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Chapter 5: Problem 182

Show that the dot product can be derived from the theorem of Pythagoras and the law of cosines.

Short Answer

Expert verified
Using the theorem of Pythagoras and law of cosines, we considered a triangle formed by vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\), where \(\mathbf{c} = \mathbf{a} - \mathbf{b}\). By expressing the magnitude of the vector \(\mathbf{c}\) using the dot product and relating it to the law of cosines, we derived the formula for the dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\): \[\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\theta)\]

Step by step solution

01

Definition of the dot product

The dot product, also known as the scalar product or inner product, is the operation defined on two vectors (let's denote them as \(\mathbf{a}\) and \(\mathbf{b}\)) and calculates a scalar quantity. It's represented by the formula: \[\mathbf{a}\cdot\mathbf{b} =|\mathbf{a}||\mathbf{b}|\cos(\theta)\] where \(|\mathbf{a}|\) and \(|\mathbf{b}|\) are the magnitudes (also called lengths) of vectors \(\mathbf{a}\) and \(\mathbf{b}\), respectively, and \(\theta\) is the angle between them.
02

The theorem of Pythagoras

The theorem of Pythagoras states that, in a right triangle, the square of the length of the hypotenuse (the longest side, opposite to the right angle) is equal to the sum of the squares of the lengths of the other two sides: \(c^2 = a^2 + b^2\) We will use this theorem to analyze the relationship between the vectors, their magnitudes and their angles.
03

The law of cosines

The law of cosines generalizes the theorem of Pythagoras for any triangle. It states that in a triangle with sides of lengths a, b, and c, and an angle opposite to the side of length c, cosine of that angle can be found using: \(c^2 = a^2 + b^2 - 2ab\cos(\theta)\) We will use the law of cosines to relate the magnitudes of the vectors and the angle between them.
04

Relating vectors to the theorem of Pythagoras and the law of cosines

Consider \(\mathbf{c} = \mathbf{a} - \mathbf{b}\), which is a vector that goes from the endpoint of \(\mathbf{b}\) to the endpoint of \(\mathbf{a}\). The vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) form a triangle, where the length of each side is the magnitude of the corresponding vector: \(|\mathbf{c}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos(\theta)\) Here, \(|\mathbf{c}|\) is the magnitude of \(\mathbf{a} - \mathbf{b}\), and \(\theta\) is the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\).
05

Expressing the magnitude of a vector difference

We can rewrite the magnitude of vector \(\mathbf{c}\) as: \(|\mathbf{c}|^2 = (\mathbf{a} - \mathbf{b})\cdot(\mathbf{a} - \mathbf{b})\) Using this expression, we can plug in the expression for the dot product from Step 1: \[(\mathbf{a} - \mathbf{b})\cdot(\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos(\theta)\]
06

Solving for the dot product

We need to work with the left side of the equation in Step 5 to derive the dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\): \[\begin{aligned} (\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{b}) &= \mathbf{a}\cdot\mathbf{a}-\mathbf{a}\cdot\mathbf{b}-\mathbf{b}\cdot\mathbf{a}+\mathbf{b}\cdot\mathbf{b} \\ &= |\mathbf{a}|^2 - 2\mathbf{a}\cdot\mathbf{b} + |\mathbf{b}|^2 \end{aligned}\] Therefore, by equating with the right side of the equation in Step 5: \(|\mathbf{a}|^2 - 2\mathbf{a}\cdot\mathbf{b} + |\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos(\theta)\)
07

final derivation

By simplifying the equation from Step 6, the dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be expressed using the theorem of Pythagoras and the law of cosines: \[\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\theta)\] Thus, we have demonstrated the derivation of the dot product from the theorem of Pythagoras and the law of cosines.

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Most popular questions from this chapter

\(\mathrm{A}=\mid \begin{array}{lll}1 & 1 \mid \text { and } \mathrm{P}=\mid 1 & 1 \mid \\ \mid \begin{array}{ll}0 & 1 \mid\end{array} & \mid 1 & -1 \mid \text { . }\end{array}\) (a) Find \(\mathrm{P}^{-1}\). (b) Find \(\mathrm{P}^{-1} \mathrm{AP}\). (c) Verify that, if \(\mathrm{B}\) is similar to \(\mathrm{A}\) then \(\mathrm{A}\) is similar to \(\mathrm{B}\). (d) Show that \(\mathrm{B}^{\mathrm{k}}=\mathrm{P}^{-1} \mathrm{~A}^{\mathrm{k} \mathrm{P}}\) if \(\mathrm{B}=\mathrm{P}^{-1}\) AP where \(\mathrm{k}\) is any positive integer.

Determine the parity of \(\sigma=542163\).

If \(\begin{array}{rrrrrrrr}\mathrm{A}= & \mid 1 & 2 & 4 ; & \mathrm{B}= & 14 & 1 & 4 & 3 \mid, \\ & 12 & 6 & 0 \mid & & 10 & -1 & 3 & 1 \\ & & & & & & 2 & 7 & 5 & 2 \mid\end{array}\) find \(\mathrm{AB}\).

Show that the matrix \(\mathrm{A}\) is not diagnalizable where $$ \mathrm{A}=\mid \begin{array}{rr} -3 & 2 \mid \\ -2 & 1 \mid \end{array} $$

Find the real eigenvalues of \(\mathrm{A}\) and their associated eigenvectors when \(\mathrm{A}=\mid \begin{array}{ll}1 & 1 \mid \\ \mid-2 & 4 \mid \text { . }\end{array}\)
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