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fix formatting issue

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Akbar Rahman 2022-02-20 14:18:43 +00:00
parent d845161287
commit 9c8de3928e
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1 changed files with 4 additions and 5 deletions
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uni/mmme/1026_maths_for_engineering/vectors.md
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@ -58,12 +58,14 @@ Let $k$, an arbitrary scalar and $\pmb a$, an arbitrary vector.
- $k(\pmb a + \pmb b) = k\pmb a + k\pmb b$ - $k(\pmb a + \pmb b) = k\pmb a + k\pmb b$
- $(k_1 + k_2)\pmb a = k_1\pmb a + k_2\pmb a$ - $(k_1 + k_2)\pmb a = k_1\pmb a + k_2\pmb a$
- $(k_1k_2)\pmb a = k_1(k_2\pmb a)$ - $(k_1k_2)\pmb a = k_1(k_2\pmb a)$
-
### The Scalar Product (Inner Product, Dot Product) ### The Scalar Product (Inner Product, Dot Product)
The scalar product of two vectors $\pmb a$ and $\pmb b$ is a scalar defined by 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 $\cos\theta = \cos(2\pi - \theta)$). 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. 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 ii. One or both of the vectorse are zero vectors
- $\pmb a \cdot \pmb a = |\pmb a|^2 = a^2$ - $\pmb a \cdot \pmb a = |\pmb a|^2 = a^2$
- 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 The base vectors are said to be *orthonormal* when $\pmb i^2 = \pmb j^2 = \pmb k^2 = 1$ and
$i\cdot j = i\cdot k = j\cdot k = 0$. $i\cdot j = i\cdot k = j\cdot k = 0$.