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	<title>
	Comments on: 花花酱 LeetCode 963. Minimum Area Rectangle II	</title>
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	<link>https://zxi.mytechroad.com/blog/geometry/leetcode-963-minimum-area-rectangle-ii/</link>
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	<lastBuildDate>Fri, 28 Dec 2018 19:32:53 +0000</lastBuildDate>
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		<title>
		By: zxi		</title>
		<link>https://zxi.mytechroad.com/blog/geometry/leetcode-963-minimum-area-rectangle-ii/#comment-26</link>

		<dc:creator><![CDATA[zxi]]></dc:creator>
		<pubDate>Fri, 28 Dec 2018 19:32:53 +0000</pubDate>
		<guid isPermaLink="false">https://zxi.mytechroad.com/blog/?p=4544#comment-26</guid>

					<description><![CDATA[In reply to &lt;a href=&quot;https://zxi.mytechroad.com/blog/geometry/leetcode-963-minimum-area-rectangle-ii/#comment-25&quot;&gt;sicheng tang&lt;/a&gt;.

向量 p1p2 和 向量 p1p3 的点积(dot product)为0的话表示它们垂直，则p1p2p3构成一个直角三角形（不一定等腰）

v1 = (x1, y1) 和 v2 = (x2, y2) 的点积 v1 * v2 = x1 * x2 - y1 * y2

是余弦定理的一个特例 cos(90) = v1 * v2 / (&#124;v1&#124; * &#124; v2&#124;) = 0 =&gt; v1 * v2 = 0]]></description>
			<content:encoded><![CDATA[<p>向量 p1p2 和 向量 p1p3 的点积(dot product)为0的话表示它们垂直，则p1p2p3构成一个直角三角形（不一定等腰）</p>
<p>v1 = (x1, y1) 和 v2 = (x2, y2) 的点积 v1 * v2 = x1 * x2 &#8211; y1 * y2</p>
<p>是余弦定理的一个特例 cos(90) = v1 * v2 / (|v1| * | v2|) = 0 => v1 * v2 = 0</p>
]]></content:encoded>
		
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		<title>
		By: sicheng tang		</title>
		<link>https://zxi.mytechroad.com/blog/geometry/leetcode-963-minimum-area-rectangle-ii/#comment-25</link>

		<dc:creator><![CDATA[sicheng tang]]></dc:creator>
		<pubDate>Thu, 27 Dec 2018 10:21:26 +0000</pubDate>
		<guid isPermaLink="false">https://zxi.mytechroad.com/blog/?p=4544#comment-25</guid>

					<description><![CDATA[hi: 
   有个地方没看懂，能提示一下吗？


 int dot = (p2[0] - p1[0]) * (p3[0] - p1[0]) +
                    (p2[1] - p1[1]) * (p3[1] - p1[1]);
          if (dot != 0) continue;
          int p4_x = p2[0] + p3[0] - p1[0];
          int p4_y = p2[1] + p3[1] - p1[1];

这段代码是确定三个点是否能组成直角三角形对吗？为什么呢]]></description>
			<content:encoded><![CDATA[<p>hi:<br />
   有个地方没看懂，能提示一下吗？</p>
<p> int dot = (p2[0] &#8211; p1[0]) * (p3[0] &#8211; p1[0]) +<br />
                    (p2[1] &#8211; p1[1]) * (p3[1] &#8211; p1[1]);<br />
          if (dot != 0) continue;<br />
          int p4_x = p2[0] + p3[0] &#8211; p1[0];<br />
          int p4_y = p2[1] + p3[1] &#8211; p1[1];</p>
<p>这段代码是确定三个点是否能组成直角三角形对吗？为什么呢</p>
]]></content:encoded>
		
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