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- Session 2: Dot Products | Part A: Vectors, Determinants and Planes | 1. Vectors and Matrices | Multivariable Calculus | Mathematics | MIT OpenCourseWare
1 Chalkboard
Figure 1: Dot Product Definition
Figure 2: What does geometic definition mean?
Figure 3: Dot product of combined vectors
2 What is the dot product of 2 vectors?
2.1 Front
What is the dot product of 2 vectors?
2.2 Back
- Is one way to combining (“multiplying”) two vectors
- The output is a scalar
- Algebraically
- \(A \cdot B = a_1b_1 + a_2b_2\)
- Geometrically
- \(A \cdot B = |A| |B| \cos(\theta)\)
3 How can we proof the dot product geometrically?
3.1 Front
How can we proof the dot product geometrically?
3.2 Back
- Law of cosines
- \(|\vb{A} - \vb{B}|^2 = |\vb{A}|^2 + |\vb{B}|^2 - 2|\vb{A}||\vb{B}| \cos(\theta)\)
- Expanding \(|A - B|^2\)
- \(|\vb{A} - \vb{B}|^2 = (\vb{A} - \vb{B}) \cdot (\vb{A} - \vb{B}) = \vb{A} \cdot \vb{A} - \vb{A} \cdot \vb{B} - \vb{B} \cdot \vb{A} + \vb{B} \cdot \vb{B} = |\vb{A}|^2 + |\vb{B}|^2 - 2\vb{A} \cdot \vb{B}\)
- Comparing the 2 equations
- \(\vb{A} \cdot \vb{B} = |\vb{A}||\vb{B}| \cos(\theta)\)
4 What does mean that \(\vb{A} \cdot \vb{B} = 0\)?
4.1 Front
What does mean that $\vb{A} \cdot \vb{B} = 0$?
4.2 Back
Two vectors are perpendicular to each other, we say they are orthogonal
- \(\cos(\pi/2) = 0\)
- \(\vb{A} \perp \vb{B} \Leftrightarrow \vb{A} \cdot \vb{B} = 0\)
5 Which is the dot product of the 1 vector by itself?
5.1 Front
Which is the dot product of the 1 vector by itself?
\(\vb{A} \cdot \vb{A}\)
5.2 Back
- Algebraically
- \(\vb{A} \cdot \vb{A} = \ev{a_1, a_2, a_3} \cdot \ev{a_1, a_2, a_3} = a_1^2 + a_2^2 + a_3^2 = \abs{\vb{A}}^{2}\)
- Geometrically
- \(\vb{A} \cdot \vb{A} = \abs{\vb{A}} \abs{\vb{A}} \cos \theta = \abs{\vb{A}}^2\)