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Author SHA1 Message Date
a5e865c29d add section angle between vectors 2022-02-20 14:40:12 +00:00
9c8de3928e fix formatting issue 2022-02-20 14:33:55 +00:00

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