题目
Problem
With respect to the origin O O O A A A B B B C C C
O A → = ( 2 1 − 3 ) , O B → = ( 0 4 1 ) \overrightarrow{OA}=\begin{pmatrix}2\\1\\-3\end{pmatrix},\quad
\overrightarrow{OB}=\begin{pmatrix}0\\4\\1\end{pmatrix} O A = 2 1 − 3 , O B = 0 4 1 and
O C → = ( − 3 − 2 2 ) . \overrightarrow{OC}=\begin{pmatrix}-3\\-2\\2\end{pmatrix}. O C = − 3 − 2 2 . (a) The point D D D A B C D ABCD A B C D D C → = 3 A B → \overrightarrow{DC}=3\overrightarrow{AB} D C = 3 A B
Find the position vector of D D D
(2)
(b) The diagonals of the trapezium intersect at the point P P P
Find the position vector of P P P
(5)
(c) Using a scalar product, calculate angle A B C ABC A B C
(4)
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 1 − 3 ) , O B → = ( 0 4 1 ) \overrightarrow{OA}=\begin{pmatrix}2\\1\\-3\end{pmatrix},\quad
\overrightarrow{OB}=\begin{pmatrix}0\\4\\1\end{pmatrix} O A = 2 1 − 3 , O B = 0 4 1 和
O C → = ( − 3 − 2 2 ) . \overrightarrow{OC}=\begin{pmatrix}-3\\-2\\2\end{pmatrix}. O C = − 3 − 2 2 . (a) 点 D D D A B C D ABCD A B C D D C → = 3 A B → \overrightarrow{DC}=3\overrightarrow{AB} D C = 3 A B
求 D D D
(2)
(b) 该梯形的两条对角线相交于点 P P P
求 P P P
(5)
(c) 使用数量积计算角 A B C ABC A B C
(4)
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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先求 A B → \overrightarrow{AB} A B D D D D D D C C C
答题过程
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We have
A B → = ( − 2 , 3 , 4 ) . \overrightarrow{AB}=(-2,3,4). A B = ( − 2 , 3 , 4 ) . Since D C → = 3 A B → \overrightarrow{DC}=3\overrightarrow{AB} D C = 3 A B
O D → = O C → − 3 A B → . \overrightarrow{OD}
=\overrightarrow{OC}-3\overrightarrow{AB}. O D = O C − 3 A B . Thus
O D → = ( − 3 , − 2 , 2 ) − 3 ( − 2 , 3 , 4 ) = 3 i − 11 j − 10 k . \begin{align*}
\overrightarrow{OD}=&\,(-3,-2,2)-3(-2,3,4)\\[4mm]
=&\,\boxed{3\mathbf i-11\mathbf j-10\mathbf k}.
\end{align*} O D = = ( − 3 , − 2 , 2 ) − 3 ( − 2 , 3 , 4 ) 3 i − 11 j − 10 k .
(b)
解法一
思路
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分别参数化两条对角线，联立分量求交点，最后检查第三个分量。
答题过程
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A point on A C AC A C
( 2 , 1 , − 3 ) + λ ( − 5 , − 3 , 5 ) , (2,1,-3)+\lambda(-5,-3,5), ( 2 , 1 , − 3 ) + λ ( − 5 , − 3 , 5 ) , and a point on B D BD B D
( 0 , 4 , 1 ) + μ ( 3 , − 15 , − 11 ) . (0,4,1)+\mu(3,-15,-11). ( 0 , 4 , 1 ) + μ ( 3 , − 15 , − 11 ) . At their intersection,
2 − 5 λ = 3 μ , 1 − 3 λ = 4 − 15 μ . \begin{align*}
2-5\lambda=&\,3\mu,\\[4mm]
1-3\lambda=&\,4-15\mu.
\end{align*} 2 − 5 λ = 1 − 3 λ = 3 μ , 4 − 15 μ . The second equation gives λ = 5 μ − 1 \lambda=5\mu-1 λ = 5 μ − 1
2 − 5 ( 5 μ − 1 ) = 3 μ ⟹ 28 μ = 7. 2-5(5\mu-1)=3\mu
\quad\Longrightarrow\quad 28\mu=7. 2 − 5 ( 5 μ − 1 ) = 3 μ ⟹ 28 μ = 7. Thus μ = λ = 1 / 4 \mu=\lambda=1/4 μ = λ = 1/4
The third components agree:
− 3 + 5 4 = − 7 4 = 1 − 11 4 . -3+\frac54=-\frac74
=1-\frac{11}{4}. − 3 + 4 5 = − 4 7 = 1 − 4 11 . Hence
O P → = 3 4 i + 1 4 j − 7 4 k . \boxed{\overrightarrow{OP}
=\frac34\mathbf i+\frac14\mathbf j-\frac74\mathbf k}. O P = 4 3 i + 4 1 j − 4 7 k .
解法二
思路
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梯形平行底边使对角线交点两侧的三角形相似，沿对角线的分段比等于底边比。
答题过程
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Since A B ∥ D C AB\parallel DC A B ∥ D C D C = 3 A B DC=3AB D C = 3 A B P A B PAB P A B P C D PCD P C D 1 : 3 1:3 1 : 3
Hence A P : P C = 1 : 3 AP:PC=1:3 A P : P C = 1 : 3 A P = A C / 4 AP=AC/4 A P = A C /4 A C AC A C
Now
A C → = ( − 5 , − 3 , 5 ) . \overrightarrow{AC}=(-5,-3,5). A C = ( − 5 , − 3 , 5 ) . Therefore
O P → = O A → + 1 4 A C → = ( 2 , 1 , − 3 ) + 1 4 ( − 5 , − 3 , 5 ) = 3 4 i + 1 4 j − 7 4 k . \begin{align*}
\overrightarrow{OP}
=&\,\overrightarrow{OA}+\frac14\overrightarrow{AC}\\
=&\,(2,1,-3)+\frac14(-5,-3,5)\\
=&\,\boxed{\frac34\mathbf i+\frac14\mathbf j-\frac74\mathbf k}.
\end{align*} O P = = = O A + 4 1 A C ( 2 , 1 , − 3 ) + 4 1 ( − 5 , − 3 , 5 ) 4 3 i + 4 1 j − 4 7 k .
(c)
解法一
思路
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角在 B B B B B B
答题过程
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The relevant vectors are
B A → = ( 2 , − 3 , − 4 ) , B C → = ( − 3 , − 6 , 1 ) . \begin{align*}
\overrightarrow{BA}=&\,(2,-3,-4),\\[4mm]
\overrightarrow{BC}=&\,(-3,-6,1).
\end{align*} B A = B C = ( 2 , − 3 , − 4 ) , ( − 3 , − 6 , 1 ) . Their scalar product is
2 ( − 3 ) + ( − 3 ) ( − 6 ) + ( − 4 ) ( 1 ) = 8 , 2(-3)+(-3)(-6)+(-4)(1)=8, 2 ( − 3 ) + ( − 3 ) ( − 6 ) + ( − 4 ) ( 1 ) = 8 , and their moduli are 29 \sqrt{29} 29 46 \sqrt{46} 46
cos ∠ A B C = 8 29 46 , \cos\angle ABC=\frac8{\sqrt{29}\sqrt{46}}, cos ∠ A B C = 29 46 8 , giving
∠ A B C = 77.3 ∘ . \boxed{\angle ABC=77.3^\circ}. ∠ A B C = 77. 3 ∘ . 