Thiết kế bài giảng hình học 12 (tập 2) phần 2

Bieu thiic tpa dp ciia cac phep toan : cdng, trii vecto; nhan vecta vdi mdt so thirc.. KT nang • Thuc hien thanh thao cac phep toan ve vecto, tfnh dp dai vecto, • Vie't dupe phuang trinh

1 Khai niem tao do vecta trong khdng gian, tpa do diem va dp dai vecto

2 Bieu thiic tpa dp ciia cac phep toan : cdng, trii vecto; nhan vecta vdi mdt so thirc

3 Bieu thiic tpa dp ciia tfch vd hudng ciia hai vecta

4 Riuong trinh mat cau

2 KT nang

• Thuc hien thanh thao cac phep toan ve vecto, tfnh dp dai vecto,

• Vie't dupe phuang trinh mat ciu

3 Thai do

• Lien he dupe vdi nhieu van de thue te' trong khdng gian

• Cd nhieu sang tao trong hinh hpc

• Humg thii trong hpc tap, tich cue phat huy tfnh dpc lap trong hpc tap

II CHUAN DI CUA GV VA H&

1, C h u a n bi c u a G V :

• Hinh ve 3.1 den 3.3

• Thudc ke, phan mau,

2 Chuan bj cua HS :

III PHAN PHOI THOI LUONG

Bai duoc chia thanh 4 tie't:

Tie't 1

Tie't 2

Tiet 3

Tie't 4

Tii dau de'n het phan I

Tie'p theo den he't phan II

Tie'p theo de'n het phan III

Phan IV va hudng dan bai tap

IV TIEN TDINH DAY HOC

a DRT VA'N Di

Cau hdi 1

Nhic lai khai niem hinh hop, hinh chdp

Cau hdi 2

Cho hinh lap phuong ABCDA'B'CD'

a) Chiing minh cac canh ciia hinh lap phuong xuit phat tii mdt dinh vudng gdc vdi nhau

b) Cho canh ciia hinh lap phuong la a, tfnh dp dai dudng cheo ciia hinh lap phuong

n isni MOI

HOATDONCl

I TOA DO CUA DIEM VA CUA VECTO

1 He toa do

GV mo ta he true toa do trong khong gian va neu cau hoi :

HI Hai vecto i, j cd vudng gdc vdi nhau hay khdng?

H2 Vecto k cd vudng gdc vdi tat ca cac vecta thudc mat phing (Oxy) khdng?

• GV sir dung hinh 3.1 trong SGK va dat van de:

H3 Hay dpc ten cac mat phing tpa dp

H4 Hay ke' ten cac vecto dan vi

H5 Cd the ed them mdt gdc tpa dp niia khac O hay khdng?

H6 Hay neu cac tfnh chit ciia mat phing tpa dp, vecto don vi?

- 2 — - 2 — - 2

H7 Tfnh i = i.i, j = j j , k = k.k

H8 Tfnh i.j,j.k, k.i

• Thue hien ^ 1 trong 4 phiit

Su dung hinh ve 3.2 GV cho HS len bang ve lai hinh va hudng din HS thuc hien

Cdu hoi 3

Tim cac mdi quan he giiia cac

vecto ON OH va OK

Cdu hdi 4

Bieu dien OM theo i, j va k

Ggi y trd Idi cdu hoi 3

OK = x.i OH = y.j,ON =

Ggi y trd Idi cdu hdi 4

OM = x.i + y.j + z.k

= z.k

2 Toa dp cua diem

GV sir dung hinh 3.2 va dat cau hdi:

H9 Cho ba sd thuc x, y va z Cd bao nhieu diem M thda man OM = x.i + y.j + z.k HIO Cho OM = x.i + y.j + z.k Cd bao nhieu bd sd sd thuc x, y va z thda man he thiie tren

• GV tra Idi va neu dinh nghia :

Bd ba sd thuc (x; y z) thda mdn OM = x.i + y.j + z.k ggi la tga do diem

M vd ki hieu M (x ; y ; z) hoac M = (x ; y ; z)

HI 1 Cho M (0 ; 0 ; 0) Hay chi ra M

HI2 Cho M(0 ; 1 ; 2) Hdi M thudc true nao ?

HI3 Cho M(l ; 0 ; 2) Hdi M thudc true nao ?

H14 Cho M(l ; 2 ; 0) Hdi M thudc true nao ?

3 Toa dp vecto

• GV neu dinh nghia :

Trong khdng gian cho vecta a Bg ba sd (x ; y ; z) thda mdn

a = x.i + y.j + z.k ggi la tga do cua vecta a Ki hieu a(x;y;z) hoac

a = (x;y;z)

HI5 Vecta OM va diem M cd ciing tpa dp khdng?

• GV neu nhan xet trong SGK:

Tga do cua OM chinh Id tga do cua M

• Thue hien A2 trong 4 phiit

Su dung hinh ve 3.2 GV cho HS len bang ve lai hinh va hudng din HS thuc hien

HOATDQNC 2

II BIEU THtrC TOA D O C U A C A C P H E P T O A N V E C T O

• GV neu dinh If:

Trong khdng gian Oxyz cho ba vecta a(ai;a2;a2) vab(bi ;b2;b3) fa cd :

a + b = (aj +bi;a2 + b2;a3+b3)

a - b = (a, - b p a 2 - b 2 ; a 3 - b 3 )

ka = (kai;ka2;ka3), trong dd k la mgt sdthuc

• GV hudng din HS chiing minh dinh If tren

H16 Hay so sanh cac tpa dp ciia a va b khi a = b

• GV neu he qua 1:

Hai vecta bdng nhau thi cdc tga do tuang img bdng nhau

HI7 Hay viet cac bieu thiic tpa dp cua he qua 1

• Trong sach GK khdng cd vf du nhung GV nen lay vf du minh hpa cho dinh If va

he qua nay Vi day la kien thiic rat quan trpng

Vfdu.ChoA(l ;1 ; 1),B(-1 ; 2; 3) va C (0 ; 4 ;-2)

a) Hay tim cac tpa dp ciia AB va AC

b) Tim tpa dp ciia vec to 3AB

III TICH VO HUONG

1 Bieu thurc toa do cua tich vd hudng

• GV ndu dinh If

Trong khdng gian Oxyz cho ba vecta aia^;32;a2) vab(b, ;h2;h-^) ta cd:

a.b = ajb2 + a2b2 + a3b3

• GV hudng din HS chiing minh dinh If tren

2 tTng dung

a) Do ddi cua vectff

• GV neu cau hdi sau :

H21 Trong boat ddng 2, hay tfnh dp dai AC

GV neu dinh nghia :

Cho a(a| ;a2 ;a3) khi do do ddi ciia vecta aki hieu va:

3.1 ~r 3,T ~r 3.-5

b) Khodng cdch gida hai diem

H22 Cho A (XA ; yA ;ZA) va B (Xg ; y^ ;z^)

Xac dinh AB

Tfnh AB

GV neu ket qua 2:

Khodng cdch gida hai diem AB Id

AB = AB = 7(XB-XA)^+(yB-yA)^+(ZB-ZA)^

c) Gdc giita hai vectff

• GV neu cdng thiic tfnh gdc giiia hai vecto :

Cho cdc vecta iij = (x,; yj; Zj), M2 = (-^2 ! }'2 > ^2) ^^ ^^ ^ ^^y >*'

H24 Vecto to 0 vudng gdc vdi mpi vecto

• Thuc hien ^ 3 trong 4 phiit

IV PHUONG TRINH MAT CAU

• GV neu each chia mdt sd khd'i da dien va dat cau hdi:

H25 Tfnh khoang each giiia hai diem M(x ; y ; z) va I (a ; b ; e)

H26 Bie't khoang each dd la r, hay lap bieu thiie mdi quan he dd

• GV neu dinh If

Mat cdu tdm I(a ; b ; c), bdn kinh r cd phuang trinh

(x-a)^ +iy-b)'^ + iz-c)^ =r^

• GV hudng din HS chiing minh dinh If tren

• Thuc hien ^ 4 trong 4 phiit

H27 Hay neu mdt dang khac ciia phuong trinh mat ciu

• GV neu nhan xet:

Phuang trinh x + y^ + z + 2ax + 2b\ + 2cz + d = 0 la phuang trinh ciia mat cdu khi vd chi khi d^ + b^ + c' > d Khi dd tdm mat cdu la diem I(-a ; -b ; -c )vd bdn kinh mat cdu la

TOM T ^ Bfit HOC

1 Cho cac vecto u^ = ( x , ; y i ; z , ) , ^2 = (A^ ; y2 : '-2) va sd k tuy y, ta cd :

1) »i = fh <=> xi = X2, yi = y2 zi = ^2

2) », - U2 = (AI + x , : y, + y , ; ?i + -2)

3) ;ii - i?2 = (-^'i - ^'2; >'i - >''2; ^i - ^2)

4) kn^ = (A-A'i; ky^ ; fe,)

5) N|.U2 = Vj.vo + 3'iy2 + z,Z2

3 Cho hai diem Aix^ ; y^ ; z^) va Bixg; VB '

khi va chi khi a + b + c > d Khi dd tam mat cau la diem I ( - a ; - b ; - c ) va

ban kfnh mat cau la

•I? + b^+c-d

HOATDONC 6

MOT SO C^U HOI TR^C NQHim

Hay dien dung (D) sai (S) vao cac khing dinh sau :

Cdul Cho a = ( l ; 2 ; 3 ) , b = ( - 2 ; 3 ; - l ) Khi dd a + b cd toa dp la

(a) a + b cd toa dp la (-1 : 5 ; 2) Lj (b) a - b c d t o a d d l a ( 3 ; - l ;4) • (c) b - a c d t o a d d l a ( 3 ; - l :4) [ ]

(d) Ca ba khing dinh tren dSv; sai \_\

CAM 2 Cho a = ( l ; 2 ; 3 ) , b = ( - 2 ; 3 ; - l ) Khi dd a + b c d t o a d d i a

Cdu 4 Cho hinh cau cd phuong trinh : (x -1)^ + (y + 2f + (z + 3)^ = 2

(a) Tam ciia hinh cau la 1(1 ; -2 ; -3)

(b) Tam cua hinh cau la I(-l ; 2 ; 3)

(c) Ban kfnh ciia hinh cau la 2

(d) Ban kfnh ciia hinh cau la yf2

Chon khang djnh diing trong cac cau sau:

Cdu 5 Trong cac cap vecto sau, cap vecto ddi nhau la

Cdu 7 Cho hinh ve :

Cdu 9 Cho hinh ve

Tilr dd ta cd ket qua

Bai 2 Hirdng ddn Dua vao tfnh chat chit XQ - - ( x ^ + Xg + x^ );

Bai 4 Hudng ddn Dua vao tfnh chat ciia tfch vo hudng hai vecto

U\.U2 =x\X2+yiy2+h^2

a) a.b = 6

b) c.d = - 2 1

Bai 5 Hirdng ddn Dua vao phuong trinh mat cau

a) Phuang trinh mat cau dupe vie't dudi dang :

( x - 4 f + ( y - l ) 2 + z2=16

Tir dd ta ed tam va ban kfnh mat cau

b) Phuang trinh mat ciu dupe vie't dudi dang :

(x-1)^ + y + - 4

3

^2 :^2

z + - 19^

Tii dd ta cd tam va ban kfnh mat cau

Bai 6 Hudng ddn Dua vao phuong trinh mat cau

a) Xac dinh tam mat cau : I = (3;-1 ;5), ban kfnh mat ciu r = 3 Tii dd ta cd tam va ban kfnh mat ciu

( x - 3 f + ( y + l ) ' + ( z - 5 f = 9

b) Xac dinh tarfi mat cau : C - ( 3 ; - 3 ; l ) , ban kfnh mat ciu r = v5

Tii dd ta cd tam va ban kfnh mat ciu

( x - 3 f + ( y + 3 ) ^ ( z - l f = 5

1 Vecto phap tuye'n ciia mdt mat phing, cap vecto chi prfiuong ciia mat phing

2 Sir xac dinh mdt mat phing

3 Biet duoc phuong trinh tdng quat va phuong trinh tham so cua mat phang

4 Xac dinh dupe didu kien de hai mat piling song scHig va hai mat piling vudng gdc

• Lien he dupe vdi nhieu van de cd trong thuc te' ve mat phing trong khdng gian

• Cd nhieu sang tao trong hinh hpc

• Hiing thii trong hpc tap, tfch cue phat huy tfnh dpc lap trong hpc tap

n CHUAN BI CUA GV VA H6

1 C h u a n bi c u a G V :

• Hinh ve 3.4 den 3.8 trong SGK

• Thudc ke, phan mau,

• Chuan bi sin mdt vai hinh anh thuc te' trong trudng ve hai mat phang vudng gdc , hai mat phing song song

2 Chuan bj cua HS :

• Dpc bai trudc d nha, dn tap lai mdt sd kien thiic da hpc

• Chuan bi thudc ke, biit chi, biit mau de ve hinh

m DHAN PHOI T H 6 I LUONG

Bai nay chia thanh 5 tiet:

Tie't 1: tii diu den het dinh nghia phan I

Tie't 2 : tie'p theo den het muc 1 phin II

Tie't 3: tie'p theo den het phan II

Tie't 4 : tie'p theo den het muc 1 phan III

Tie't 5: tie'p theo den het phin IV

IV TIEN TDINH DAY HOC

n DRT VAN ff>€

Cau hdi 1

Cho hinh lap phuong ABCD.A'B'CD' cd A trung vdi gdc toa dp AB triing vdi Ox, AD triing vdi Oy, AA' triing vdi Oz

a) Tim toa dp tat ca cac dinh cua hinh vuong

b) Tim toa dp vecto AM vdi M la trung diem C C

Cau hdi 2

Neu mdt so tfnh chit co ban cua phep toan ve vecto

B Bni MOI

HOATDONCl

I VECTO PHAP TUYEN CUA MAT P H A N G

GV neu mot so eau hdi sau day:

HI cd bao nhieu dudng thing vudng gdc vdi mat phang

H2 Mot mat phing xac dinh khi nao ?

• GV neu dinh nghia :

Cho mat phdng (a) Ne'u vecta n khdc vecta 0 cd gid vudng gdc vdi mat phdng (a) ggi la vecta phdp tuyen cua mat phdng (a)

H3 Cho n la vecto phap tuyen ciia (a), hdi k n cd la vecto phap tuyen cua (a) khdng?

Hoat ddng ciia GV

Cdu hdi 1

De chiing minh n la vec to phap

tuyen ciia (a) ta can chiing minh

va'n de gi ?

Cdu hoi 2

Hay chiing minh nhan dinh tren

Hoat ddng cua HS

Ggi y trd Idi cdu hdi 1

Ta chiing minh n.a = n.b = 0

Ggi y trd Idi cdu hdi 2

HS tu chiing minh

GV neu nhan xet:

Vecta n thod mdn n.a = n.b - 0 ggi Id tich cd hudng cua hai vecta

HOATDONC 2

II PHUONG TRINH TONG QUAT CUA MAT P H A N G

• GV neu va cho HS giai bai toan 1 (Sii dung hinh 3.5)

• GV neu va cho HS giai bai toan 2 (Sir d ling hinh 3.5)

Hoat ddng cua GV r Hoat ddng cua HS

Cdu hdi I Ggi y trd Idi cdu hdi I

Hay chpn diem MQ thoa man

phuong trinh da cho

nhan n(A;B;C) lam vecto p\rd~i

tuye'n Chiing min-^ M ihup'? (a)

khi Ax + By + Cz +D = 0 1

1 Djnh nghia

• GV neu dinh nghia sau :

Ax + By + Cz + D = 0 trong dd A^ + B^ + C^ > 0 ggi la phuang trinh

tdng qudi ciia mat phdng (or)

H4 Tim vecto phap tuyen eiia mat phing : Ax +By + Cz + D-0

• GV neu nhan xet a) :

(a) cd phuang trinh Ax + By + Cz + D = 0 thi vecta phdp tuye'n cua (a)

la n = ( A ; B ; C )

H5 Lap phuang trinh mat phing di qua Mo(x,); yo; Z;,) va nhan n = ( A ; B ; C ) la vecto phap tuyen

• GV neu nhan xet b) :

Mat phdng (a) di qua diem MQ (XO,>'O,ZO) ^''^' "^^ vecta phdp tuye'n n (A ;

B Old A(x-Xo) +B(y-yo) + C(z-Zo) = 0

• Tliuc hien ^ 2 trong 5 phut

GV gpi ba HS len bang dien vecta phap tuye'n vao d trdng sau:

2 Cac trudng hop rieng

a) Mat phdng di qua gdc tog dg:Sit dung hinh 3.6

H6 Diem O (0 ; 0 ; 0) thudc mat phing (a) Tim D

• GV ke't luan :

Mat phdng (a) di qua gdc tog do O khi vd chi khi D = 0

b) Mol Irong cdc he sd: A, B hgc C bang 0 ; Sii dung hinh 3 7

H7 A = 0, mat phing (a) va Ox cd quan he nhu the' nao ?

c) Mat phdng (a) triing vdi mot trong cdc mat phdng tog do: Su dung hinh 3 8

H9 A = 0, B = 0 mat phing (a) va mp(Oxy) cd quan he nhu the' nao ?

• GV ke't luan :

Mat phdng (a) song song hoac triing vdi mat phdng (Oxy) khi vd chi khi

A = B = 0

HIO Phat bieu trong trudng hop : B = 0, C= 0 hoac C = 0, A = 0

• Thuc hien ^ 5 trong 5 phiit

GV neu tdng quat:

Mat phdng (a) song song vdi mat phdng tog do ndo dd khi vd chi khi he

sd tuang img cda cdc bie'n sd'bdng 0

• GV neu cac cau hdi

H l l Tii phucmg trinh : Ax + By + Cz +D = 0 <=>- + - + - = 1 bing each nao?

Cdu hoi 2

Xac dinh vecto phap tuye'n ciia (P)?

Cdu hoi 3

Hai vecto tren quan he nhu the' nao?

Ggi y trd Idi cdu hoi 2

ir-(2;-4;6)

Gffi y trd loi cdu hoi 3

Hai vecto tren cdng tuyd'n

• GV dat vin de :

Trong khdng gian toa dp Oxyz, cho hai mat phing (a) va (a') lin luprt cd

phuong trinh :

ia):Ax + By + Cz + D = 0 ia') :A'x + B'y + C'z + D' = Q;

chiing lan lupt cd eac vecto phap tuyen la «(A ; 5 ; C) va n\A ;B'; C)

Khi nao (a) va (a') song song ?

Khi nao (a) va (a') vudng gdc ?

1 Dieu kien de hai mat phang song song

• GV ke't luan : (a) // (a') c^ hai vecto phap mye'n ciia hai mat phing dd cdng tuye'n H14 Hay vie't bieu thiic toan hpc de hai mat phing (a) va ( a ' ) song song

• GV cd the neu each khac cua ke't luan tren cho de hieu hon:

Cho hai mat phdng (a) vd (a') Idn lu0 cd phuang trinh :

(a) :Ax + By + Cz + D = 0 (a') :A'x + B'y + C'z + D' = 0 a) Hai mat phdng dd song song khi vd chi khi

Hai mat phdng (a) vd (a') cdt nhau o(A ; B ; C) i^ k(A ; B'; C)

GV cd the neu each khac cho d i nhd:

Cho hai mat phdng (a) vd (a') Idn lu0 cd phuang trinh :

(a) :Ax + By + Cz + D^O (a') A'x + B'y + C'z + D' = 0 Hai mat phdng dd cdt nhau khi vd chi khi A B • C ^A': B': C

• GV neu vf du trong SGK va giai GV cd the neu vf du khac

• Sau day la vf du khac :

Cho hai mat phing (a) : x -my + 4z + m = 0

i^):x-2y + (m + 2)z-4 = 0

Hay tim gia tri ciia m de :

a) Hai mat phing dd song song

b) Hai mat phing dd trdng nhau

c) Hai mat phing dd cit nhau

1 -m 4 m

1 - 2 m + 2 -4

Khdng cd m

m ^ 2

2 Dieu kien de hai mat phang vudng gdc

GV sii dung hinh 3.12 va dat cac cau hdi:

H15 Nhan xet ve hai vecto Uj va n2

• GV neu dieu kien :

(a,) ± ( 0 2 ) 0 n,'.n^ = O o A A ' + B B ' + C C = 0

• GV neu vf du trong SGK va giai :

Cdu hoi 1

Mat phing da cho cd cap vecto

chi phuong nao ?

Cdu hdi 2

Xac dinh vecto phap tuye'n ciia

mat phing cin lap

Cdu hdi 3

Xac dinh mat phing can lap

Gffi y trd Idi cdu hdi 1

AB va n

Ggi y trd Idi cdu hoi 2

Mat phing cd vecto phap tuyen la

n = AB, n = (-I;13;5)

Ggi y trd loi cdu hdi 3

X - 13y-5z+5 = 0

HOATDONC 4

IV KHOANG CACH TtTMOT D I £ M DEN M O T MAT P H A N G

• GV neu dinh If:

Trong khdng gian Oxyz cho mat phdng (a) cd phuang trinh :

• De chiing minh dinh If tren, GV can dua ra cac budc sau

Gpi M,(x, ; y, ; z,) la hinh chieu ciia M,, tren (a)

Tfnh dd dai MJMQ.U

Tfnh dp dai : M,Mo

» Thuc hien vf du 1 trong 4'

• Thuc hien vf du 2 trong 4'

Ggi y trd Idi cdu hoi 1

GV cho HS chpn diem bat ki

Gffi y trd Idi cdu hdi 2

Hai mat phing nay cdng song

song vdi mat phing nao?

1 Vecto n^O gpi la vecta phdp tuye'n ciia mat phing ( a ) neu gia eiia vecta n

vudng gdc vdi mat phing (or)

2 Trong khdng gian Oxyz, cho mat phing id) di qua diem A/Q(XQ,>^0'^O) ^^ ^^ vecto phap tuyen niA; B ;C):

ia) : Aix - XQ) + Biy- yo) + Ciz - Zg) = 0

Dat D = -iAxQ + ByQ + CZQ) thi phuong trinh ciia mat phing (or) dupe viet dudi dang :Ax + fiy + Cz + D = 0 trong dd A^ + B^ + C^ > 0

3 Cac trudng hop rieng:

Phuang trinh cua (a)

b) Hai mat phing dd song song khi va chi khi

HOATDONC 5

MQT SO C6U HOI TRAC NGHI|M

Cdu 1 Hay dien diing, sai vao cac d trdng sau day:

(a) Mat phing x + 3 y - z + 2 = 0 c d vecto phap tuyen la (1 ; 3 ; -1)

(b) Mat phing x + 3 y - z + 2 = 0 c d vecto phap tuyen la (1 ; 3 ; 2)

(c) Mat phing x 3 y - z + 2 = 0 c d vecto phap tuyen la (1 ; -3 ; -1)

(d) Mat phing -x + 3y - z + 2 == 0 cd vecto phap tuye'n la (-1 ; 3 ; -1)

Cdu 2 Hay dien diing, sai vao cac d trdng sau day:

(a) Mat phing 3y - z + 2 = 0 cd vecto phap tuyen la (0 ; 3 ; -1)

(b) Mat phing x + 3y + 2 = 0 cd vecto phap tuyd'n la (1 ; 3 ; 2)

(c) Mat phing x - z + 2 = 0 c d vecto phap tuyen la (1 ; 0 ; -1)

(d) Mat phang - x + 3 y - z =:0cd vecto phap tuyen la (-1 ; 3 ; -1)

Hay dien dung, sai vao cac d trdng sau day:

(a) Mat phing ABCD cd phuang trinh la z = 0

(b) Mat phing BB'CC cd phuong trinh la x = 1

(c) Mat phing A'B'C'D' cd phuong trmh la z = 1

(d) Mat phing CCD'D cd phuang trinh la y = 1

Phuong trinh ciia (a) Dgc diem ciia (a)

(a) song song hoac chiia Ox (a) song song hoac chira Oz (a) song song hoac chda Oy (a) song song hoac triing Oxy (a) song \i.ng hoac triing Oxz (a) song song hoac triing Oyz

Chgn cdu trd Idi dung trong cdc bdi tap sau:

Cdu 5 Cho hinh ve Hinh lap phuang ABCD.A'B'CD' cd canh 1

Mat phing (A'BD cd phuang trinh nao sau:

Cdu 7 Cho mat phing cd phuang trinh (P) : x + 2y + 3z - 1 = 0

Mat phing nao sau day song song vdi (P)

(a) 2x +4y + 6x 1 = 0 ; (b) 2x +4y - 6z -2 = 0;

(c) 2x - 4y + 6z -2 = 0; (d) - 2x +4y + 6z -2 = 0

Trd Idi (a)

Cdu 8 Cho mat phing cd phuong trinh (P): x + 2y + 3z - 1 = 0

Mat phing nao sau day triing vdi (P)

(a) 2x +4y + 6z -2 = 0; (b) 2x +4y - 6z -2 = 0;

(c) 2x - 4y + 6z -2 = 0; (d) - 2x +4y + 6z -2 = 0

Trd Idi (a)

Cdu 9 Cho mat phing cd phuong trinh (P): x + 2y + 3z - 1 = 0

Mat phang nao sau day vudng gdc vdi (P)

(a) 2x +4y + 6x -2 = 0; (b) 2x +4y - 6x -2 = 0;

(c) 2x - 4y + 6x -2 = 0; (d) -3x + z -2 = 0

Trd Idi (d)

Cdu 10 Cho mat phing cd phuong trinh (P) : x + 2y + 3z - 1 = 0

Khoang each tilr M(l, 2, -1) den (P) la :

( a ) ^ ; ( b ) i - ; (c) ^ ; (d) ^

Vl4 14 6 7

Trd Idi (a)

HOATDONC 6

HaQNG DfiN GIfil Bfil TfiP SfiCH GIfiO KHOfi

Bai 1 Sii dung phuong trinh ciia mat phing

a) Hudng ddn Six dung cong thiic : Aix -XQ) + Biy -yo) + Ciz - ZQ) = 0

b) Hudng ddn Vecta phap tuye'n cua mat phang dd la : [u, vl = (2; 6; 6)

Su dung cdng thiie : Aix -XQ) + Biy -yo) + Ciz - ZQ) = 0 \

Bai 2 Sii dung tfnh chit cua trung diem va mat phang trung true

•Trung diem eiia AB la M = (3 ; 2 ; -5)

AB = ( 2 ; - 2 ; - 4 )

Ddp so (a) : X - y - 2z + 9 = 0

Bai 3 Sii dung cac trudng hpp rieng cua mat phang

a) Hudng ddn Six dung hoat ddng 4

Ddp sd mp(Oxy): z = 0, mp(Oxz): zy= 0, mp(Oyz): x = 0

b) mp(a) //(Oxy) nhan k(0;0;l) lam vecto phap tuye'n

Ddp sd ( a ) : z + 3 = 0

Tuong tu : (p) //(Oyz): x -2 = 0 ; (y) //(Oxz): x - 6 = 0

Bai 4 Su dung phuong trinh tdng quat cua mat phing

a) Hudng ddn mp(a) chiia true Ox va di qua P se nhan i va OP lira cap vecto

c) Hudng ddn mpiy) chiia true Oz va di qua R se nhan k va OR lam cap vecto

chi phuang

" Y = k,ORj = (4;3;0)

(y): 4x + 3y = 0

Bai 5 Sii dung phuang trinh tdng quat ciia mat phing

Nen cd hinh ve de HS trung binli de tudng tupng

Bai 6 Sii dung phuong trinh tdng quat ciia mat phing

Nen cd hinh ve de HS trung binh de tudng tupng

Hudng ddn mp(a) se nhan xxn = (2;-l;3) lam vecto phap tuye'n

( a ) : 2 x - y + 3 z - l l = 0

Bai 7 Su dung phuang trinh tdng quat ciia mat phing

Mat phing (a) can lap nhan: AB va nn n lam cap vecto chi phuong Do dd