题目
Problem
Relative to the origin O O O A A A B B B C C C
O A → = ( 1 3 1 ) , O B → = ( 3 1 2 ) \overrightarrow{OA}=\begin{pmatrix}1\\3\\1\end{pmatrix},\quad\overrightarrow{OB}=\begin{pmatrix}3\\1\\2\end{pmatrix} O A = 1 3 1 , O B = 3 1 2 and
O C → = ( 5 3 − 2 ) . \overrightarrow{OC}=\begin{pmatrix}5\\3\\-2\end{pmatrix}. O C = 5 3 − 2 . (a) Using a scalar product, find the cosine of angle B A C BAC B A C
(4)
(b) Hence find the area of triangle 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 → = ( 1 3 1 ) , O B → = ( 3 1 2 ) \overrightarrow{OA}=\begin{pmatrix}1\\3\\1\end{pmatrix},\quad\overrightarrow{OB}=\begin{pmatrix}3\\1\\2\end{pmatrix} O A = 1 3 1 , O B = 3 1 2 和
O C → = ( 5 3 − 2 ) . \overrightarrow{OC}=\begin{pmatrix}5\\3\\-2\end{pmatrix}. O C = 5 3 − 2 . (a) 使用数量积求角 B A C BAC B A C
(4)
(b) 由此求三角形 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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角的顶点是 A A A A B → \overrightarrow{AB} A B A C → \overrightarrow{AC} A C
答题过程
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The vectors are
A B → = ( 2 , − 2 , 1 ) , A C → = ( 4 , 0 , − 3 ) . \overrightarrow{AB}=(2,-2,1),\qquad
\overrightarrow{AC}=(4,0,-3). A B = ( 2 , − 2 , 1 ) , A C = ( 4 , 0 , − 3 ) . Their scalar product is 2 ( 4 ) + ( − 2 ) ( 0 ) + 1 ( − 3 ) = 5 2(4)+(-2)(0)+1(-3)=5 2 ( 4 ) + ( − 2 ) ( 0 ) + 1 ( − 3 ) = 5 3 3 3 5 5 5
cos ∠ B A C = 5 3 ( 5 ) = 1 3 . \boxed{\cos\angle BAC=\frac5{3(5)}=\frac13}. cos ∠ B A C = 3 ( 5 ) 5 = 3 1 .
(b)
解法一
思路
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承接(a)，先用 sin 2 A + cos 2 A = 1 \sin^2A+\cos^2A=1 sin 2 A + cos 2 A = 1
答题过程
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Let θ = ∠ B A C \theta=\angle BAC θ = ∠ B A C
sin θ = 1 − 1 9 = 2 2 3 , \sin\theta=\sqrt{1-\frac19}=\frac{2\sqrt2}{3}, sin θ = 1 − 9 1 = 3 2 2 , since 0 < θ < π 0<\theta<\pi 0 < θ < π
Area of A B C = 1 2 ∣ A B → ∣ ∣ A C → ∣ sin θ = 1 2 ( 3 ) ( 5 ) 2 2 3 = 5 2 . \begin{align*}
\text{Area of }ABC
=&\,\frac12|\overrightarrow{AB}|\,
|\overrightarrow{AC}|\sin\theta\\[4mm]
=&\,\frac12(3)(5)\frac{2\sqrt2}{3}\\[4mm]
=&\,\boxed{5\sqrt2}.
\end{align*} Area of A B C = = = 2 1 ∣ A B ∣ ∣ A C ∣ sin θ 2 1 ( 3 ) ( 5 ) 3 2 2 5 2 .
解法二
思路
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承接(a)已经求出的边向量，向直线 B C BC B C
答题过程
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Let H H H A A A B C BC B C B C → = ( 2 , 2 , − 4 ) \overrightarrow{BC}=(2,2,-4) B C = ( 2 , 2 , − 4 ) B H → = λ ( 2 , 2 , − 4 ) \overrightarrow{BH}=\lambda(2,2,-4) B H = λ ( 2 , 2 , − 4 )
A H → = ( 2 + 2 λ , − 2 + 2 λ , 1 − 4 λ ) . \overrightarrow{AH}=(2+2\lambda,-2+2\lambda,1-4\lambda). A H = ( 2 + 2 λ , − 2 + 2 λ , 1 − 4 λ ) . The perpendicularity condition gives
A H → ⋅ ( 2 , 2 , − 4 ) = 0 , 2 ( 2 + 2 λ ) + 2 ( − 2 + 2 λ ) − 4 ( 1 − 4 λ ) = 0 , 24 λ − 4 = 0. \begin{align*}
\overrightarrow{AH}\cdot(2,2,-4)=&\,0,\\[4mm]
2(2+2\lambda)+2(-2+2\lambda)
-4(1-4\lambda)=&\,0,\\[4mm]
24\lambda-4=&\,0.
\end{align*} A H ⋅ ( 2 , 2 , − 4 ) = 2 ( 2 + 2 λ ) + 2 ( − 2 + 2 λ ) − 4 ( 1 − 4 λ ) = 24 λ − 4 = 0 , 0 , 0. Thus λ = 1 / 6 \lambda=1/6 λ = 1/6
A H → = ( 7 3 , − 5 3 , 1 3 ) , A H = 49 + 25 + 1 9 = 5 3 3 . \begin{align*}
\overrightarrow{AH}=&\,\bigg(\frac73,-\frac53,\frac13\bigg),\\[4mm]
AH=&\,\sqrt{\frac{49+25+1}{9}}=\frac{5\sqrt3}{3}.
\end{align*} A H = A H = ( 3 7 , − 3 5 , 3 1 ) , 9 49 + 25 + 1 = 3 5 3 . Also B C = 2 2 + 2 2 + ( − 4 ) 2 = 2 6 BC=\sqrt{2^2+2^2+(-4)^2}=2\sqrt6 B C = 2 2 + 2 2 + ( − 4 ) 2 = 2 6
Area of A B C = 1 2 ( 2 6 ) 5 3 3 = 5 2 . \text{Area of }ABC
=\frac12(2\sqrt6)\frac{5\sqrt3}{3}
=\boxed{5\sqrt2}. Area of A B C = 2 1 ( 2 6 ) 3 5 3 = 5 2 . 