fix formatting issue

uni/mmme/1026_maths_for_engineering/vectors.md

@@ -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
 $i\cdot j = i\cdot k = j\cdot k = 0$.