Chapter 8: Problem 529
For the following exercises, find the magnitude and direction of the vector, \(0 \leq \theta<2 \pi\). Given \(u=\langle- 1,6\rangle\) and \( v=\langle 6,-1\rangle,\) calculate \(u \cdot v\).
Short Answer
Expert verified
The dot product of vectors \(u\) and \(v\) is \(-12\).
Step by step solution
01
Understand the Dot Product Formula
The dot product (or scalar product) of two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\) is given by the formula: \(a_1 \cdot b_1 + a_2 \cdot b_2\). We will use this formula to calculate the dot product of vectors \(u\) and \(v\).
02
Identify Components of Given Vectors
Vector \(u\) is \(\langle -1, 6 \rangle\) and vector \(v\) is \(\langle 6, -1 \rangle\). Hence, the components are \(u_1 = -1\), \(u_2 = 6\), \(v_1 = 6\), and \(v_2 = -1\).
03
Substitute the Components into the Dot Product Formula
Use the components identified in the previous step to substitute into the dot product formula: \((-1) \cdot 6 + 6 \cdot (-1)\).
04
Perform the Calculation
Calculate the dot product: \(-1 \times 6 = -6\) and \(6 \times -1 = -6\). Thus, the dot product is \(-6 + (-6) = -12\).
05
Interpret the Result
The dot product of the vectors \(u\) and \(v\) is \(-12\). This scalar quantity indicates the cosine of the angle between the vectors when considered with their magnitudes.
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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 calculus. It combines two vectors and results in a scalar quantity. Given two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is calculated using:
For example, with vectors \( u = \langle -1, 6 \rangle \) and \( v = \langle 6, -1 \rangle \), the calculation is:
- \( a_1 \cdot b_1 + a_2 \cdot b_2 \)
For example, with vectors \( u = \langle -1, 6 \rangle \) and \( v = \langle 6, -1 \rangle \), the calculation is:
- \( (-1) \cdot 6 + 6 \cdot (-1) = -6 - 6 = -12 \)
Magnitude
The magnitude of a vector can be thought of as its length or size in space. It's a measure of how long the vector is regardless of its direction. For a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), the magnitude is given by:
- \( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2} \)
- Magnitude of \( u = \sqrt{(-1)^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} \)
- Magnitude of \( v = \sqrt{6^2 + (-1)^2} = \sqrt{36 + 1} = \sqrt{37} \)
Vector Direction
Vectors possess both magnitude and direction. The direction of a vector gives information about where it points in space. It is commonly represented using an angle \( \theta \). To find the direction of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), you can use trigonometric functions:
- \( \theta = \tan^{-1}\left(\frac{a_2}{a_1}\right) \)
- \( \theta = \tan^{-1}\left(\frac{6}{-1}\right) \)
Scalar Product
The scalar product refers to another name for the dot product. It is called scalar because, unlike a cross product which results in a vector, the outcome of a dot product is a single number – a scalar. This product provides insight into the angle between two vectors as well as their parallelity.
- A dot product of zero indicates perpendicular vectors.
- A positive dot product implies the vectors tend to point in the same direction.
- A negative dot product, like \(-12\) from vectors \( u \) and \( v \), suggests the vectors point in somewhat opposite directions, forming an obtuse angle.
- Calculating work done when force is applied at an angle
- Finding projections in vector analysis