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Magazine Chapter 17: Problem 1611
if \(\underline{a}\) and \(\underline{b}\) are vector and \(\underline{a} \cdot \underline{b}<0,|\underline{a} \cdot \underline{b}|=|\underline{a} \times \underline{b}|\) then angle between \(\underline{a}\) and \(\underline{b}\) is = (a) \(\pi\) (b) \((7 \pi / 4)\) (c) \((\pi / 4)\) (d) \((3 \pi / 4)\)
Short Answer
Expert verified
The angle between the vectors \(\underline{a}\) and \(\underline{b}\) is \(\frac{3\pi}{4}\) (option d).
Step by step solution
01
Recall the formulas for dot product and cross product
The dot product of two vectors \(\underline{a}\) and \(\underline{b}\) is defined as:
\(\underline{a} \cdot \underline{b} = |\underline{a}| |\underline{b}| \cos{\theta}\), where \(\theta\) is the angle between the two vectors.
The cross product of two vectors \(\underline{a}\) and \(\underline{b}\) is defined as:
\(\underline{a} \times \underline{b} = |\underline{a}| |\underline{b}| \sin{\theta} \cdot \hat{n}\), where \(\hat{n}\) is the unit vector perpendicular to both \(\underline{a}\) and \(\underline{b}\).
The magnitude of the cross product is given by:
\( |\underline{a} \times \underline{b}| = |\underline{a}| |\underline{b}| \sin{\theta} \).
02
Use the given conditions to solve for 胃
We know that \(\underline{a} \cdot \underline{b} < 0\) and \(|\underline{a} \cdot \underline{b}| = |\underline{a} \times \underline{b}|\). Therefore, we have:
\( | \underline{a} \cdot \underline{b} | = | \underline{a} \times \underline{b} | \)
\[ | \underline{a}| |\underline{b}| |\cos{\theta}| = |\underline{a}| |\underline{b}| |\sin{\theta}| \]
Since both magnitudes of \(\underline{a}\) and \(\underline{b}\) are non-negative, we can divide both sides by \(|\underline{a}| |\underline{b}|\) (assuming neither vector is the zero vector, in which case the angle is undefined):
\(|\cos{\theta}| = |\sin{\theta}|\)
03
Determine possible angle values
We know that \(\theta\) is the angle between the vectors, so it must be in the range \(0 \leq \theta \leq \pi\). Since \(\underline{a} \cdot \underline{b} < 0\), the angle between the vectors must be greater than \(\frac{\pi}{2}\). We will examine the possible values of \(\theta\) in the range \(\frac{\pi}{2} < \theta \leq \pi\):
(a) \(\theta = \pi\) (180掳): In this case, the vectors are opposed, which means their dot product is negative, satisfying one of the conditions. However, in this case, \(\sin{\theta} = 0\) and \(\cos{\theta} = -1\), so the other condition is not satisfied.
(b) \(\theta = \frac{7\pi}{4}\) (315掳): This angle is not valid because it is outside of the range \(0 \leq \theta \leq \pi\).
(c) \(\theta = \frac{\pi}{4}\) (45掳): This angle is not valid because it is smaller than \(\frac{\pi}{2}\).
(d) \(\theta = \frac{3\pi}{4}\) (135掳): In this case, \(\sin{\theta} = \cos{\theta} = \frac{\sqrt{2}}{2}\), so the condition \(|\cos{\theta}| = |\sin{\theta}|\) is satisfied. Since the angle is greater than \(\frac{\pi}{2}\), the dot product is negative, satisfying the other condition.
Thus, the answer is (d) \(\theta = \frac{3\pi}{4}\).
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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, measures how much two vectors point in the same direction. It is a fundamental operation in vector calculus and geometry, defined by the equation \[\begin{equation}\underline{a} \cdot \underline{b} = |\underline{a}| |\underline{b}| \cos{\theta}\end{equation}\]where \(\underline{a}\) and \(\underline{b}\) are vectors, \(|\underline{a}|\) and \(|\underline{b}|\) are their magnitudes (lengths), and \(\theta\) is the angle between them. The result of a dot product is a single number, not a vector. When the dot product is less than zero, as in our original exercise, it signifies that the vectors are pointing in opposite directions, indicating an obtuse angle between them.In practical terms, if the dot product is positive, the angle is acute; if negative, the angle is obtuse; if zero, the vectors are perpendicular. Understanding the sign of the dot product is crucial when solving vector angle problems, as it narrows down the possible range for the angle between the vectors.
Cross Product
The cross product, or vector product, stands in contrast to the dot product by providing a vector that is perpendicular to both original vectors. It is expressed mathematically as \[\begin{equation}\underline{a} \times \underline{b} = |\underline{a}| |\underline{b}| \sin{\theta} \cdot \hat{n}\end{equation}\]where \(\hat{n}\) is a unit vector in the direction perpendicular to the plane containing \(\underline{a}\) and \(\underline{b}\), following the right-hand rule. The magnitude of this cross product vector gives us valuable information about the original vectors, particularly relating to the sin of the angle between them.By comparing the absolute values of the dot and cross products, as in the given exercise, we leverage the respective cosine and sine functions to deduce information regarding the angle, while the right-hand rule aids in determining the direction of the resulting vector. This is pivotal for establishing vector orientations in three-dimensional space.
Vector Magnitude
Vector magnitude refers to the size or length of a vector. It is a measure of how far the vector extends in space and is computed by taking the square root of the sum of the squares of the vector's components. For a vector \(\underline{a}\) with components \((a_x, a_y, a_z)\), the magnitude is \[\begin{equation}|\underline{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}\end{equation}\]In our context, knowing the magnitudes of \(\underline{a}\) and \(\underline{b}\) is essential because these quantities directly influence the results of both the dot and cross products. A deeper grasp of vector magnitude is beneficial not just for solving vector angle problems but also for understanding motion and force vectors in physics or gradients in multivariable calculus.
Angle Between Vectors
Determining the angle between two vectors is a pivotal concept in vector calculus. The angle can range from 0 to \(\pi\) radians (or 0 to 180 degrees) and provides insight into the directional relationship of the vectors. As we saw in the exercise, given the magnitude of both dot and cross products, the absolute values of cosine and sine of the angle were equal, suggesting an angle where these trigonometric functions overlap, which is specifically at \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) radians.To solve for the angle in the original exercise, understanding the behavior of sine and cosine in different quadrants is essential. Sine is positive while cosine is negative in the second quadrant, fitting the criteria provided where the vectors have a negative dot product (implying an obtuse angle). Consequently, the correct angle is \(\frac{3\pi}{4}\) radians, highlighting how the integration of vector magnitudes with trigonometric functions aids in determining vector orientations relative to each other.
