求平面2x-2y+z+5=0与各坐标面的夹角的余弦.
求平面$2x-2y+z+5=0$与各坐标面的夹角的余弦.
题目解答
答案
【答案】
平面$2x-2y+z+5=0$与$xOy$面的夹角余弦值为$dfrac{1}{3}$，
与$yOz$面的夹角余弦值为$dfrac{2}{3}$，与$xOz$面的夹角余弦值为$dfrac{2}{3}$.
【解析】
$because $平面$2x-2y+z+5=0$的法向量为$overrightarrow{n}=left(2,-2,1right)$，
设平面与坐标平面$xOy$、$yOz$、$xOz$的夹角分别为${theta }_{1}$、${theta }_{2}$、${theta }_{3}$.
$because overrightarrow{i}=left(0,0,1right)$是平面$xOy$的一个法向量，
$therefore $平面$2x-2y+z+5=0$与$xOy$面的夹角余弦值为：$cos left({theta }_{1}right)=dfrac{left|overrightarrow{n}cdot overrightarrow{i}right|}{mid overrightarrow{n}mid cdot mid overrightarrow{i}mid }=dfrac{left|left(2,-2,1right)cdot left(0,0,1right)right|}{sqrt{{2}^{2}+{left(-2right)}^{2}+{1}^{2}}cdot 1}=dfrac{1}{3}$
同理平面$2x-2y+z+5=0$与$yOz$面的夹角余弦值为$cos left({theta }_{2}right)=dfrac{left|left(2,-2,1right)cdot left(1,0,0right)right|}{sqrt{{2}^{2}+{2}^{2}+{1}^{2}}cdot 1}=dfrac{2}{3}$，
平面$2x-2y+z+5=0$与$xOz$面的夹角余弦值为$cos left({theta }_{3}right)=dfrac{left|left(2,-2,1right)cdot left(0,1,0right)right|}{sqrt{{2}^{2}+{2}^{2}+{1}^{2}}cdot 1}=dfrac{2}{3}$，
故平面$2x-2y+z+5=0$与$xOy$面的夹角余弦值为$dfrac{1}{3}$，
与$yOz$面的夹角余弦值为$dfrac{2}{3}$，与$xOz$面的夹角余弦值为$dfrac{2}{3}$.
解析
平面$2x-2y+z+5=0$的法向量为$overrightarrow{n}=left(2,-2,1right)$。
步骤 2：计算与$xOy$面的夹角余弦值
$xOy$面的法向量为$overrightarrow{i}=left(0,0,1right)$，则平面$2x-2y+z+5=0$与$xOy$面的夹角余弦值为：
$cos left({theta }_{1}right)=dfrac{left|overrightarrow{n}cdot overrightarrow{i}right|}{mid overrightarrow{n}mid cdot mid overrightarrow{i}mid }=dfrac{left|left(2,-2,1right)cdot left(0,0,1right)right|}{sqrt{{2}^{2}+{left(-2right)}^{2}+{1}^{2}}cdot 1}=dfrac{1}{3}$
步骤 3：计算与$yOz$面的夹角余弦值
$yOz$面的法向量为$overrightarrow{j}=left(1,0,0right)$，则平面$2x-2y+z+5=0$与$yOz$面的夹角余弦值为：
$cos left({theta }_{2}right)=dfrac{left|overrightarrow{n}cdot overrightarrow{j}right|}{mid overrightarrow{n}mid cdot mid overrightarrow{j}mid }=dfrac{left|left(2,-2,1right)cdot left(1,0,0right)right|}{sqrt{{2}^{2}+{2}^{2}+{1}^{2}}cdot 1}=dfrac{2}{3}$
步骤 4：计算与$xOz$面的夹角余弦值
$xOz$面的法向量为$overrightarrow{k}=left(0,1,0right)$，则平面$2x-2y+z+5=0$与$xOz$面的夹角余弦值为：
$cos left({theta }_{3}right)=dfrac{left|overrightarrow{n}cdot overrightarrow{k}right|}{mid overrightarrow{n}mid cdot mid overrightarrow{k}mid }=dfrac{left|left(2,-2,1right)cdot left(0,1,0right)right|}{sqrt{{2}^{2}+{2}^{2}+{1}^{2}}cdot 1}=dfrac{2}{3}$