题目
Problem
6.
The plane Π \Pi Π
r = ( − 1 5 2 ) + λ ( 2 0 7 ) + μ ( 3 − 2 3 ) \mathbf{r} = \begin{pmatrix} -1 \\ 5 \\ 2 \end{pmatrix}
+ \lambda \begin{pmatrix} 2 \\ 0 \\ 7 \end{pmatrix}
+ \mu \begin{pmatrix} 3 \\ -2 \\ 3 \end{pmatrix} r = − 1 5 2 + λ 2 0 7 + μ 3 − 2 3 where λ \lambda λ μ \mu μ
(a) Determine a vector perpendicular to Π \Pi Π
(2)
The line l l l A ( 1 , 7 , 3 ) A(1, 7, 3) A ( 1 , 7 , 3 ) Π \Pi Π B ( − 1 , 5 , 2 ) B(-1, 5, 2) B ( − 1 , 5 , 2 ) Π \Pi Π l l l α \alpha α
(b) Determine the value of α \alpha α
(4)
(c) Determine the exact shortest distance from A to Π \Pi Π
(4)
题目中文翻译
6.
平面 Π \Pi Π
r = ( − 1 5 2 ) + λ ( 2 0 7 ) + μ ( 3 − 2 3 ) \mathbf{r} = \begin{pmatrix} -1 \\ 5 \\ 2 \end{pmatrix}
+ \lambda \begin{pmatrix} 2 \\ 0 \\ 7 \end{pmatrix}
+ \mu \begin{pmatrix} 3 \\ -2 \\ 3 \end{pmatrix} r = − 1 5 2 + λ 2 0 7 + μ 3 − 2 3 其中 λ \lambda λ μ \mu μ
(a) 求一个垂直于 Π \Pi Π
(2)
直线 l l l A ( 1 , 7 , 3 ) A(1,7,3) A ( 1 , 7 , 3 ) Π \Pi Π B ( − 1 , 5 , 2 ) B(-1,5,2) B ( − 1 , 5 , 2 ) l l l Π \Pi Π α \alpha α
(b) 求 α \alpha α
(4)
(c) 求 A 到 Π \Pi Π
(4)
解答
(a)
解法一
思路
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平面的两个参数向量都平行于平面，因此它们的叉积同时垂直于这两个方向，也就是平面的一个法向量。
答题过程
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The direction vectors of the plane are
u = ( 2 0 7 ) and v = ( 3 − 2 3 ) . \mathbf u=
\begin{pmatrix}
2\\
0\\
7
\end{pmatrix}
\quad\text{and}\quad
\mathbf v=
\begin{pmatrix}
3\\
-2\\
3
\end{pmatrix}. u = 2 0 7 and v = 3 − 2 3 . Their cross product is
u × v = ∣ i j k 2 0 7 3 − 2 3 ∣ = [ 0 ( 3 ) − 7 ( − 2 ) ] i − [ 2 ( 3 ) − 7 ( 3 ) ] j + [ 2 ( − 2 ) − 0 ( 3 ) ] k = 14 i + 15 j − 4 k . \begin{align*}
\mathbf u\times\mathbf v
=&\,
\begin{vmatrix}
\mathbf i&\mathbf j&\mathbf k\\
2&0&7\\
3&-2&3
\end{vmatrix}\\
=&\,[0(3)-7(-2)]\mathbf i\\
&\,-[2(3)-7(3)]\mathbf j\\
&\,+[2(-2)-0(3)]\mathbf k\\
=&\,14\mathbf i+15\mathbf j-4\mathbf k.
\end{align*} u × v = = = i 2 3 j 0 − 2 k 7 3 [ 0 ( 3 ) − 7 ( − 2 )] i − [ 2 ( 3 ) − 7 ( 3 )] j + [ 2 ( − 2 ) − 0 ( 3 )] k 14 i + 15 j − 4 k . Therefore, one vector perpendicular to Π \Pi Π
( 14 15 − 4 ) . \boxed{
\begin{pmatrix}
14\\
15\\
-4
\end{pmatrix}}. 14 15 − 4 .
解法二
思路
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也可以设一般法向量 ( p , q , r ) (p,q,r) ( p , q , r )
答题过程
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Let a normal vector be
n = ( p q r ) . \mathbf n=
\begin{pmatrix}
p\\
q\\
r
\end{pmatrix}. n = p q r . Since n \mathbf n n
2 p + 7 r = 0 , 3 p − 2 q + 3 r = 0. \begin{align*}
2p+7r=&\,0,\\
3p-2q+3r=&\,0.
\end{align*} 2 p + 7 r = 3 p − 2 q + 3 r = 0 , 0. Choose r = − 4 r=-4 r = − 4 p = 14 p=14 p = 14
42 − 2 q − 12 = 0 , 42-2q-12=0, 42 − 2 q − 12 = 0 , so q = 15 q=15 q = 15
n = ( 14 15 − 4 ) . \boxed{
\mathbf n=
\begin{pmatrix}
14\\
15\\
-4
\end{pmatrix}}. n = 14 15 − 4 .
(b)
解法一
思路
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直线方向向量可由 A − B A-B A − B α \alpha α sin α = ∣ d ⋅ n ∣ ∣ d ∣ ∣ n ∣ \sin\alpha=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|} sin α = ∣ d ∣∣ n ∣ ∣ d ⋅ n ∣
答题过程
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A direction vector of l l l
d = B A → = ( 2 2 1 ) . \mathbf d=
\overrightarrow{BA}=
\begin{pmatrix}
2\\
2\\
1
\end{pmatrix}. d = B A = 2 2 1 . Also,
∣ d ∣ = 3 , ∣ n ∣ = 14 2 + 15 2 + ( − 4 ) 2 = 437 , d ⋅ n = 2 ( 14 ) + 2 ( 15 ) + 1 ( − 4 ) = 54. \begin{align*}
|\mathbf d|=&\,3,\\
|\mathbf n|=&\,\sqrt{14^2+15^2+(-4)^2}\\
=&\,\sqrt{437},\\
\mathbf d\mathbin{\boldsymbol\cdot}\mathbf n
=&\,2(14)+2(15)+1(-4)\\
=&\,54.
\end{align*} ∣ d ∣ = ∣ n ∣ = = d ⋅ n = = 3 , 1 4 2 + 1 5 2 + ( − 4 ) 2 437 , 2 ( 14 ) + 2 ( 15 ) + 1 ( − 4 ) 54. For the acute angle between a line and a plane,
sin α = ∣ d ⋅ n ∣ ∣ d ∣ ∣ n ∣ = 54 3 437 = 18 437 . \begin{align*}
\sin\alpha
=&\,\frac{
|\mathbf d\mathbin{\boldsymbol\cdot}\mathbf n|}
{|\mathbf d||\mathbf n|}\\
=&\,\frac{54}{3\sqrt{437}}\\
=&\,\frac{18}{\sqrt{437}}.
\end{align*} sin α = = = ∣ d ∣∣ n ∣ ∣ d ⋅ n ∣ 3 437 54 437 18 . Thus
α = sin − 1 ( 18 437 ) = 59.435 … ∘ . \alpha=\sin^{-1}\big(\frac{18}{\sqrt{437}}\big)
=59.435\ldots^\circ. α = sin − 1 ( 437 18 ) = 59.435 … ∘ . Therefore, to the nearest degree,
α = 59 ∘ . \boxed{\alpha=59^\circ}. α = 5 9 ∘ .
解法二
思路
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也可以先求直线与平面法向量的锐角 β \beta β cos β \cos\beta cos β α = 90 ∘ − β \alpha=90^\circ-\beta α = 9 0 ∘ − β
答题过程
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Let β \beta β l l l Π \Pi Π
cos β = ∣ d ⋅ n ∣ ∣ d ∣ ∣ n ∣ = 18 437 . \begin{align*}
\cos\beta
=&\,\frac{
|\mathbf d\mathbin{\boldsymbol\cdot}\mathbf n|}
{|\mathbf d||\mathbf n|}\\
=&\,\frac{18}{\sqrt{437}}.
\end{align*} cos β = = ∣ d ∣∣ n ∣ ∣ d ⋅ n ∣ 437 18 . Hence
β = 30.564 … ∘ , α = 90 ∘ − β = 59.435 … ∘ . \begin{align*}
\beta=&\,30.564\ldots^\circ,\\
\alpha=&\,90^\circ-\beta\\
=&\,59.435\ldots^\circ.
\end{align*} β = α = = 30.564 … ∘ , 9 0 ∘ − β 59.435 … ∘ . Therefore,
α = 59 ∘ . \boxed{\alpha=59^\circ}. α = 5 9 ∘ .
(c)
解法一
思路
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先用 (a) 的法向量和已知平面上的点 B B B
答题过程
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Using the normal vector from part (a), write the plane as
14 x + 15 y − 4 z = d . 14x+15y-4z=d. 14 x + 15 y − 4 z = d . Since B = ( − 1 , 5 , 2 ) B=(-1,5,2) B = ( − 1 , 5 , 2 ) Π \Pi Π
d = 14 ( − 1 ) + 15 ( 5 ) − 4 ( 2 ) = 53. \begin{align*}
d=&\,14(-1)+15(5)-4(2)\\
=&\,53.
\end{align*} d = = 14 ( − 1 ) + 15 ( 5 ) − 4 ( 2 ) 53. Thus the equation of Π \Pi Π
14 x + 15 y − 4 z = 53. 14x+15y-4z=53. 14 x + 15 y − 4 z = 53. The perpendicular distance from A = ( 1 , 7 , 3 ) A=(1,7,3) A = ( 1 , 7 , 3 ) Π \Pi Π
D = ∣ 14 ( 1 ) + 15 ( 7 ) − 4 ( 3 ) − 53 ∣ 14 2 + 15 2 + ( − 4 ) 2 = 54 437 . \begin{align*}
D
=&\,\frac{|14(1)+15(7)-4(3)-53|}
{\sqrt{14^2+15^2+(-4)^2}}\\
=&\,\boxed{\frac{54}{\sqrt{437}}}.
\end{align*} D = = 1 4 2 + 1 5 2 + ( − 4 ) 2 ∣14 ( 1 ) + 15 ( 7 ) − 4 ( 3 ) − 53∣ 437 54 .
解法二
思路
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利用 (b) 的线面角。因为 B B B A B AB A B ∣ A B ∣ sin α |AB|\sin\alpha ∣ A B ∣ sin α
答题过程
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From part (b),
sin α = 18 437 . \sin\alpha=\frac{18}{\sqrt{437}}. sin α = 437 18 . Also,
∣ A B ∣ = ( 1 + 1 ) 2 + ( 7 − 5 ) 2 + ( 3 − 2 ) 2 = 3. \begin{align*}
|AB|
=&\,\sqrt{(1+1)^2+(7-5)^2+(3-2)^2}\\
=&\,3.
\end{align*} ∣ A B ∣ = = ( 1 + 1 ) 2 + ( 7 − 5 ) 2 + ( 3 − 2 ) 2 3. Therefore, the perpendicular distance is
D = ∣ A B ∣ sin α = 3 ( 18 437 ) = 54 437 . \begin{align*}
D
=&\,|AB|\sin\alpha\\
=&\,3\big(\frac{18}{\sqrt{437}}\big)\\
=&\,\boxed{\frac{54}{\sqrt{437}}}.
\end{align*} D = = = ∣ A B ∣ sin α 3 ( 437 18 ) 437 54 .
解法三
思路
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把 B A → \overrightarrow{BA} B A A A A
答题过程
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The unit normal vector is
n ^ = 1 437 ( 14 15 − 4 ) . \widehat{\mathbf n}
=\frac1{\sqrt{437}}
\begin{pmatrix}
14\\
15\\
-4
\end{pmatrix}. n = 437 1 14 15 − 4 . Since
B A → = ( 2 2 1 ) , \overrightarrow{BA}=
\begin{pmatrix}
2\\
2\\
1
\end{pmatrix}, B A = 2 2 1 , the perpendicular distance is
D = ∣ B A → ⋅ n ^ ∣ = ∣ 2 ( 14 ) + 2 ( 15 ) + 1 ( − 4 ) ∣ 437 = 54 437 . \begin{align*}
D
=&\,\left|
\overrightarrow{BA}
\mathbin{\boldsymbol\cdot}\widehat{\mathbf n}
\right|\\
=&\,\frac{|2(14)+2(15)+1(-4)|}
{\sqrt{437}}\\
=&\,\boxed{\frac{54}{\sqrt{437}}}.
\end{align*} D = = = B A ⋅ n 437 ∣2 ( 14 ) + 2 ( 15 ) + 1 ( − 4 ) ∣ 437 54 .
解法四
思路
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从 A A A X X X A X AX A X
答题过程
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The plane has Cartesian equation 14 x + 15 y − 4 z = 53 14x+15y-4z=53 14 x + 15 y − 4 z = 53 X X X A A A Π \Pi Π A X AX A X
O X → = ( 1 7 3 ) + t ( 14 15 − 4 ) . \overrightarrow{OX}=
\begin{pmatrix}1\\7\\3\end{pmatrix}
+t\begin{pmatrix}14\\15\\-4\end{pmatrix}. O X = 1 7 3 + t 14 15 − 4 . Substituting X = ( 1 + 14 t , 7 + 15 t , 3 − 4 t ) X=(1+14t,7+15t,3-4t) X = ( 1 + 14 t , 7 + 15 t , 3 − 4 t )
14 ( 1 + 14 t ) + 15 ( 7 + 15 t ) − 4 ( 3 − 4 t ) = 53 , 107 + 437 t = 53 , t = − 54 437 . \begin{align*}
14(1+14t)+15(7+15t)-4(3-4t)=&\,53,\\
107+437t=&\,53,\\
t=&\,-\frac{54}{437}.
\end{align*} 14 ( 1 + 14 t ) + 15 ( 7 + 15 t ) − 4 ( 3 − 4 t ) = 107 + 437 t = t = 53 , 53 , − 437 54 . Thus X X X Π \Pi Π A X AX A X Π \Pi Π
A X = ∣ t ∣ 14 2 + 15 2 + ( − 4 ) 2 = 54 437 437 = 54 437 . \begin{align*}
AX=&\,|t|\sqrt{14^2+15^2+(-4)^2}\\
=&\,\frac{54}{437}\sqrt{437}\\
=&\,\boxed{\frac{54}{\sqrt{437}}}.
\end{align*} A X = = = ∣ t ∣ 1 4 2 + 1 5 2 + ( − 4 ) 2 437 54 437 437 54 . 