题目
Problem
With respect to the origin O O O A A A B B B C C C
O A → = 2 i − j − 6 k , O B → = b i − 2 j + 3 k , O C → = − 4 i + 5 j − 2 k . \begin{align*}
\overrightarrow{OA}=&\,2\mathbf{i}-\mathbf{j}-6\mathbf{k},\quad \overrightarrow{OB}=b\mathbf{i}-2\mathbf{j}+3\mathbf{k},\\
\overrightarrow{OC}=&\,-4\mathbf{i}+5\mathbf{j}-2\mathbf{k}.
\end{align*} O A = O C = 2 i − j − 6 k , O B = b i − 2 j + 3 k , − 4 i + 5 j − 2 k . (a) It is given that ∣ A B → ∣ = ∣ B C → ∣ |\overrightarrow{AB}|=|\overrightarrow{BC}| ∣ A B ∣ = ∣ B C ∣
Find the value of b b b
(3)
(b) A A A B B B C C C D D D
Find the position vector of D D D
(2)
(c) Calculate angle A B C ABC A B C
(3)
题目中文翻译
相对于原点 O O O A A A B B B C C C
O A → = 2 i − j − 6 k , O B → = b i − 2 j + 3 k , O C → = − 4 i + 5 j − 2 k . \begin{align*}
\overrightarrow{OA}=&\,2\mathbf{i}-\mathbf{j}-6\mathbf{k},\quad \overrightarrow{OB}=b\mathbf{i}-2\mathbf{j}+3\mathbf{k},\\
\overrightarrow{OC}=&\,-4\mathbf{i}+5\mathbf{j}-2\mathbf{k}.
\end{align*} O A = O C = 2 i − j − 6 k , O B = b i − 2 j + 3 k , − 4 i + 5 j − 2 k . (a) 已知 ∣ A B → ∣ = ∣ B C → ∣ |\overrightarrow{AB}|=|\overrightarrow{BC}| ∣ A B ∣ = ∣ B C ∣
求 b b b
(3)
(b) A A A B B B C C C D D D
求 D D D
(2)
(c) 计算角 A B C ABC A B C
(3)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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分别求两条边的向量，长度相等可等价比较平方，以避免根号。
答题过程
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The side vectors are
A B → = ( b − 2 , − 1 , 9 ) , B C → = ( − 4 − b , 7 , − 5 ) . \begin{align*}
\overrightarrow{AB}=&\,(b-2,-1,9),\\[4mm]
\overrightarrow{BC}=&\,(-4-b,7,-5).
\end{align*} A B = B C = ( b − 2 , − 1 , 9 ) , ( − 4 − b , 7 , − 5 ) . Equating their squared lengths,
( b − 2 ) 2 + 1 + 81 = ( b + 4 ) 2 + 49 + 25. (b-2)^2+1+81=(b+4)^2+49+25. ( b − 2 ) 2 + 1 + 81 = ( b + 4 ) 2 + 49 + 25. Thus
b 2 − 4 b + 86 = b 2 + 8 b + 90 , b^2-4b+86=b^2+8b+90, b 2 − 4 b + 86 = b 2 + 8 b + 90 , giving − 12 b = 4 -12b=4 − 12 b = 4 b = − 1 3 \boxed{b=-\frac13} b = − 3 1
(b)
解法一
思路
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菱形也是平行四边形，按顶点顺序有 A D → = B C → \overrightarrow{AD}=\overrightarrow{BC} A D = B C
答题过程
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For the rhombus A B C D ABCD A B C D
O D → = O A → + B C → = O A → + O C → − O B → . \overrightarrow{OD}
=\overrightarrow{OA}+\overrightarrow{BC}
=\overrightarrow{OA}+\overrightarrow{OC}-\overrightarrow{OB}. O D = O A + B C = O A + O C − O B . Using b = − 1 / 3 b=-1/3 b = − 1/3
O D → = ( 2 , − 1 , − 6 ) + ( − 4 , 5 , − 2 ) − ( − 1 / 3 , − 2 , 3 ) = − 5 3 i + 6 j − 11 k . \begin{align*}
\overrightarrow{OD}
=&\,(2,-1,-6)+(-4,5,-2)\\[4mm]
&\,\hspace{2pt}-(-1/3,-2,3)\\[4mm]
=&\,\boxed{-\frac53\mathbf i+6\mathbf j-11\mathbf k}.
\end{align*} O D = = ( 2 , − 1 , − 6 ) + ( − 4 , 5 , − 2 ) − ( − 1/3 , − 2 , 3 ) − 3 5 i + 6 j − 11 k .
(c)
解法一
思路
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角在 B B B B B B A B → \overrightarrow{AB} A B B C → \overrightarrow{BC} B C
答题过程
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With b = − 1 / 3 b=-1/3 b = − 1/3
B A → = ( 7 / 3 , 1 , − 9 ) , B C → = ( − 11 / 3 , 7 , − 5 ) . \begin{align*}
\overrightarrow{BA}=&\,(7/3,1,-9),\\[4mm]
\overrightarrow{BC}=&\,(-11/3,7,-5).
\end{align*} B A = B C = ( 7/3 , 1 , − 9 ) , ( − 11/3 , 7 , − 5 ) . Their scalar product is
− 77 9 + 7 + 45 = 391 9 . -\frac{77}{9}+7+45=\frac{391}{9}. − 9 77 + 7 + 45 = 9 391 . Also,
B A 2 = B C 2 = 787 9 . BA^2=BC^2=\frac{787}{9}. B A 2 = B C 2 = 9 787 . Therefore
cos ∠ A B C = 391 / 9 787 / 9 = 391 787 . \cos\angle ABC=\frac{391/9}{787/9}=\frac{391}{787}. cos ∠ A B C = 787/9 391/9 = 787 391 . Hence
∠ A B C = 60.2 ∘ . \boxed{\angle ABC=60.2^\circ}. ∠ A B C = 60. 2 ∘ . 