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Chapter 1: Problem 92

Vector \(\vec{A}\) has magnitude 6.00 \(\mathrm{m}\) and vector \(\vec{\boldsymbol{B}}\) has magni- tude 3.00 \(\mathrm{m} .\) The vector product between these two vectors has magnitude 12.0 \(\mathrm{m}^{2} .\) What are the two possible values for the scalar product of these two vectors? For each value of \(\vec{A} \cdot \vec{B},\) draw a sketch that shows \(\vec{A}\) and \(\vec{B}\) and explain why the vector products in the two sketches are the same but the scalar products differ.

Short Answer

Expert verified
The two possible scalar product values are \( 6\sqrt{5} \, \mathrm{m}^2 \) and \( -6\sqrt{5} \, \mathrm{m}^2 \).

Step by step solution

01

Understanding the Problem

Given vectors \( \vec{A} \) and \( \vec{B} \), we need to find the two possible values for their scalar product. We know their magnitudes and the magnitude of their vector (cross) product.
02

Formula for Vector Product

The magnitude of the vector (cross) product \( \vec{A} \times \vec{B} \) is given by \( |\vec{A}| |\vec{B}| \sin \theta = 12.0 \, \mathrm{m}^2 \), where \( \theta \) is the angle between the vectors.
03

Calculate Sine of Angle

Substitute the magnitudes into the cross product formula: \[6.00 \times 3.00 \times \sin \theta = 12.0.\]Solve for \( \sin \theta \):\[\sin \theta = \frac{12.0}{18.0} = \frac{2}{3}.\]
04

Formula for Scalar Product

The scalar product \( \vec{A} \cdot \vec{B} \) is given by \( |\vec{A}| |\vec{B}| \cos \theta \). Use this formula with the known magnitudes to compute the dot product.
05

Calculate Cosine of Angle for Both Configurations

Recall \( \theta \) ranges from 0 to 180 degrees. Find \( \cos \theta \) for two configurations: \( \sin \theta = \frac{2}{3} \) implies \( \theta_1 \) and \( \theta_2 = 180^\circ - \theta_1. \)For \( \theta_1 \), use \( \cos^2 \theta = 1 - \sin^2 \theta \) to find \( \cos \theta_1 = \frac{\sqrt{5}}{3} \), and for \( \theta_2 = 180^\circ - \theta_1 \), \( \cos \theta_2 = -\frac{\sqrt{5}}{3}. \)
06

Calculate Scalar Product for Both Angles

Use \( |\vec{A}| = 6.00 \) m and \( |\vec{B}| = 3.00 \) m. For \( \theta_1 \):\[ \vec{A} \cdot \vec{B} = 6.00 \times 3.00 \times \frac{\sqrt{5}}{3} = 6\sqrt{5} \, \mathrm{m}^2.\]For \( \theta_2 \):\[ \vec{A} \cdot \vec{B} = 6.00 \times 3.00 \times -\frac{\sqrt{5}}{3} = -6\sqrt{5} \, \mathrm{m}^2.\]
07

Conclusion

The two possible values for \( \vec{A} \cdot \vec{B} \) are \( 6\sqrt{5} \, \mathrm{m}^2 \) and \( -6\sqrt{5} \, \mathrm{m}^2 \). The vectors in both configurations have the same vector product magnitude due to the same sine value, while the scalar product differs in sign due to cosine.

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Key Concepts

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

Scalar Product
In vector algebra, the scalar product, also known as the dot product, is a fundamental way of measuring how much one vector acts in the direction of another. It is calculated by taking the product of the magnitudes of the two vectors and the cosine of the angle between them. For vectors \( \vec{A} \) and \( \vec{B} \), the formula is:
  • \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \)
The scalar product is crucial in projecting force vectors onto various axes or measuring work done by a force when moving an object. It is scalar, meaning it results in a single numerical value, unlike vector multiplication which results in a vector. There are often two possible results when calculating a scalar product, particularly when different angles can be considered, leading to values that differ in sign, as seen from positive and negative cosine values.
Vector Product
The vector product, commonly referred to as the cross product, provides information about the area spanned by two vectors and the direction perpendicular to the plane they form. In terms of magnitude and direction, it is unique compared to the scalar product. The magnitude of the cross product \( \vec{A} \times \vec{B} \) is calculated as:
  • \( |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \)
where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \). This gives a vector orthogonal to both original vectors, defining a plane. The vector product remains constant with the same sine value due to its reliance only on the angle, regardless of the cosine component's sign. In practical applications, the cross product is widely used in physics for calculating torque and angular momentum.
Trigonometric Functions
Trigonometric functions such as sine, cosine, and tangent are critical in analyzing angles and distances in both scalar and vector products. They relate the angle between vectors to their products. Understanding these functions is essential in vector algebra:- Sine (\( \sin \theta \)): Determines the component of the angle used in the vector product, expressing the perpendicular aspect of one vector relative to another.- Cosine (\( \cos \theta \)): Used in scalar products to find components of one vector in the direction of another, reflecting how much one vector aligns with the other.In the exercise, using \( \sin \theta \) helps find the magnitude of the vector product, while \( \cos \theta \) discovers the scalar product's two possible values. Understanding their complementary roles in defining vector relationships helps solve complex physics and engineering problems more intuitively.

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Most popular questions from this chapter

The following conversions occur frequently in physics and are very useful. (a) Use \(1 \mathrm{mi}=5280 \mathrm{ft}\) and \(1 \mathrm{h}=3600 \mathrm{s}\) to convert 60 \(\mathrm{mph}\) to units of \(\mathrm{ft} / \mathrm{s}\) . (b) The acceleration of a freely falling object is 32 \(\mathrm{ft} / \mathrm{s}^{2} .\) Use \(1 \mathrm{ft}=30.48 \mathrm{cm}\) to express this acceleration in units of \(\mathrm{m} / \mathrm{s}^{2}\) . (c) The density of water is 1.0 \(\mathrm{g} / \mathrm{cm}^{3} .\) Convert this density to units of \(\mathrm{kg} / \mathrm{m}^{3} .\)

Stars in the Universe. Astronomers frequently say that there are more stars in the universe than there are grains of sand on all the beaches on the earth. (a) Given that a typical grain of sand is about 0.2 \(\mathrm{mm}\) in diameter, estimate the number of grains of sand on all the earth's beaches, and hence the approximate number of stars in the universe. It would be helpfult an atlas and do some measuring. (b) Given that a typical galaxy contains about 100 billion stars and there are more than 100 billion galaxies in the known universe, estimate the number of stars in the universe and compare this number with your result from part (a).

A graphic artist is creating a new logo for her company's website. In the graphics program she is using, each pixel in an image file has coordinates \((x, y),\) where the origin \((0,0)\) is at the upper left corner of the image, the \(+x\) -axis points to the right, and the \(+y\) -axis points down. Distances are measured in pixels. (a) The artist draws a line from the pixel location \((10,20)\) to the location \((210,200) .\) She wishes to draw a second line that starts at \((10,20),\) is 250 pixels long, and is at an angle of \(30^{\circ}\) measured clockwise from the first line. At which pixel location should this second line end? Give your answer to the nearest pixel. (b) The artist now draws an arrow that connects the lower right end of the first line to the lower right end of the second line. Find the length and direction of this arrow. Draw a diagram showing all three lines.

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

The most powerful engine available for the classic 1963 Chevrolet Corvette Sting Ray developed 360 horsepower and had a displacement of 327 cubic inches. Express this displacement in liters ( L) by using only the conversions \(1 \mathrm{L}=1000 \mathrm{cm}^{3}\) and 1 in. \(=2.54 \mathrm{cm} .\)
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