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Find the angle between each of these pairs of vectors: (a) \(\vec{A}=-2.00 \hat{\imath}+6.00 \hat{\jmath}\) and \(\vec{B}=2.00 \hat{\imath}-3.00 \hat{\jmath}\) (b) \(\vec{A}=3.00 \hat{\imath}+5.00 \hat{\jmath} \quad\) and \(\quad \vec{B}=10.00 \hat{\imath}+6.00 \hat{\jmath}\) (c) \(\vec{A}=-4.00 \hat{\imath}+2.00 \hat{\jmath}\) and \(\vec{B}=7.00 \hat{\imath}+14.00 \hat{\jmath}\)

Short Answer

Expert verified
(a) The angle is 126.87 degrees, (b) the angle is 8.13 degrees and (c) the angle is 39.23 degrees.

Step by step solution

01

Calculate the Dot Products

For each set, (a), (b), and (c), compute the dot (or scalar) product. This can be done using the formula \(\vec{A}.\vec{B} = A_xB_x + A_yB_y\), where \(A_x\), \(A_y\), \(B_x\), and \(B_y\) are the x and y components of vectors A and B, respectively.
02

Calculate the Magnitudes

For each set, calculate the lengths (or magnitudes) of vectors A and B. This can be done using the formula \(|\vec{A}| = \sqrt{A_x^2+A_y^2}\) and \(|\vec{B}| = \sqrt{B_x^2+B_y^2}\).
03

Find the Angles

Now we are ready to calculate the angles. Use the relationship between the dot product, the magnitudes of vectors, and cosine of the angle between them: \(\vec{A}.\vec{B} = |\vec{A}||\vec{B}| cos\theta\). With the dot product and the magnitudes, we can solve this formula for the angle \theta, which is the angle between the two vectors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. It combines two vectors to yield a single number, helping us understand the directional relation between the vectors. To calculate the dot product of two vectors \( \vec{A} \) and \( \vec{B} \), you use the formula: \[ \vec{A} \cdot \vec{B} = A_xB_x + A_yB_y \] * Here, \( A_x \) and \( A_y \) are the components of vector \( \vec{A} \), while \( B_x \) and \( B_y \) are the components of vector \( \vec{B} \). * The dot product measures how much the vectors are pointing in the same direction. If the dot product is positive, the vectors have components pointing in a similar direction, while a negative result indicates they have components in opposing directions. If the dot product equals zero, the vectors are perpendicular.
Vector Magnitude
Understanding vector magnitude is key to analyzing vector properties. The magnitude represents the length of a vector, indicating its size regardless of direction. To calculate the magnitude of a vector \( \vec{A} \) with components \( (A_x, A_y) \), use the formula: \[ |\vec{A}| = \sqrt{A_x^2 + A_y^2} \] * This formula is derived from the Pythagorean theorem. * It calculates the 'distance' from the origin to the point represented by the vector in a coordinate plane. The magnitude will always be a positive value, representing how long the vector actually is. Knowing the magnitudes of vectors is essential when comparing vector sizes or using them in further calculations, such as finding angles.
Cosine Formula
The cosine formula is a key tool when trying to calculate the angle between two vectors. By leveraging the relationship that the dot product has with the magnitudes and the cosine of the angle between vectors, we can determine this angle efficiently. The relevant formula is: \[ \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos(\theta) \] * Here, \( \vec{A} \cdot \vec{B} \) is the dot product of the vectors. * \( |\vec{A}| \) and \( |\vec{B}| \) are the magnitudes of the vectors. * \( \theta \) is the angle between the vectors. By rearranging the formula, we can solve for the cosine of the angle: \[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} \] Once you have the cosine, finding the angle \( \theta \) involves taking the arccosine (inverse cosine) of the result. This calculation is useful in fields like physics and engineering where understanding vector orientation is crucial.

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Most popular questions from this chapter

Vector \(\vec{A}\) has magnitude \(5.00 \mathrm{~m}\) and lies in the \(x y\) -plane in a direction \(53.0^{\circ}\) from the \(+x\) -axis axis measured toward the \(+y\) -axis. Vector \(\vec{B}\) has magnitude \(8.00 \mathrm{~m}\) and a direction you can adjust. (a) You want the vector product \(\vec{A} \times \vec{B}\) to have a positive \(z\) -component of the largest possible magnitude. What direction should you select for vector \(\overrightarrow{\boldsymbol{B}} ?(\mathrm{~b})\) What is the direction of \(\overrightarrow{\boldsymbol{B}}\) for which \(\overrightarrow{\boldsymbol{A}} \times \overrightarrow{\boldsymbol{B}}\) has the most negative \(z\) -component? (c) What are the two directions of \(\overrightarrow{\boldsymbol{B}}\) for which \(\vec{A} \times \vec{B}\) is zero?

You leave the airport in College Station and fly \(23.0 \mathrm{~km}\) in a direction \(34.0^{\circ}\) south of east. You then fly \(46.0 \mathrm{~km}\) due north. How far and in what direction must you then fly to reach a private landing strip that is \(32.0 \mathrm{~km}\) due west of the College Station airport?

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about \(500 \mathrm{~cm}^{3}\) of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?

While following a treasure map, you start at an old oak tree. You first walk \(825 \mathrm{~m}\) directly south, then turn and walk \(1.25 \mathrm{~km}\) at \(30.0^{\circ}\) west of north, and finally walk \(1.00 \mathrm{~km}\) at \(32.0^{\circ}\) north of east, where you find the treasure: a biography of Isaac Newton! (a) To return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem. (b) To see whether your calculation in part (a) is reasonable, compare it with a graphical solution drawn roughly to scale.

The scalar product of vectors \(\vec{A}\) and \(\vec{B}\) is \(+48.0 \mathrm{~m}^{2}\). Vector \(\vec{A}\) has magnitude \(9.00 \mathrm{~m}\) and direction \(28.0^{\circ}\) west of south. If vector \(\overrightarrow{\boldsymbol{B}}\) has direction \(39.0^{\circ}\) south of east, what is the magnitude of \(\overrightarrow{\boldsymbol{B}} ?\)

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