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@@ -58,12 +58,14 @@ Let $k$, an arbitrary scalar and $\pmb a$, an arbitrary vector.
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- $k(\pmb a + \pmb b) = k\pmb a + k\pmb b$
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- $(k_1 + k_2)\pmb a = k_1\pmb a + k_2\pmb a$
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- $(k_1k_2)\pmb a = k_1(k_2\pmb a)$
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-
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### The Scalar Product (Inner Product, Dot Product)
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The scalar product of two vectors $\pmb a$ and $\pmb b$ is a scalar defined by
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$$\pmb a \cdot \pmb b = |\pmb a||\pmb b|\cos\theta$$
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$$\pmb a \cdot \pmb b = |\pmb a||\pmb b|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3$$
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where $\pmb a = (a_1, a_2, a_3)$ and $\pmb b = (b_1, b_2, b_3)$
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where $\theta$ is the angle between the two vectors (note that $\cos\theta = \cos(2\pi - \theta)$).
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This definition does not depend on a coordinate system.
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@@ -78,9 +80,6 @@ This definition does not depend on a coordinate system.
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ii. One or both of the vectorse are zero vectors
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- $\pmb a \cdot \pmb a = |\pmb a|^2 = a^2$
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- If $\pmb a = (a_1, a_2, a_3)$ and $\pmb b = (b_1, b_2, b_3)$ then
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$$\pmb a \cdot \pmb b = a_1b_1 + a_2b_2 + a_3b_3$$
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The base vectors are said to be *orthonormal* when $\pmb i^2 = \pmb j^2 = \pmb k^2 = 1$ and
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$i\cdot j = i\cdot k = j\cdot k = 0$.
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@@ -188,6 +187,13 @@ which has a magnitude and direction given by the line $\overrightarrow{OP}$.
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A point $(x, y, z)$ in Cartesian space has the position vector $r = x\pmb i + y\pmb j + z\pmb k$.
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## Angle Between Vectors
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By rearranging the [dot product equation](#the-scalar-product-inner-product-dot-product) you can get
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an equation to find the angle, $\theta$, between two vectors:
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$$\cos\theta = \frac{\pmb a \cdot \pmb b}{|\pmb a||\pmb b|} = \frac{a_1b_1 + a_2b_2 + a_3b_3}{|\pmb a ||\pmb b|}$$
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# Applications of Vectors
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## Application of Vectors to Geometry
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