given by: $$ \theta = \arctan \left(\frac{R_y}{R_x}\right) $$ For our vector, this will be: $$ \theta = \arctan \left(\frac{91.0}{-113.0}\right) $$ Computing the arctangent: $$ \theta \approx -39.08^\circ $$ If we want to find the angle in the standard polar notation (positive angle measured counterclockwise from the positive x-axis), we can add \(360^\circ\) to our result: $$ \theta = -39.08^\circ + 360^\circ = 320.92^\circ $$ Our final answer is that this vector has a magnitude of approximately 145.18 units and a direction of approximately 320.92 degrees.

Key concepts

resultant vector

In vector mathematics, combining two or more vectors involves the process of vector addition. In the given exercise, you need to find the resultant vector from the operation -\vec{A}+\vec{B}. The first step involves identifying each vector's components and then applying addition or subtraction component-wise.

The original vectors given are:

• \(\vec{A} = (23.0, 59.0)\)
• \(\vec{B} = (90.0, -150.0)\)

To subtract \( \vec{A} \) from \( \vec{B} \), we first negate the components of \( \vec{A} \), giving \( -\vec{A} = (-23.0, -59.0) \). Then, the sum of the negated components of \( \vec{A} \) and those of \( \vec{B} \) provides the resultant vector:

• Resultant vector components: \( -\vec{A}+\vec{B} = (-113.0, 91.0) \).

Finding the resultant vector this way ensures that both direction and magnitude changes introduced by the vector operations are accurately captured.

magnitude calculation

Once you have determined the resultant vector, the next step is to calculate its magnitude, which tells you the vector's length or size. For a vector \( \vec{R} = (R_x, R_y) \), the magnitude \(|\vec{R}|\) can be found using the Pythagorean theorem:

\[|\vec{R}| = \sqrt{R_x^2 + R_y^2}\]
This formula arises from treating the vector as the hypotenuse of a right-angled triangle, where the lengths of the two sides are the vector's components.

In the example, for the vector with components \(-113.0\) and \(91.0\), the magnitude is:

• \( |\vec{R}| = \sqrt{(-113.0)^2 + (91.0)^2} \approx 145.18\).

This magnitude signifies the "straight-line" distance from the vector's origin point to its terminal point.

direction finding

After finding the magnitude, the direction of the resultant vector specifies its alignment concerning the coordinate axes. The direction is usually expressed as an angle \( \theta \), derived using the arctangent function:

\[\theta = \arctan \left(\frac{R_y}{R_x}\right)\]
This angle is initially calculated with reference to the positive x-axis. For example, if the vector components are \(91.0\) and \(-113.0\), then:

• \(\theta = \arctan \left(\frac{91.0}{-113.0}\right) \approx -39.08^\circ \)

To express the angle in standard polar notation, adjust if necessary. If the angle is negative, as in this instance, add \(360^\circ\) to find the positive equivalent:

• \(\theta = -39.08^\circ + 360^\circ = 320.92^\circ\)

This final angle provides a complete picture of the vector's orientation in its plane, representing the direction in which the vector is pointing.