fix generation errors for chp08

Author: longfei-chen

chp08/subsections/8.1.html

@@ -883,18 +883,16 @@ <h4>习题1 <span class="difficulty medium">中等难度</span></h4>
                     <h4>习题2 <span class="difficulty hard">较难</span></h4>
                     <p>已知向量 $\vec{a} = (1, 2, -2)$，$\vec{b} = (2, -1, 2)$。</p>
                     <ol>
-                        <li>证明 $\vec{a} \perp \vec{b}$</li>
+                        <li>判断 $\vec{a}$ 与 $\vec{b}$ 是否垂直</li>
                         <li>求向量 $\vec{c} = 2\vec{a} + 3\vec{b}$ 在坐标轴上的投影</li>
                         <li>求 $\vec{c}$ 与各坐标轴的夹角</li>
                     </ol>
 
                     <div class="solution">
                         <div class="solution-header">解：</div>
-                        <p><strong>1) 证明垂直：</strong></p>
+                        <p><strong>1) 由：</strong></p>
                         <p>$\vec{a} \cdot \vec{b} = 1 \times 2 + 2 \times (-1) + (-2) \times 2 = 2 - 2 - 4 = -4 \neq 0$</p>
-                        <p>所以 $\vec{a}$ 与 $\vec{b}$ 不垂直。重新计算：</p>
-                        <p>$\vec{a} \cdot \vec{b} = 1 \times 2 + 2 \times (-1) + (-2) \times 2 = 2 - 2 - 4 = -4$</p>
-                        <p>实际上 $\vec{a}$ 与 $\vec{b}$ 不垂直，题目可能有误。</p>
+                        <p>所以 $\vec{a}$ 与 $\vec{b}$ 不垂直。</p>
 
                         <p><strong>2) 求 $\vec{c}$ 的坐标和投影：</strong></p>
                         <p>$\vec{c} = 2(1, 2, -2) + 3(2, -1, 2) = (2, 4, -4) + (6, -3, 6) = (8, 1, 2)$</p>

chp08/subsections/8.3.html

@@ -744,7 +744,7 @@ <h4>两平面的夹角</h4>
 
                     <p>两平面的夹角 $\theta$ 满足：</p>
                     <div class="math-expression">
-                        $$\cos \theta = \frac{|A_1A_2 + B_1B_2 + C_1C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}}$$
+                        $$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|A_1A_2 + B_1B_2 + C_1C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}}$$
                     </div>
 
                     <div class="highlight">