From 570594cffb0570eaa3d3dbdc578e7dcbbf843fbb Mon Sep 17 00:00:00 2001
From: 24OI-bot <15963390+24OI-bot@users.noreply.github.com>
Date: Fri, 13 Sep 2019 12:16:26 -0400
Subject: [PATCH] style: format markdown files with remark-lint

---
 docs/math/vector.md | 10 +++++-----
 1 file changed, 5 insertions(+), 5 deletions(-)

diff --git a/docs/math/vector.md b/docs/math/vector.md
index e7298716..3afde18a 100644
--- a/docs/math/vector.md
+++ b/docs/math/vector.md
@@ -148,23 +148,23 @@ $$
 
 #### 判定两向量垂直
 
- $\boldsymbol a \perp \boldsymbol b$  $\Leftrightarrow$  $\boldsymbol a\cdot \boldsymbol b=0$
+ $\boldsymbol a \perp \boldsymbol b$  $\Leftrightarrow$  $\boldsymbol a\cdot \boldsymbol b=0$ 
 
 #### 判定两向量共线
 
- $\boldsymbol a = \lambda \boldsymbol b$  $\Leftrightarrow$  $\boldsymbol a\cdot \boldsymbol b=|\boldsymbol a||\boldsymbol b|$
+ $\boldsymbol a = \lambda \boldsymbol b$  $\Leftrightarrow$  $\boldsymbol a\cdot \boldsymbol b=|\boldsymbol a||\boldsymbol b|$ 
 
 #### 数量积的坐标运算
 
-若 $\boldsymbol a=(m,n),\boldsymbol b=(p,q),$ 则 $\boldsymbol a\cdot \boldsymbol b=mp+nq$
+若 $\boldsymbol a=(m,n),\boldsymbol b=(p,q),$ 则 $\boldsymbol a\cdot \boldsymbol b=mp+nq$ 
 
 #### 向量的模
 
- $|\boldsymbol a|=\sqrt {m^2+n^2}$
+ $|\boldsymbol a|=\sqrt {m^2+n^2}$ 
 
 #### 两向量的夹角
 
- $\cos \theta=\cfrac{\boldsymbol a\cdot\boldsymbol b}{|\boldsymbol a||\boldsymbol b|}$
+ $\cos \theta=\cfrac{\boldsymbol a\cdot\boldsymbol b}{|\boldsymbol a||\boldsymbol b|}$ 
 
 ### 扩展
 
-- 
2.11.0