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C h u o n g I I I : P H U O N G P H A P T O A
T R O N G K H O N G

D O

G I A N

§1. H E T O A D O T R O N G

K H O N G

G I A N

A. KIEN THUC C O B A N
Toa do trong khong gian
Ba vecto don v i i , j , k tren 3 true Ox, Oy, Oz :
i = ( 1 ; 0 ; 0 ) , ] = ( 0 ; 1;0), k = (0; 0; 1)

M

M(x, y, z) hay M = (x, y, z):

>T>~

OM = x . i + y . j + z . k
a(x, y, z) hay a = (x, y, z):
a = x. i + y. j + z. k
• Hai vecta: u = (x, y, z) va v = (x', y', z') thi:
u ± v = (x ± x' ; y ± y'; z ± z') ; k u = (kx; ky, kz)
u . v = xx' + yy' + zz' ; | u

Vx + y + z
2

2

2

x.x + y.y + z.z'

cos(u, v)
>/x

2
+

y

2

z . Vx' +y' z'
2

2

+

2

2

+

Hai diem A ( x i , y i , z\) va B(x , y , z ) thi:
2

2

2

AB = (x - x i ; y - y i , z - z\)
2

AB = yj(

2

2

2

- x ^ + ( y - y i ) +(z - z r
2

2

x

M chia AB theo ti so k * 1: M

x

i - k x _y - k y
l-k
l-k
2

1

kz.,

2

l - k

Tich co hirotig: cua a = (x, y, z) va b = (x', y', z') la vecto:
n = [a , b ] =

y

z

z

y'

z'

z

x

X y
x' J X ' y'

• Ket qua:
- Vecto [ a , b ] vuong goc vdi a , b
- Do dai cua vecto [a , b ] : | [a , b ] | = | a | . | b | .sin(a , b )
- 2 vecto a , b cung phuong:
- 3 vecto a , b , c dong phang:

[a, b ] = 0
[a, b ] .c =0

- 3 vecto a', b , c khong dong phang:
154

[a , b ] . c * 0


Dien tich va the tich
Dien tich tam giac ABC:

S=-|[AB,AC]|

The tich tii dien ABCD: V = — | [ AB. AC 1. AD |
6
The tich hinh hop ABCD.A'B'C'D': V = | [ A B , AD ] . AA~' |
The tich hinh lang try A B C . A ' B ' C : V = | | [ A B , ADJ.AAT' |
Phirong trinh mat cau:
Mat cau (S) tam I(a, b, c) ban kinh R:

(x - a) + (y - b) + (z - c) = R hay:
x + y + z + 2Ax + 2By + 2Cz + D = 0, A + B + C - D > 0
2

2

2

2

2

2

2

2

2

2

co tam I ( - A , - B , - C ) va ban kinh R = \ / A + B + C - D
Chu y: - Tam duong tron I ngoai tiep tam giac ABC trong khong gian:
f l A = IB = IC
2

2

2

[ I e (ABC)
Tam K mat cau ngoai tiep tii dien ABCD: KA = KB = KC = KD.
D A N G 1: TOA D O DIEM, VECTO
Ba vecto don v i tren 3 true Ox, Oy, Oz :
i = ( 1 ; 0 ; 0 ) , j = ( 0 ; 1;0),

k = (0; 0; 1)

Hai diem A ( x i , y i , Z\) va B(x , y , z ) thi:
2

2

2

AB = (x - x i ; y - y i , z - Zi)
2

2

AB = f x - x )
2

2

1

2

+(y - y ^
2

2

+ (z - z ^

2

2

^ 2 . yi - y . i k

- Diem M chia A B theo ti k * 1: M ,

z

k z

2

2

l-k
l-k
l-k
Toa do trung diem M cua doan A B , trong tam G cua tam giac ABC.
trong tam E cua t i i dien ABCD voi toa dp A(x y zO, B(x , y , z ),
ys, Z ) D(X4 y4 , z ):
+x

+ x +x +x
i + x + x
X = l
X =
4
3
2
+ y + y
yi + y
E y = y i + y + y + y.4
G. v M
y =
y 4
3
2
+ z +z + z
+z f z
Zl + z
zz= i
z=
Z—
4
2
3
1;

1(

2

2

2

4

3

x

x

X l

2

2

2

2

2

3

2

3

3

2

Y l

Z l

3

2

3

z

2

3

4

4

Phep toan ciia hai vecto: u = (x, y, z) va v = (x , y' z') thi:
1

u + v = (x + x' ; y + y'; z + z')
k u = (kx; ky, kz),

u - v = (x - x' ; y - y'; z - z')
u . v = xx' + yy' + zz' ;
155


[u , v ] =

f y
Vy '

z
z' j

Z X
z x'

x
x'

y s
y' >

- Quan he cac vecta
Vecta [ a, b ] vuong goc v o i a, b
Vecto a vuong goc voi b » a . b = 0
Hai vecto a, b cung phuong <=> b = k a
Hai vecto a, b cung phuong <=> [ a, b ] = 6
Ba 3 vecta a, b , c dong phang < > [ a , b ] .c = 0
=
B6n diem A , B , C, D dong phang <=> [ A B , A C ] . A D = 0

Ba 3 vecto a, b , c khong dong phang <=> [ a , b ] . c # 0
Bon diem A , B, C, D khong dong phang <=> [ A B , A C ] . A D * 0.
Chu y: Ung dung toa dp trong khong gian de giai cac bai toan hinh khong
gian co dien, quan he song song, vuong goc, dp dai, goc, khoang each, vi
tri tuong doi,...
V i du 1: Cho ba vecto a = (2; - 5 ; 3), b = (0; 2; - 1 ) , c = (1; 7; 2)
a) T i m toa dp ciia vecto e = a - 4b - 2 c .
b) Tim toa do ciia vecta f = 4 a - — b + 3c.
3
Giai
a) e = a - 4b - 2c = (2 - 0 - 2; -5 - 8 - 14; 3 + 4 1 2
b) f = 4 a - - b +3c = (8 + 0 + 3 ; - 2 0 - - + 21; 12 +
3
3
V i du 2: Tim toa dp cua vecto m cho biet:
a ) a + m = 4 a v a a = ( 0 ; - 2 ; 1).

4) = (0; -27; 3)
1
1 K
K
- +6) = ( 1 1 ; - : — ) .
3
3 3
V

b) a + 2 m = b va a = ( 5 ; 4 ; - l ) , b = ( 2 ; - 5 ; 3).
Giai
a) a + m = 4 a => m = 3 a = (0; -6; 3)
b)

a + 2m = b = > m =

a + — b = —;
;2 .
2
2
{ 2 2 )
V i du 3: Cho hai bp ba didm: A ( l ; 3; 1) B(0; 1; 2), C(0; 0; 1) va
A ' ( l ; 1; 1;), B'(-4; 3; 1), C'(-9; 5; 1). Hoi bp ba diem nao thang hang?
Giai
Taco CA = ( 1 ; 3; 0), CB = ( 0 ; 1; 1)
V i cac toa dp khong tuong xiing ti le nen khong co s6 k nao de
CA = kCB , suy ra A , B, C khong thang hang.
Ta co t T T = (10; - 4 ; 0), G B = (5; - 2 ; 0) => C^A' = 2 C^"'
Do do A', B', C thing hang.
1

156


V i du 4: Tinh tich vo huong ciia hai vecto trong moi truong hop sau:
a) a = (3; 0; -8), b = (2; - 7 ; 0).
b) a = (1; - 5 ; 2), b = (4; 3; -5).
c) a = (0;
a)

N/2

; S ) , b = (4; S ; - 7 2 ) .

Giai

a.b = 3.2 + 0.(-7) + (-8).0 = 6

b) a . b = 1.4 + (-5).3 + 2(-5) = - 2 1
c) a .b = 0.4 + N/2 .73 + V3(-%/2) = 0.
V i du 5: Cho ba vecto: a = ( 1 ; - 1 ; 1), b = ( 4 ; 0 ; - l ) , c = (3; 2 ; - 1 ) . Tinh:
a) (a . b ). c , a .(b . c ).
b) a b + b c + c a , 4 a . c + b - 5 c
Giai
2

2

:

2

2

a) Taco: a.b = 1.4 + ( - l ) . 0 + l . ( - l ) = 3
D o d 6 : ( a . b ) c = 3 c = (9; 6;-3)
Ta co b . c = 4.3 + 0.2 + (-1)(-1) = 13
Do do a ( b . c ) = 13a = ( 1 3 ; - 1 3 ; 13)
b) Taco i = 3, b = 17, c = 14 nen
a . b + b . c + c . a = 3b + 1 7 c + 1 4 a = (77; 20; - 6 ) va a . c = 0
2

2

2

2

2

2

b4a.c + b -5c
2

2

= -53.

Vi du 6: Cho I u 1=2. | v I = 5 , goc giua hai vecto u va v bang — Tim
k de vecto p = k u + 17 v vuong goc voi vecto q = 3 u - v
Giai
l-i
^ ,_ |
- _
2n
Ta co | u | = 2, I v | = 5, cos(u . v ) - cos—

1
— -

Do do p lq <=>p.q = 0<=>(ku +17v)(3u - v) = 0
o 3 k u - 17 v
2

2

+ (51 - k)u .v - 0

< > 3k.4 - 17.25 + (51 - k)2.5— = 0
=
2
< > 17k - 680 = 0 < > k - 40.
=
=
Vi du 7: Cho ba diem A(2; 0; 4). B(4; S ; 5) va C(sin5t; cos3t; sin3t). Tim t
de A B vuong goc voi OC.
Giai
Ta co AB = (2; 73 ; 1), OC = (sin5t; cos3t; sin3t)
Hai dudng thang A B va OC vuong goc vdi nhau khi va chi khi:
A B . OC = 0 < > 2sin5t + 73 cos3t + sin3t = 0.
=
< > sin5t + — cos3t + — sin3t = 0 < > sin5t + sin(3t + — ) = 0
=
=
157


Jl

o sin5t = sin(-3t

)
3

Vay t = -— + -k*.t=—
+ kn, k e Z.
24
4
3
V i du 8: Xet su dong phang cua ba vecto
a) a = (-3; 1; - 2 ) b = (1; 1; 1), c' = (-2; 2; 1).
b) a = (4; 3; 4), b = (2; - 1 ; 2), c = (1; 2; 1).
Giai
a) T a c 6 [ a , b ] = ^

3
]

J ~ j = (3; 1;-4)
2

Do do [a, b ]. c = -6 + 2 - 4 = -8 * 0
Vay 3 vecto khong dong phang
b) Taco [ a , b ] = (10; 0;-10) => [ a , b ] . c = 0
Vay 3 vecto dong phang
V i du 9: Cho a = ( 1 : - 1 ; 1), b = ( 0 ; 1;2), c = ( 4 ; 2 ; 3 ) v a d = (2; 7; 7).
"
a) Chung minh cac vecto a , b , c khong dong phang.
b) Hay bieu thi vecto d theo cac vecto a , b , c .
Giai
a) Taco [a", b ] = ( - 3 ; - 2 ; 1)=> [a", b ] c = - 1 3 * 0
Vay 3 vecto khong dong phang.
b) Gia sir d = m a + nb + pc
m + 4p = 2

f m = -2
<=> < - m + n + 2p = 7 <=>m + 2n + 3p = 7

[p = 1

Vay d = - 2 a + 3 b + c.
V i du 10: Trong khong gian cho bon diem:
A(0; 0; 3), B ( l ; 1; 5), C(-3; 0; 0), D(0; - 3 ; 0)
a) Tinh ( A B . B C ) C A + C D . AB.
b) Chung minh bon diem A, B, C, D dong phang.
Giai
2

a)

AB = ( 1 ; 1; 2), BC (—4; —1;-5), CA = ( 3 ; 0; 3), CD = ( 3 ; - 3 ; 0)
Do do AB
nen(AB

BC = -15 , C D = 18
2

BC)CA + C D AB =-15CA +18AB =(-27; 18;-9)
2

b) AB = (1; 1;2), AC = (-3; 0;-3) => [AB

A C ] = (-3;-3; 3)

Va AD = ( 0 ; - 3 ; - 3 ) = > [ A B , A C ] AD = 0: dpcm.
V i du 11: Chung minh cac tinh chat sau day cua tich vo huong
a)[a,a]=6
c) [ k a , b ] = k [ a , b ] = [ a , k b ]
158

b) [a , b ] = - [ b , a ]


Giai
Yi
y

a) [ a , a ]

i
Zj

z

1

b) [a, b]

z

i
z

x

1

i
x'

x
y
Xj y
:

l

Zj Xj

yi

2

i

x

2

= (0; 0; 0) = 0

1

:

yi

y )
= (yi 2 - Y2 i ; zix - z xi ; xiy - x y )
= - ( y z i - yiz ; z Xi - Z!X ; x y - x y )
z

z

x

x

2

2

z

2

2

2

2

y
.yi

2

x

2

A
x

2

2
z

2

2

'"2
|z
i' k

z

2

2

x

i

x

x

2
i

2

:

x

2

y
yi

2

= - [ b , a]
Ket qua [ a , a ] = - [ a , a ] => [ a , a ]
c) k [ a , b ] =

yi
Ik'"
y

i

z

2

ky
y

z

i

z

2

,k

z
2

kz

x

z
2

kz

x

2

z

i

x

x

, k

x

x

2

2

yi
y )
kxj k y j
2

2

kx

x

i

x

x

2

x
2

y

2

= [ka, b].
Tucmg tu: k [ a , b ] = [ a , k b ]
V i du 12: Chung minh cac tinh chat sau day cua tich vo huang
a) [ c , a + b ] = [ c , a ] + [ c , £ ] .
b) a [b , c ] = [a , b ]. c.
c) I [a . b ] | = | a

b | -(a.b)
Giai

2

2

y
yi + y

a) [c, a + b]

2

3

y
yi

z

3
z

2

3i

z
l

3

Z

y z.
- [a.b]c
2

3
z
2

2

j
:

z

x
3

z

l
x

3
z

= [c, a]
y z.

3

z

3

z

3

y

i

Jy
i |y
3

y
yi

b) a . [ b . c ] = Xj

z
2

i

x

+

3
x

3
i

l

z

3

z

2

x

-1
+ [c, b]
x y
2 x
+yi
Z, 3
3 y
l
l
l

+y
+z
2 x
X, y
Z

2

2

x

X

2

X

3

Z

3

3

Z

2

2

i

3 |z
2 '|z

Z

2

x

x

2

3

3

3

x ;

yf + y
y j , Jy

y
yi

x
3

Z

3

x

2

X


c) VP = |a | | b | - ( a . b ) = | a | | b | - | a P I b | W a
= Ia |
| b 1 (1 - cos a) = | a |
| b 1 sin a
2

2

2

2

2

2

= |[a.b]|

2

2

2

2

2

2

= VT.

V i du 13: Trong khong gian cho ba vecto a , b , c tirng doi khong cimg
phuong. Chung minh rang dieu kien can va dii de vecto tong:
a + b + c = 6 la [ a , b ] = [ b , c ] = [ c , a ]
Giai
Tira + b + c = 6 - ^ a = - ( b + c ) = > [ a , - b - c ] = 0
D o d 6 [ a , - b - c ] = [ a . - b ] - [ a . c ] = 0 =>[c, a ] = [a . b J
Tuong tu ta cung c o [ b , c ] = [ a , b ]
Vay: [ a , b ] = [ b , c ] = [ c , a ] .
Nguoc lai, tir [ a , b ] = [ b , c ] => [ b , a + c ] = 0
Mat khac, [ b , b ] = 0 = > [ b . a + b + c ] = 0 = > b cimg phuong voi
vecto a + b

+ c.

Chung minh tuong tu ta cung co vecto a ciing phuong voi vecto a + b + c.
Nhung a va b khong cimg phuong, vay a + b + c = 0.
V i du 14: Cho diem M(a; b; c)
a) Tim toa do hinh chieu ciia M tren cac mat phang toa do va tren cac
true toa do.
b) Tim khoang each tir diem M den cac mat phang toa do, den cac true
toa do.
Giai
a) Goi M i ( x ; y; 0) la hinh chieu ciia diem M(a; b; c) tren mp(Oxy) thi:
M M j = (x - a; y - b; -c) V i MM . I = 0 va M M ^ T = 0 nen x - a = 0.
l

y - b = 0.
VayM,(a;b;0)
Tuong tu, hinh chieu ciia M tren mp(Oyz) la Mi(0; b; c) hinh chieu ciia
M tren mp(Oxz) la M (a: 0; c)
Gia sir M ( x ; 0; 0) la hinh chieu ciia M(a; b; c) tren true Ox thi
3

x

MM

X

= (x - a; - b ; - c ) . V i M M . i = 0 nen x = a, do do M (a; 0: 0)
V

x

Tuong tu, hinh chieu ciia M(a; b; c) tren true Oy la M ( 0 ; b; 0), hinh
chieu ciia M(a; b; c) tren true Oz la M (0; 0; c).
y

z

b) d(M; (Oxy)) = M M , = V ( a - a ) + ( b - b ) + ( c - 0 )
2

2

= |c |

2

Tuong tu d(M; (Oyz)) = | a | , d(M; (Ozx)) = | b |
Ta co d(M; Ox) = M M , = x/(a - a) + (b - 0) + (c - 0) =x/b +c
2

Tuong t u d(M; Oy) = Va + c
2

160

2

2

2

. d(M; Oz) = Va + b
2

2

2

2


V i du 15: Cho hai diem A ( x i ; y i ; z\) va B(x ; y ; z ). T i m toa do diem M
chia doan thang A B theo ti s6 k * 1.
Giai
Voi diem M(x; y; z) ta co: MA = (xi - x; y, - y; z\ - z),
2

2

2

MB = (x - x; y - y; z - z)
Diem M chia doan A B theo ti so k * 1 khi va chi khi
2

2

2

x = —l-k

yi - ^
MA =kMB<=> Yi ~ Y = k ( y - y) o
l-k
Zj - z = k(z - z)
Z =—
l-k
V i du 16: Trong khong gian Oxyz cho ba diem A ( l ; 2; 4), B(2; - 1 ; 0),
C(-2;3;-l).
a) Gpi (x; y; z) la cac toa dp cua diem M nam tren mat phang (ABC).
Tim su lien he giua x, y, z.
b) Tim toa dp cua diem D biet rang hinh ABCD la hinh binh hanh.
Giai
a) AB = (1; - 3 ; - 4 ) , AC (-3; 1; -5), A M = (x - 1; y - 2; z - 4)
Xj - x = k(x - x)
2

k

y

=

2

2

Ta co M nam uen mat phang (ABC) <=> [ A B , AC]. A M = 0
< > 19(x - 1) + 17(y - 2) - 8(z - 4) = 0 < > 19x + 17y - 8z - 21 = 0.
=
=

b) V i ABCD la hinh binh hanh nen AB = DC :
1 = -2 - x
D

-3 = 3 - y

» y =6

D

D

-4 = - 1 - z

Vay D(-3; 6; 3).

n

V i du 17: Cho A(2; l ; 3), B(4; 0; 1), C(-10; 5; 3)
a) Chung minh rang: A, B, C la ba dinh ciia mpt tam giac
b) Tim chan duong phan giac ngoai ciia goc B ciia tam giac ABC.
Giai
r

a) BA = ( - 2 ; - l ; 2 X B C = ( - 1 4 ; 5; 2)
Ta co BA va BC khong ciing phuong nen A . B, C la ba dinh ciia mpt
tam giac.
b) Gpi BE la ducmg phan giac ngoai ciia goc B, khi do:
BA EA
EA

3
1
BC " EC ^ EC ~ 15 " 5
V i 2 vecto E A , EC ciing huong nen E chia doan AC theo t i k = —
5
E(5;~;3)

161


V j du 18: Cho hinh hop ABCD.A'B'C'D* biet A ( x y ; z ), C(x ; y ; z ),
B'(x' ; y' ; z' ), D'(x' ; y' ; z' ). Tim toa do cua cac dinh con lai.
Giai
Goi Q = AC n BD, Q' = A ' C n B'D' thi Q,
Q' la trung diem cua AC, B'D' nen:
f i + . yi+y ,z + z
l
2 '
2 "* 2
i;

2

2

2

x

4

4

x

x

3

3

3

4

x

n

3

3

1

3

V

/x' +x

I
2
2

_ y + y , z +z 4
'
2
'
2

4

2

4

2

Tu AA*' = BB ' = CC"' = DTI' = QQ?, suy ra
( x + x ' + x ' " 3 yi + y ' + y 4 - y
2
2
J
I
2
f x ' + x - - HX _ y' + y - y i + y . z ' + z ' - z + z )
)
2
2
J

I
2
z +z +z — z
l 3
2~ 4 . yi + y + y '2-y'4
J
2
2
{
2
7

A
C
B

x

2

2

X

+

4

4

X

+

X

2

X l

3

2

3

4

2

3

X

X

4

1

3

:

3

3

2

4

( i+ -x' -f x
y -y2+y4 . l
3~ 2 4
2
2
I
2
V i du 19: Cho hinh hop ABCD.A'B'C'D' co cac diem A ( l ; 0; 1), B(2; 1; 2),
D ( l ; - 1 ; 1) va C'(4; 5; -5). Tim cac diem con lai.
Giai
Ta co ABCD la hinh binh hanh nen:
x - 2=0
x =2
y - 1 = - 1 » < y = 0 Do do: C(2; 0; 2)
BC = AD < >
=
x

x

D

z

4

2

3

+ z

z

+ z

3

c

c

c

c

z - 2=0
c

L

z

=2

c

Va A A ' = BB' = DD' = CC' = (2; 5; -7)
Nen A'(3; 5; -6), B'(4; 6; -5), D'(3; 4; -6).
V i du20:
a) Tim toa do diem M thuoc true Ox sao cho M each deu hai diem
A(l;2;3)vaB(-3;-3;2)'
b) Tren mat phang (Oxz) thi diem M each deu ba diem A ( l ; 1; 1),
B ( - l ; 1; 0), C(3; 1; -1).
Giai
a) M thuoc Ox nen M(x; 0; 0). Ta co MA = MB o MA = M B
o (1 - x) + 2 + 3 = (-3 - x) + (-3) + 2 o x - - 1 .
Vay M ( - l ; 0; 0)
b) M thuoc (Oxz) tren M(x; 0; z). Ta co: M A = M B = M C
2

2

2

(AM = B M
2

2

2

A M = CM
2

162

2

2

((x-l)

2

+ 1 + (Z-1)

2

2

2

2

= ( X + 1)

2

+l + z

2

(x - 1 ) +1 + (z - 1 ) = (x - 3) + 1 + (z + l )
2

2

2

2


»

{

4x + 2z = 1
4x-4z = 8

x=

VayNlf-jO;-6
6

!

o

7

- J

v

Vi du 21: Cho tam giac ABC co A(-2; 1; 0), B(0; 3; - 1 ) , C ( - l ; 0; 2).
a) Chung minh tam giac ABC co goc B nhon.
b) T i m toa do diem H la hinh chieu ciia A tren canh BC.
Giai
a) Taco BA = ( - 2 ; - 2 ; 1), BC = ( - l ; - 3 ; 3)
BA.BC
11
Nen cos B =
> 0 => goc B nhon.
BA . BC 3V19
b) H(x; y; z) thuoc BC nen B H = tBC
Do do x = - t , y - 3 = -3t, z + 1 = 3t
=> x = - t , y = 3 - 3t, z = - 1 + 3t
Ta co A H 1 BC nen A H BC = 0.
(-t + 2 ) ( - l ) + (-3t + 2)(-3) + (-1 + 3t).3 = 0

9

Vay hinh chieu H f - — ; - —; —
9
3 3
Vi du 22: Cho b6n diem A(-3; 5; 15), B(0; 0; 7), C(2; - 1 ; 4), D(4; - 3 ; 0).
Hoi hai duong thang A B va CD co cat nhau hay khong Neu chiing cat

nhau, hay tim toa do giao diem.
Giai
Ta co: AB = (3; - 5 ; - 8 ) , AC = (5; - 6 ; -11)
v

9

AD - (7; - 8 ; -15), CD = (2; - 2 ; -4)
Do do [ A B . A C ] = (7; - 7 ; 7) => [ A B

A C ] AD = 0 nen A B . CD dong

phang, hon nua A B , CD khong ciing phuong. do do 2 duong thang A B
va CD cat nhau.
Goi M ( X M ; >M, ZM) la giao diem ciia A B va CD.
Dat MA = k M B , MC = k'MD Taco:
k x
A
kx
2 - 4k
<>
=
l-k
1-k'
l-k'
i-k
5
- l + 3k
y - y
_y y

<>
=
i-k
i-k'
l-k'
l-k
15-7k
4
A ~ B_ y ~
<>
=
l-k
l-k'
l-k
l-k'
7
[—3 5
Giai ra duoc k' = — nen M — ; —; 11 I .
11
12 2
X

T

B

k

A

Z

k

B

c

k z

D

k

c

163


V i du 23:
a) Cho hai diem A(2; 5; 3), B(3; 7; 4). Tim diem C(x; y; 6) dk A . B, C
thang hang.
b) Cho hai diem A ( - l ; 6; 6), B(3; - 6 ; -2). Tim diem M thuoc mp(Oxy)
sao cho M A + M B nho nhat.
Giai
a) A, B, C thang hang <=> AC = k AB
2: k
5 = 2k
= 1 1 . Vay diem C(5; 11; 6).
k

= 3

b) V i ZA = 6. ZB = -2 => z . ZB < 0 => A, B 6 hai phia cua mp(Oxy).
Vay M A + M B nho nhat khi A, B, M thang hang
A

<=> A M , AB ciing phuong <=> [ A M , AB ] = 0
Gpi M(x; y; 0) e mp(Oxy)
•=> A M = (x + 1; y - 6; - 6 ) , AB = ( 4 ; - 1 2 ; - 8 )
Taco: [ A M , A B ] = (-8y - 24 ; 8x - 16; -12x - 4y + 12) = 0
-8y - 24 = 0
<=>i8x-16 = 0

<=>-T
y = -3

- 1 2 x - 4 y + 12 = 0
Vay M A + M B ngin nhat khi M(2; - 3 ; 0)
V i du 24: Cho t i i dien ABCD co: A(2; 1; - 1 ) , B(3; 0; 1), C(2; - 1 ; 3) va D
thuoc true Oy. Biet VABCD = 5. Tim toa dp dinh D.
Giai
Gpi D(0; y; 0) thuoc true Oy. Ta co:
l J

AB = (1; - 1 ; 2), AD = (-2; y - 1; 1), AC = (0; - 2 ; 4)
=> [ AB , A C ] = (0; - 4 ; -2)
=> [ A B , AC] AD = -4(y - 1) - 2 = - 4 y + 2.
Theo gia thiet VABCD = 5 <=> — I [ AB, AC ] AD |=5
6

< > | - 4y + 2 | = 30 < > y = - 7 ; y = 8
=
=
Vay co 2 diem D tren true Oy: (0; - 7 ; 0) va (0; 8; 0).
V i du 25: Gpi G la trpng tam ciia tu dien ABCD
Chimg minh rang duong thang di qua G va mpt dinh ciia tir dien ciing di
qua trpng tam ciia mat doi dien voi dinh do. Gpi A' la trpng tam tam giac
GA
3.
BCD. Chiing minh rang GA'
Giai
Ta giai bang phuong phap toa dp.
Trong khong gian toa dp Oxyz, gia sir A(x ; y ; zi), B(x ; y ; z ), C(x3; y , z ),
D(x ; y ; z ) thi trpng tam A' ciia tam giac BCD, trpng tam tii dien G:
x

4

164

4

4

x

2

2

2

3

3


x +x +x .y +y +y . z +z + 4
z

2

v

3

4

X, + X2 z
4

3

; 4

4

2

3

yi + y + y + y4. i +
4

x

3

A

2

z

2

z

3

2

+

+4

z

z

3

4

Do do:
cX_f

i

3 x

- x

2 - x -x .3yi - y - y - y 4 . i -z - ~ 4
4
4
4
3 z

3

GA'-f

+

x

2

4

2

2

3 + 4 . ~ y i + y +ys + y . 12
12
+

x

x

z

3

3

2

3 z

4

Suy ra: GA = -3 G A ' => G, A , A' thang hang va

z

3

i+

z

+3
12
z

2

GA
GA'

+ z

4

3.

Tuong tu thi co dpcm.
V i du 26: Cho tu dien noi tiep trong mat cau tam O va co A B = AC = A D .
Goi G la trong tam AACD, E, F la trung diem BG, A E .
Chung minh: OF _L BG O OD 1 AC.
Giai
AB = AC = A D va OB = OC = OD
=> OA 1 (BCD) tai chan duong cao H voi HB = HC = HD.
Chon H lam goc toa do, voi he true Hx, Hy, Hz sao cho H A la true Hz,
HB la true Hy, H D la true Hx.
A ( 0 ; 0 ; a ) , B(0; b; o j ; C ( c , ; c ; 0),

D(d; 0; 0) va O(0; 0; z) suy ra
( i + . 2 . a \ |'c +d b c a
{ 3
' 3 '3J'
I 6
'2
6 '6,
2

c

d

c

c

£

Cj + d b
12 ' 4
& ^
va OF _ f

c

1

|

2

c 7a
12'l2
2

+

i
12
+

.

d

c 7a
+ —;
4 12 12
b

2

z

BG = f ^ i - A - b ; i
3
3
3
AC = (c,; c ; -a), OD = (d; 0; -z)

Theo gia thiet OA = OB = OC = OD < > OA = OB = OC = OD
=
v

2

2

< > (a - z) = b + z = c +
=
2

2

2

o a - 2az = b = c + c = d
2

2

+z = d + z

2

2

2

2

2

2

2

2

2

(1)

2

Taco: OF.BG = 0 o(ci + d) + c - 9b + 7a - 12az = 0
2

2

2

2

(2)

Khai trien (2) va thay the (1) ta duoc:
(2) < > az + cid = 0 < > OD . AC = 0 : dpcm.
=
=

165


D A N G 2: G O C , K H O A N G C A C H , DI$N T I C K T j H | j l C l i
- Do dai cua vecto: u = (x, y, z) : I u | = \ / x + y + ^
Khoang each giua hai diem A ( x i , y i , zi) va B(X2, y , Z2):
2

2

2

AB = 7(x 2

) + ( y - y ) + (z 2

X l

2

2

x

2

Z l

)

2

- Goc giua hai vecto: u = (x,y,z) v a v = (x',y',z'):
.
,
x.x'+y.y'+z.z'
cos(u, v) =
x/x + y + z .yjx' + y ' + z'
2

2

2

2

2

2

- Goc cua tam giac A B C : cos A = cos(AB, AC)
Chii y: Goc giua 2 vecto tir 0° den 180° va cac goc con lai giira duong
thang, mat phang deu tir 0° den 90°
Dien tich va the tich:
Dien tich tam giac ABC: S = - | I [ A B , A C ] I
The tich tii dien ABCD: V = - | [AB, AC ]. AD |
6
The tich hinh hop A B C D . A ' B ' C ' D ' : V = | [ A B , AD ] . A A ' |
The tich hinh lang tru A B C . A ' B ' C : V = i | [ A B , A D ] . A A ' |

Z
V i du 1: Tinh cosin ciia goc giira hai vecto u va v trong moi trudng hop
sau:
a) u = ( 1 ; 1; 1); v = ( 2 ; 1;-1).
b) u = 3F + 4 f

; v =-2j

+3k.
Giai

a) cos(u, v) =

xx'+ y.y'+ z.z
^x

2
+

y

2

z Vx'
2

+

2
+

y'

_ V2
2
+

z'

2

3

b)Taco u =(3;4;0), v = (0; -2; 3) => cos(u , v ) =
V i du 2: Cho cac vecto: u = i - 2 j
w=2i

- k);

— k + 3j

a) T i m cosin ciia cac goc ( v

i ), ( v

b) Tinh cac tich vo hudng u v
a)

;v=3i+5(j

•8Vl3

j )va(v

k ).

u . w, v w
Giai

u = ( l ; - 2 ; 0 ) , v = ( 3 ; 5;-5), w = ( 2 ; 3 ; - l ) ^
Va cac vecto don vi X = (1; 0; 0), j = (0; 1; 0), k = (0; 0; 1) nen

166


v.i
cos(v, i ) = 7=

3

cos(v,k) =

cos(v, j ) =

759

1
v.k

M T

R

pj

59

-5
759

Iv 1. k

b) Ta co u . v = x.x' + y.y' + z.z' = - 7 .
Tuong tu thi duoc u w = - 4 , v w = 26.
Vi du 3: Cho hinh binh hanh ABCD voi A(-3; - 2 ; 0), B(3; -3; 1), C(5; 0; 2)
Tim toa do dinh D va tinh goc giua hai vecto AC va BD .
Giai
Ta co BA = (-6; 1; - 1 ) , BC = (2; 3; 1). V i toa do ciia hai vecto do
khong ti le nen ba diem A, B, C khong thang hang.
Gpi D(x; y; z). Tii giac ABCD la hinh binh hanh khi va chi khi
x +3=2
x = -1
AD = BC <=> y + 2 = 3 < > \y = l
VayD(-l;l;l)
=
|z

=1

1

Ta co AC = (8; 2; 2), BD = (-4; 4; 0), do do:
1
cos(AC, B D )
2
772.732
Vay ( A C , B D ) = 120°
Vi du 4: Cho vecto u tuy y khac 6. Chiing minh rang:
cos ( u , i ) + cos ( u , j ) + cos ( u , k ) = 1.
2

2

2

Giai
u.i

Gia sii u = (x; y; z) ta co: cos(u, i) =
Do do cos (u, i ) = • + y + z
x

v/x

2
+

y

2

+

z

2

2

2

2

Tuong tu: cos (u, j )

2

2

; cos (u,k)

2

x +y +z
2

2

2

x +y +z

2

2

2

Tir do suy ra dieu phai chimg minh.
Vi du 5: Cho tam giac ABC vuong 6 A biet A(4; 2; -1), B(3; 0; 2), C(x; - 2 ; 1)
a) Tim tam va ban kinh duong Uon ngoai tiep tam giac ABC.
b) Tim dp dai duong cao ciia tam giac ABC "e tir dinh A.
Giai
a) Tam giac ABC vuong tai A nen A B + AC = BC
Ma AB = 1 + 4 + 9 = 14, AC = (x - 4) + 16 + 4 = (x - 4) + 20
BC = (x - 3) + 4 + 1 = (x - 3) + 5 => x = 18 ^> C(18; - 2 ; 1)
Tam duong tron ngoai tiep I la trung diem ciia BC.
2

2

2

2

2

::

2

2

2

2

167


21
2

3
2

BC
/230
va ban kinh R = — = 2
2
b) Tam giac ABC vuong tai A , dudng cao A H nen
AH.BC = A B . A C
AB.AC
674830
=> A H =
=
BC
115
V i du 6: Cho ba diem A ( l ; 0; 0), B(0; 0; 1), C(2; 1; 1)
a) Chung minh A , B, C khong thang hang.
b) Tinh chu vi, dien tich va do dai duong cao ciia tam giac ABC ve tii dinh A.
c) Tinh cac goc cua tam giac ABC.

Giai
Nen I

;

;

A T T

a) Ta co BA = (1; 0; - 1 ) , BC = (2; 1; 0), toa do hai vecto do khong ti le
hen chiing khong ciing phuong. Vay ba diem A, B, C khong thang hang.
b) A B = V l + 0 + (-1) = 72
2

2

BC = V2 + l

2

2

2

+ 0 = 75
2

AC = 7 l + l + l = 73 .
Vay chu v i tam giac A B C bang 72 + N/3 + 75
V i B C = A B + A C nen tam giac A B C vuong tai A do do dien tich:

1
/R
S = -AB.AC = —
2
2
. 2S
30
Ta co S B C = - B C . h
BC
c) V i tam giac A B C vuong tai A nen:
AB 72
_ AC 73
COSD =
= —=• , C S C =
O
= —=
=
BC
75
BC 75
Cach khac: Tinh cosB theo cong thiic:
BC.BA _ 2.1 + 0 + 0 _ 72
cosB
BC.BA
N/5.72
"75
2

2

2

2

2

2

ABC

A

A

D

V i du 7: Trong khong gian toa do Oxyz, cho tam giac ABC co A ( l ; 2; -1),
B(2; - 1 ; 3), C(-4; 7; 5).
a) Tinh do dai duong cao h ciia tam giac ve tir dinh A.
b) Tinh do dai duong phan giac trong ciia tam giac ve tir dinh B.
Giai
a) Ta co AB = (1; - 3 ; 4) AC = (-5; 5; 6), BC = (-6; 8; 2)
A

[AB
Vay

AC] = (-38;-26;-10)

SABC

= - [AB, AC] I = - 738 + 26 + 10

2S

2

2

2

2N/555
/555
AC
B
BC
V104
726
b) Gpi D(x; y; z) la chan dudng phan giac ve tir B:

168.

/555


, DA
Ta co
DC

BA

BC

726

1

/104

V i D nam giua A, C nen DA = — DC
2
5

2
x = —
2(1 - x) = x + 4
3
11
O CD = 2DA < > \2 ( 2 - y ) = y - 7
=
o
T
2 ( - l - z) = z - 5
z=1
y

=

Vi du 8: Cho hinh binh hanh ABCD co 3 dinh A(3; 0; 4), B ( l ; 2; 3), C(9; 6; 4).
a) Tinh goc B cua tam giac ABC
b) Tinh dien tich hinh binh hanh ABCD.

Giai
a) Ta co BA = (2; - 2 ; 1), BC = (8; 4; 1)
Ta co cosB = cos( B A , BC) = -j=-.
75
b) Hinh binh hanh ABCD co dien tich: S =

2S BC
A

= 2.-

1
2

I [ B A , BC ] |

Taco [ B A , B C ] = (-6; 6; 24) => S = 18 72
Vi du 9: Cho b6n diem co toa do A(2; 5; -4), B ( l ; 6; 3), C(-4; - 1; 12),
D(-2; - 3 ; -2)
a) Chung minh ABCD la mpt hinh thang.
b) Tinh dien tich hinh thang ABCD.
Giai
a) AB = ( - 1 ; 1; 7), AC = (-6; - 6 ; 16), hai vecto nay khong ciing phuong v i
toa dp khong ti le suy ra A, B, C khong thang hang va co: A_
DC = (-2; 2; 14) = 2 AB => A B // CD.
Vay ABCD la hinh thang
")

SABCD = SABC + SADC

D"

= - | [ A B , A C ] | + - | [ A D , A C ] | = 371046
2
2
V i d u 10: Cho hai diem A(2; 0; -1), B(0; - 2 ; 3)
a) Tim toa dp diem C e Oy de tam giac ABC co dien tich bang T i l va
thoa man OC > 1.
b) Tim toa do diem D e (Oxz) &k ABCD la hinh thang co canh day A B .
Giai
a) Gpi C(0; y; 0) => AB = (-2; - 2 ; 4), AC = (-2; y; 1).
169


Ta co: S BC =

VU

A

O - I [ A B , AC] |

11 o - v / ( 2 + 4y) +36 + (2y + 4r = V l l
2

20y + 32y + 12 = 0 <=> y = -1 hoac y = -— (loai)
2

VayC(0;-l;0)
b) Goi D(x; 0; z) e (Oxz) => DC = (-x; - 1 ; -z)

A B C D la hinh thang khi va chi khi A B , DC cung huong
- 1 -z
.
<=> — = — = — < = > x = l , z = - 2 .
- 2 - 2
4
VayD(l;0;-2)
Vi du 11: Cho b6n diem A ( l ; 0; 0), B(0; 1; 0), C(0; 0; 1) va D(-2; 1; -2)
a) Chung minh rang A. B, C, D la bon dinh cua mot hinh tu dien.
b) Tinh goc giua cac duong thang chua cac canh doi cua tu dien do.
c) Tinh the tich tu dien ABCD va do dai duong cao cua tu dien ve tir
dinh A.
Giai
a) Taco: AB = ( - 1 ; 1;0), AC = ( - l ; 0 ; 1), AD = ( - 3 ; l ; - 2 )
/ 1 0 0 -1 -1 1 \
nen [ A B . A C ]
0;i;i)
V0 1 1 - 1 - 1 0 /
_

x

D o a i [ A B , A C ] . AD = - 3 + 1 - 2 = - 4 * 0, suy ra ba vecto A B , AC, AD
kho lg dong phang. Vay A, B, C, D la bon dinh ciia mot hinh tir dien.
Taco CD = ( - 2 ; l ; - 3 ) , BC = ( 0 ; - l ; 1), BD = ( - 2 ; 0 ; - 2 ) .
Goi a, p, y lan luot la goc tao boi cac cap duong thang: A B va CD, AC
va BD, A D va BC thi ta co:
,
—. — . ,
2 + 1 •01 3N/7

cos a cos(AB,CD)
14
V2.N/14
cosp = |cos(AC,BD)|
cosy = |cos(AD,BC)| =

2 + 0-21

0

V2.V8
1-1-21

377

yfe.yfU

14

"A
A =
Th£ tich tii dien ABCD la V = -1[AB AC]B.AD |C ] . A D I = 6IL
J
I 3
Do dai duong cao ve tir dinh A la:
3V
3V
2^3
"i—^Zi
^BCD i | | [ B C , B D ] |

21
V i du 12: Cho tir dien ABCD co: A ( - l ; 2; 0), B(0; 1; 1), C(0; 3; 0), D(2; 1; 0)
a) Tinh dien tich tam giac ABC va t h i tich t i i dien A B C D .
b) T i m hinh chiiu ciia D len mat phang (ABC).
A

—

=

r

:


Giai
a) Taco AB = ( ! ; - ! ; 1), AC = ( 1 ; 1;0), AD = (3; - 1 ; 0)
TS
-r-x -, I
V6
n e n [ A B , A C ] = ( - 1 ; 1;2)
S c=-| l[AB,
1
A

C

A B

Va[AB, AC].AD = - 4 ^ > V

I [ A B . A C ] . AD | = 6
i
b) Goi H(x; y; z) la hinh chieu D tren mat phang (ABC) thi:
AH = (x + 1; y - 2; z ) , D H = ( x - 2 ; y - 1; z). Taco:
4
x =—
DH.AB = 0
x -y +z
3
5
DH.AC = 0
x +y = 3
<=> i
3
x - y 2z
AB,AC] A H = 0
4
z=—
3
4 5 4
Vay H
.3 3 3,
Vi du 13:Trong khong gian Oxyz, cho 4 diem A ( l ; 0; 3), B(-3; 1; 3), C(l; 5; 1)
A B

CD=

±

Y

=

va M(x; y; 0). Tim gia tri nho nhat T = 2 | M A | + |MB + MC|
Giai
Goi I la trung diem cua BC:
=> I ( - l ; 3 ; 2 ) = > M B + MC = 2MI =>T = 2(MA + M I )
z = 3 > 0 v a z i = 2 > 0 = > A v a I nam ve ciing 1 phia doi vdi mp(Oxy)
va M(x; y; 0) thuoc mp(Oxy) nen lay doi xirng I ( - l ; 3; 2) qua mp(Oxy)
A

thanh J ( - l ; 3; -2) => M I = MJ => T = 2(MA + MJ) > 2AJ - 2 738
Dau = xay ra khi M la giao diem ciia doan MJ vdi mp(Oxy) la
1 9
M
0
5 ' 5
Vay minT = 2738.
Vi du 14: Cho hinh lap phuong ABCD.A'B'C'D' co canh bang a. Goi I , J lan
luot la trung diem ciia A'D' va B'B.
a) Chiing minh rang IJ _L A C . Tinh do dai doan thang IJ.
b) Chung minh rang D'B 1 mp(A'C'D), mp(ACB').
c) Tinh goc giira hai dudng thang IJ va A'D.
Giai
a) Chon he toa do Axyz sao cho
A ( 0 ; 0 ; 0 ) , D(a; 0; 0), B(0; a; 0),
A'(0; 0;a). Ta co C (a; a; a),
B'(0; a; a), D'(a; 0; a) nen:
l ( | ; 0 ; a ) ; J(0; a; | )

171


Taco: LJ = ( 0 - - ; a - 0 ; ~2
-2
2
2
AC' = ( a - 0 ; a - 0 ; a - 0 ) = (a; a; a)
a
nen LJ. A C ' = - - .a + a.a - - .a = - a + a = 0.
2
=

(

; a ;

2

2

\f
\
r
V a y l J l A C . Doan IJ = 1— 1 + a2 + 1
I
2)
1
2

a

a

}

\2
1
2j

b) DS chung minh D'B 1 mp(A'C'D), ta chung minh
D ' B I A ' C . D ' B l A ' D o D ' B . A ' C = 0, D ' B . A ' D = 0
Taco D ' B = (-a; a;-a), A ' C = (a; a; 0); A ' D = (a; 0;-a)
Do do D ' B A ' C = 0, DTJ.AT) = 0
Tuong tu, ta cung chung minh duoc D'B 1 mp(ACB')
c)

A ' D = (a; 0; -a). Goi cp la goc giua hai dudng thang IJ va A ' D thi:

•— .a + a.O (-a)
coscp = cos(JJ, A ' D )

IJ.A'D

2

IJ.A'D

176

2
iV2

Vay c = 90°
p
V i d u 15- Cho hinh lap phuong ABCD.A'B'C'D' canh bang a. Tren cac canh
BB', CD, A'D' lan luot lay cac diem M , N , P sao cho B ' M = CN = D'P = ka
: 0 < k < 1)
a) Tinh dien tich tam giac MNP theo k va a.
b) Xac dinh v i tri M tren BB' de dien tich tam giac MNP co gia tri be
nhat.
Giai:
Chon he true toa do Axyz nhu hinh ve
A(0; 0; 0), B(a; 0; 0)
C(a; a; 0), D(0; a; 0)
A'(0; 0; a), B'(a; 0; a)
C(a; a; a), D'(0; a; a)
a) ECM = k B T l => M(a ; 0; a - k a )
CN = kCD

=> N(a - ka; a ; 0)

D ' P = k D ' A ' = > P ( 0 ; a - k a ; a)
=> M N = (-ka; a; -a + ka), MP = (-a; a - ka; ka) nen:
[ MN

MP ] = ( k a - ka + a ; k a - ka + a ; k a - ka + a )
2

2

2

2

2

2

SMNP=i |[MN, MP] | =^(k -k+l)vdike (0; 1)
z
z
2

172

2

2

2

2

2

2


b) Taco: k - k + 1 = ( k - - ) + - > |

2
4 4
2

2

Dau = khi k = - e (0; 1) nen SMNP be nhat khi M la trung diem BB'.
2
V i d u 16: Cho hinh lap phuong ABCD.A,B,C,Di canh a, tren BC, lay diem
M sao cho DTM, DA^, AB[ dong phang. Tinh dien tich S cua A M A B j .
Giai
Chon he Oxyz sao cho B = O, B,(a; 0; 0), C,(a; a; 0), C(0; a; 0), A(0; 0; a),
A i ( a ; 0 ; a ) , D i ( a ; a; a), D(0; a; a).
Vi M

G

D
/ i
/ i
/ ' il
/
i

BC| nen goi M(x; x; 0)

Ta co L \ M = (x - a; x - a; -a)
DA = (a; -a; 0)
X

A B = (a;0;-a)
V i D j M , DA , ABj dong phang nen
X

3a
[ D ^ D A ^ A B j = 0 => x = —2
3a

nen MA
MB =

A,

c>X. _ _
/
»
_—
\—
/
\
B

X

^ --(3a 3a
Do do M 1 2 , 2 0
—

D,

Ci
X

3a

22

:

a Vl9
2

Vay: S

[MA, M B ]

V i du 17: Lang tru t i i giac d i u ABCD.A1B1C1D1 co chieu cao bang nua canh
day
Diem M thay doi tren canh A B . Tim gia tri Ion nhat ciia goc
ATMC

Giai

D

A

Chon he true nhu hinh ve (Aixyz)
Dat A M = x, 0 < x < 2 .
Ta co: M(x; 0; 1 ) ; Ai(0; 0; 0); Ci(2; 2; 0)

\

/

\
\

\

\

nen M A i = (-x; 0; - 1 ) ;
MCj = ( 2 - x ; 2 ; - l )

A '

\

Dat a = A^MCTi thi:
cosa = c o s ( M A i , M C )
X

(x-l)

x -2x + l
2

>/x +l.>/(2-x) +5
2

2

2

Vx +l.v/(2-x) +5
2

>0

2

173


Do do a < 90° Vay goc a =
diem A B .
V

i

A7MC

lan nhat khi x = 1 tuc M trung

I

P S.ABC co duong cao SA = h, day la tam giac ABC
vuong tai C, AC = b. BC = a. Goi M la trung diem ciia AC va N la diim

d

u

1 8 :

C

h

o

h i n h

c h 6

sao cho SN = - S B
3
a) Tinh do dai doan thang M N .
b) Tim su lien he giua a, b, h d l M N vuong goc vdi SB.
Giai
Ta chon he true toa do Oxyz co g6c O triing vdi A, tia Ox triing vdi tia
AC, tia Oz triing vdi tia AS sao cho d i i m B nam trong goc xOy. Khi do:
A(0; 0; 0), C(b; 0; 0), B(b; a; 0), S(0; 0; h). M ( - ; 0; 0)
SB = (b; a; - h )
Goi N(x; y; z) thi SN = (x; y; z - h)
Tir dieu kien SN = - SB nen
3
_ b
a ,

-h
3'
3
~
T
b a 2h
3
3'3'li
X

=

a) Taco M N

Y

=

V

a

Z

h

=

b _ b a 2h l - f _ b . a 2h^
3

2 ' 3 ' TJ~l
6'3
b_
36

4h

2

5

- V b +4a +16h
6
b) M N vuong goc vdi SB khi va chi khi MN.SB = 0
-b
-2h
0 cs> 4h = 2a
6
3
3
2

2

»

d

u

1 9

2

2

2

V

2

2

: Cho tir dien SABC co SC = CA = A B = a V 2 , SC 1 (ABC), ta

giac ABC vuong tai A. Cac d i i m M e SA, N e BC sao cho:
A M = CN = t ( 0 < t < 2a).
a) Tinh do dai doan M N . Tim gia tri t d l M N ngan nhat.
b) K h i doan M N ngan nhat, chung minh M N la dudng vuong goc chung
ciia BC va SA.
Giai
a) Ta chpn he true Oxyz sao cho g6c toa dp O = A. True Ox chua AC, true

174

Oy chira A B va true Oz 1 (ABC). Khi do canh SC song song vdi true Oz
va ta co:
r'," J •- * $
A ( 0 ; 0 ; 0 ) B(0; aV2 ; 0), C(aV2 ; 0; 0), S(aV2 ; 0; aV2 )

M

t\/2 „ tj2]
22){22

(

r-

tJ2

t

t

tJ2

)

M N V2(a-t,;^;-^)
MN

2(a - 2 a t + t 2,
)
2

\/3t

2

4at + 2a
2a

2

2

2a<

Vay M N ngan nhat bang
b) Khi M N ngan nhat thi: M
^

2

f aV2 aV2

2a
3
aV2 \
va N

khi t
as/2

Q

2ax/2 aV2

aV2^

_
, |MN.SA = 0
ia co {
[MN.BC = 0
=> MN la duong vuong goc chung ciia SA va BC.
'nh chop tii giac deu S.ABCD canh day a, mat ben tao voi
day goc a. Tim tana dk SA vuong goc SC.
Vi du 20: Cho h

Giai
Chon he true Oxyz co O la tam day ABCD. tia Ox
chiia A, tia Oy chiia B. tia Oz chiia S. Ta co*V2 ^'aV2
; 0;0 , B 0:
;0
2
as/2

>
0;0 , D O ; - ^ ; o ]
2
J
)
I
a

va S 0 ; 0 ; - t a n a
2

aV2
.
lV2
nen SA
0;
— tan a
— — U — tan a , SB
— ,
2'2
2
2
aV2
a
SC =
U,
,
tan a
, U — tan a ,SD
2
2
v.2
2
Ta co S A ' . SC
a

175


2
2

2
tan a - 1 = 0
« SA.SC = 0 o - — + — tan a = 0 « —
2
4
2 2
< > tan a = 2 < > tan a = \J2
=
=
2

2

v

2

Vay neu tana = %/2 thi hai canh ben doi dien ciia hinh chop vuong goc
voi nhau.
D A N G 3: MAT C A u
Phirong t r i n h mat cau: (S) tam I(a, b, c) ban kinh R:
(x - a) + (y - b ) + (z - c) = R hay:
x + y + z + 2Ax + 2By + 2Cz + D = 0, A + B + C - D > 0
2

2

2

2

2

2

2

2

2

2

co tam I ( - A , - B , - C ) va ban kinh R = %/A + B + C - D
Chii y : - Tam duong tron I(x; y; z) ngoai tiep tam giac A B C trong khong gian:
f l A = IB = IC
[ I G (ABC)
2

2

2

Tam I(x; y; z) ciia mat cau ngoai tiep tii dien A B C D la diem each deu 4
IA = IB
dinh: I A = IB = IC = I D <=>

fA I = B I

IA = IC

IA = ID

o AI = CI

2

2

2

AI = DI
2

2

2

Ung dung toa dp trong khong gian de giai cac bai toan hinh khong gian
co dien, quan he song song, vuong goc, dp dar, goc , khoang each, v j tri
tuong doi,...
V i du 1: Tim toa dp tam va tinh ban kinh ciia moi mat cau sau day:
a) x + y + z - 8x + 2y + 1 = 0.
b) 3x + 3y + 3z + 6x - 3y + 15z - 2 = 0.
b) 9x + 9y + 9z - 6x + 18y + 1 = 0.
d) x + y + z - x + y - 2z + 100 = 0.
Giai
a) Taco: a = - 4 , b = l , c = 0, d = 1.
nen a + b + c - d = 16 + 1 - 1 = 16 > 0
Mat cau (S) co tam I(-a; - b ; - c ) nen 1(4; - 1 ; 0) va ban kinh
2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

R = V a + b + c - d = Vl6 = 4.

b) 3x + 3y + 3z + 6x - 3y + 15z - 2 = 0
2

2

2

2

2

2

< > x + y + z + 2x - y + 5z
=

= 0
3
1
5
_7V6
va co ban kinh R
Do do, mat cau co tam 11 -1;
2' 2
2

2

c) 9x + 9y + 9z

2

2

176'

2

2

6x + 18y + 1 = 0


<> x + y + z
=
2

2

2

3

x + 2y+ - = 0
^ 9

Do do mat cau co tam I(—; -1; 0) va co ban kinh R= 1.
d) Taco: a = --,b = -,c = -l,d = 100
'
2

2
nen a + b + c - d = - + - + 1 - 100 < 0
4
4
Vay do khong la phuong trinh mat cau.
V i du 2: Lap phuong trinh mat cau trong cac truong hop sau:
a) Co tam 1(5; - 3 ; 7) va co ban kinh R = 2.
b) Co tam 1(4; - 4 ; 2) va di qua goc toa do
c) Di qua diem A(2; - 1 ; - 3 ) va co tam 1(3; - 2 ; 1).
Giai
a) Phuong trinh mat cau tam I(a; b; c), ban kinh R:
(x - a) + (y - b) + (z - c) = R
nen phuong trinh mat cau can tim la:
(x - 5) + (y + 3) + (z - 7) = 4.
b) Mat cau tam I di qua goc toa do nen co ban kinh
2

2

2

2

2

2

2

2

2

2

R = 10 = \/l6 + 16 + 4 = 6.
Vay phuong trinh mat cau: (x - 4) + (y + 4) + (z - 2) = 36.
2

2

2

c) Ban kinh R = I A = 3 V2 nen co phuong trinh:
(x - 3) + (y + 2) + (z - l ) = 18.
V i du 3: Lap phuong trinh mat cau:
a) Nhan doan A B lam duong kinh voi A(6; 2; -5), B(-4; 0; 7).
b) Di qua ba diem A(0; 8; 0), B(4; 6; 2), C(0; 12; 4) va co tam nam tren
mp(Oyz).
Giai
a) Mat cau nhan A B lam duong kinh nen tam I la trung diem cua A B , do do
AR 1
1(1; 1; 1) va co ban kinh R =
= - V l 0 0 + 4 + 144 = V62
2
2
Phuong trinh mat cau: (x - l ) + (y - l ) + (z - l ) = 62.
hay: x + y + z - 2x - 2y - 2z - 59 = 0
b) Tam I ciia mat cau nam tren mp(Oyz) nen 1(0; b; c).
Ta co I A = IB = IC nen:

2

2

2

2

2

2

JlA = I B
2

2

f(8-b) + c = 4 + ( 6 - b ) + ( 2 - c )

2

2

[IA = IC °
2

2

2

2

2

2

2

[(8 - b) + c = (12 - b) + (4 - c)
2

2

2

Giai ra duoc b = 7 va c = 5.
Vay 1(0; 7; 5), R = I A = VO + 1 + 25 = V26
Mat cau co phuong trinh: x + (y - 7) +(z - 5)
2

2

2

2

2

= 26.
177

Y j du 4: Cho A(0; - 2 ; 1), B ( - l ; 0; 1), C(0; 9; -1). Lap phuong trinh mat cau
co duong tron Ion la duong tron ngoai tiep tam giac A B C .
Giai:
Goi I(x; y; z) la tam duong tron ngoai tiep tam giac A B C .
Ta cc^AB_= ( - 1 ; 2; 0), AC = (0; 2; -2), A I =(x; y+ 2; z - 1)
=> [ AB , AC ] = (-4^-2; -2)
nen I e(ABC) « [ A B , A C ] A l = 0 o 2 x + y + z + l = 0
_1
AI = BI
2

Ta co

AI = CI
2

2

2

I e (ABC)

2x + y + z = - 1

1
z=—
6
1

5 1
5^3
nen tam 11 — ; — ; — I va ban kinh R = A I
6
6
6 6,
75
Vay PT mat cau la
T; + y + 7
+
~36'
du 5: Lap phuong trinh mat cau:
a) Co ban kinh bang 2, tiep xuc voi mat phang (Oyz) va co tam nam tren tia Ox.
b) Co tam 1(1; 2; 3) va tiep xuc voi mp(Oyz).
Giai
V i tam I ciia mat cau nam tren tia Ox va mat cau tiep xiic voi mp(Oyz)
nen diem tiep xiic phai la O, do do ban kinh mat cau la:
R = IO = 2 v a I ( 2 ; 0 ; 0 ) .
Mat cau co phuong trinh: (x - 2) + y + z = 4.
V i mat cau co tam 1(1; 2; 3) va tiep xiic voi mp(Oyz) nen ban kinh R ciia
mat cau bang khoang each tir I toi mp(Oyz), vay R = I x j = 1.
Mat cau co phuong trinh: (x - l ) + (y - 2) + (z - 3) = 1
du 6: Cho A ( l ; 2; - 4 ) , B ( l ; - 3 ; 1), C(2; 2; 3), D ( l ; 0; 4). Lap phuong
trinh mat cau ngoai tiep tii dien ABCD.
Giai
Goi I(a ; b; c) la tam mat cau ngoai tiep tii dien
_6_
~ IT
AI = BI
-y + z = -1

riA = IB
i_
I A = IC
• A I = C I » < x + 7z = -2 a~~ TT
j - 4z = 1
IA - I D
AI = Dl
X

Vi

a)

z

+

2

b)

6
5
6

-2x + 4y = -3
o y - z = -1
< > -j
=

2

2

T

2

Vi

2

2

2

2

2

2

TT
6_ 1_
vaR = IA=V26 .
• I ' l l ' 11
Vay (S): (x + 2) + (y - l ) + z = 26.

Do doi

2

178

2

2

2

2