Hình giải tích ôn thi đại học

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1. BÀI TOÁN 1:
“Lập phương trình đường thẳng đi qua A,
vuông góc với đường thẳng (d1) và cắt (d2)”
Cách 1:
Bước 1: Chuyển phương trình về dạng tham số.
uuur
-Giả sử (d) là đường thẳng cần dựng và cắt (d2) tại B, khi đó B(...) ⇒ AB ( ...) .
ur
ur
-Gọi a1 là vtcp của (d1), ta có a1 ( ...) .
Bước 2:
uuur
ur
uuur ur
uuur
Vì (d) (d1) nên : AB a1 ⇔ AB.a1 = 0 (nhớ tích vô hướng) ⇒ AB ( ...)
Bước 3: Phương trình đường thẳng (d) được cho bởi:
 x = ...
qua A ( ...)
( d ) :  uuur ⇔ ( d ) :  y = ..., t ∈ R.
vtcp AB ( ...)
 z = ...


Cách 2:
-Giả sử (d) là đường thẳng cần dựng, khi đó (d) chính là giao tuyến của hai mặt
phẳng (P1) và (P2), trong đó:
qua A ( ...)

qua A ( ...)
P
:
(
)
và 2 
( P1 ) ( d1 )
( d 2 ) ∈ ( P2 )

( P1 ) : 

* Phương trình mặt phẳng (P1)
qua A ( ...)

( P1 ) : 

( P1 )

( d1 )

qua A ( ...)
⇔ ( P1 ) : 
⇔ ( P1 ) :....
r ur
vtptn = a1 ( ...)

* Phương trình mặt phẳng (P2) (mặt phẳng đi qua một điểm và chứa một đường
thẳng)
Viết phương trình mặt phẳng (P2) bằng 2 cách:
Cách 1: Chuyển phương trình (d2) về dạng tổng quát, sau đó sử dụng chùm mặt

phẳng.
uuuu
r
Cách 2: Chọn điểm M(...) tùy ý thuộc (d2) ⇒ AM ( ...)
qua
r
 uu
qua A ( ...)
⇔ ( P2 ) : n2
( P2 ) : 
r
( d 2 ) ∈ ( P2 )
 uu
n
 2
qua A ( ...)
( P2 ) :  uur ⇔ ( P2 ) :...
vtptn2

A ( ...)
uu
r uuuu
r uu
r
uuuu
r
AM ⇔ n2 =  AM .a2  = ....
uu
r
a2

 ptmp ( P1 )
 ptmp ( P2 )

Kết luận: Phương trình giao tuyến (d) của (P1) và (P2) có dạng: ( d ) : 
2. BÀI TOÁN 2:
“Lập phương trình đường thẳng đi qua A,
cắt hai đường thẳng (d1) và (d2)”

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Bước 1:
Cách 1: Sử dụng pp chùm mặt phẳng :
-Gọi (P) là mặt phẳng qua A chứa (d 1), ta có (P) thuộc chùm tạo bởi (d1), có
dạng :
(P) : m(pt(1) của (d1)) + n(pt2 của (d1)) = 0
⇔ ( P ) :......

Cách 2: Chọn điểm M(...) tùy ý thuộc (d1)

uuuu
r
⇒ AM ( ...)

qua A ( ...)
r
r
uuuu
r ur

 r uuuu
qua A ( ...)
⇔ ( P2 ) : n AM ⇔ n =  AM .a1  = ....
( P) : 
( d 2 ) ∈ ( P )
 r ur
n a1
qua A ( ...)
⇔ ( P ) : ...
( P ) :  r
vtptn


Bước 2:
Gọi B là giao điểm của (P) và (d2). Khi đó tọa độ của B là nghiệm của hệ:
 pt1 of ( d 2 )
x =


 pt2 of ( d 2 ) ⇒  y = ⇒ B (...)

z =

 pt ( P )

Chú ý: nếu không tồn tại B. Kết luận bài toán vô nghiệm
Nếu có vô số nghiệm. Kết luận bài toán có vô số nghiệm đó chính là chùm
đường thẳng chứa (d) đi qua A.
Bước 3:
Gọi (d) là đường thẳng qua A, B, ta có:

 x = ...
qua A ( ....)

⇔ ( d ) :  y = ..., t ∈ R.
( d ) :  uuur
vtcp AB ( ....)
 z = ...

ur
ur
Gọi a1 là vtcp của (d1), ta có a1 ( ...)
ur
uuu
r
Từ đó, dễ thấy a1 không cùng phương với AB.

Vậy, (d): ... là đường thẳng cần dựng.
3. BÀI TOÁN 3:
“Lập phương trình đường thẳng (d1) qua A,
vuông góc với (d) và nằm trong mặt phẳng (P)”
Bước 1:
- Kiểm tra (d) có cắt (P) tại A không.
- Lập phương trình mặt phẳng (Q) thỏa mãn:
qua A ( ....)
qua A ( ....)
⇔ ( Q) 
uu
r
( Q ) : 
( Q ) ( d )

vtptad ( ....)

Bước 2: Khi đó đường thẳng (d1) chính là giao tuyến của (P) và (Q).
4. BÀI TOÁN 4:

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“Lập phương trình đường thẳng (d1) qua A,
vuông góc với (d) và cắt (d)”
Gọi (d1) là đường thẳng qua A vuông góc với (d) và cắt (d), vậy (d1) qua A và H
(H là hình chiếu vuông góc của A lên (d).
* Xác định H:
r
r
Gọi a là vtcp của (d), ta có a ( ...)
Chuyển phương trình (d) về dạng tham số:
Vì H ∈ ( d ) , nên H (theo t)
AH

uuur r

uuur
⇒ AH ( ...)

 x = ...
( d ) :  y = ..., t ∈ R.
 z = ...


( d ) ⇔ AH .a = 0 ⇔ ... ⇔ t = ... ⇒ H ( ...)

Phương trình (d1), được xác định bởi:
qua A ( ....)
⇔ ( d1 ) :...
uuur
vtcp AH ( ....)

( d1 ) : 

Dựng (P1) và (P2) thỏa mãn:

qua A ( ...)
qua A ( ...)
( P1 ) : 
và ( P2 ) : 
( P1 ) ( d )
( d ) ∈ ( P2 )
Khi đó ( d1 ) = ( P1 ) ∩ ( P2 )

5. BÀI TOÁN 5:
“Xác định tọa độ hình chiếu vuông góc của điểm A
lên mặt phẳng (P)
r

Mặt phẳng (P) có vtpt n ( ...)
Gọi (d) là đường thẳng qua A và vuông góc với (P), ta được:

qua A ( ....)


⇔ ( d ) :  ,t ∈ R
( d) : r

vtcpn ( ....)


Vì hình chiếu vuông góc H của A lên (P) chính là giao điểm của (d) và (P), do
đó:
thay các tọa độ của (d) vào (P) ⇔ t = ... ⇒ H ( ...)
6. BÀI TOÁN 6:
“Xác định tọa độ điểm A1 đối xứng với A qua mặt
phẳng (P)

Bước 1: Xác định tọa độ hình chiếu vuông góc H của A lên mặt phẳng (P).
Bước 2: Suy ra tọa độ điểm A1 từ điều kiện H là trung điểm của AA1.
7. BÀI TOÁN 7:
“Xác định tọa độ hình chiếu vuông góc của điểm A
lên đường thẳng (d)

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Cách 1:
r
r
Gọi a là vtcp của (d), ta có a ( ...)
Chuyển phương trình (d) về dạng tham số:
Vì
AH

H ∈( d ) ,

nên H (theo t)

uuur r

uuur
⇒ AH ( ...)

 x = ...
( d ) :  y = ..., t ∈ R.
 z = ...


( d ) ⇔ AH .a = 0 ⇔ ... ⇔ t = ... ⇒ H ( ...)

Cách 2:
r
r
Gọi a là vtcp của (d), ta có a ( ...)
Gọi H(x,y,z) là hình chiếu vuông góc của A lên đường thẳng (d), suy ra:
 H ∈ ( d )
 H ∈ ( d )
 H ∈ ( d )
⇔  uuur r ⇔  uuur r
⇒

 AH ( d )
AH a
 AH.a = 0

8. BÀI TOÁN 8:
“Xác định tọa độ điểm A1 đối xứng với A qua
đường thẳng (d)

Bước 1: Xác định tọa độ hình chiếu vuông góc H của A lên đường thẳng (d).
Bước 2: Suy ra tọa độ điểm A1 từ điều kiện H là trung điểm của AA1.
Bài 1: Cho (d1) là đường thẳng:

x +1 y −1 z − 3
x y −1 z − 3
=
=
=
và đường thẳng (d2): =
.
3
2
−2
1
1
2

Lập phương trình mặt phẳng chứa (d1) và (d2). ĐS: 6x-8y+z+11=0
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Bài 3: Tìm phương trình đường thẳng đi qua điểm (1; 2; -2) và song song với đường thẳng:

ĐS:

x −1 y − 2 z + 2
=
=
4
−7
−3

x + y − z + 2 = 0

2 x − y + 5 z − 1 = 0

−21
24

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Bài 5: Trong không gian Oxyz cho điểm A(2, -1, 1); và đường thẳng
y + z − 4 = 0
2 x − y − z + 2 = 0

( ∆) : 

a) Viết phương trình mặt phẳng ( α ) đi qua A và vuông góc với ( ∆ ) .

b) Xác định tọa độ điểm B đối xứng với A qua ( ∆ ) .
ĐS: ( α ) : y − z + 2 = 0 ; B(0; 3; 5)
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Bài 6: Trong không gian Oxyz cho điểm A(3, 2, 1); và đường thẳng:

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( d) :

x y
= = z +3
2 4

a)Viết phương trình mặt phẳng ( P ) đi qua A và chứa (d) .

b) Viết phương trình đường thẳng ( ∆ ) đi qua A, vuông góc với (d) và cắt (d).
14 x − 5 y − 8 z − 24 = 0
2 x + 4 y + z − 15 = 0

ĐS: (P): 14x-5y-8z-24=0; ( ∆ ) : 

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Bài 7: Viết phương trình đường thẳng ( ∆ ) đi qua M(1; 1; 2) và song song với đường thẳng:
3x − y + 2 z − 7 = 0
x + 3 y − 2z + 3 = 0

( d) :
ĐS: ( ∆ ) :

x −1 y −1 z − 2
=
=
−2
4
5

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Bài 8: Trong không gian Oxyz cho đường thẳng (d) và mặt phẳng (P) có phương trình:
 x = 1 + 2t
( d ) :  y = 2 − t , t ∈ R.
 z = 3t


( P ) : 2x − y − 2z +1 = 0

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a) Tìm tọa độ các điểm thuộc đường thẳng (d) sao cho khoảng cách từ mỗi điểm đó đến
mặt phẳng (P) bằng 1.
b) Gọi K là điểm đối xứng của điểm I(2; -1; 3) qua đường thẳng (d). Hãy xác định tọa độ
điểm K.
ĐS: M1(-3; 4; -6) và M2(9; -2; 12); K(4; 3; 3).
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Bài 9: Trong không gian Oxyz cho ba điểm A(1; 3; 2), B(1; 2; 1), C(1; 1; 3). Hãy viết
phương trình tham số của đường thẳng (d) đi qua trọng tâm tam giác ABC và vuông góc với
mặt phẳng chứa tam giác đó.
x = 1+ t

ĐS: ( d ) :  y = 2 , t ∈ R.
z = 2


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Bài 11: Trong không gian Oxyz cho mặt phẳng (P): 3x+6y-z-2=0, và đường thẳng:
 x + y − 7 z − 14 = 0
x − y − z − 2 = 0

( d) :

a) Tìm tọa độ giao điểm A của (P) và (d).
b) Tìm phương trình mặt phẳng ( β ) qua B(1; 2; -1) và vuông góc với (d).
ĐS: A(0; 0; -2); ( β ) : 4x+3y+z-9=0
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8

ĐS: A1  ; ; ÷; A2  ; − ; − ÷; N ( 0; −2;1)
5 5 5
5 5 5
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x − 2z = 0
và vuông góc với
3 x − 2 y + z − 3 = 0

Bài 13: Lập phương trình mặt phẳng chứa đường thẳng: 

mặt phẳng: x – 2y + z + 5 = 0.
ĐS: ( β ) : 11x – 2y -15z – 3 = 0

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x +1 y −1 z − 3
x y −1 z − 3
=
=
=
Bài 14: Cho đường thẳng ( d1 ) :
và đường thẳng ( d 2 ) : =
. Tìm

3
2
−2
1
1
2

tọa độ giao điểm của (d1) và (d2).
ĐS: A(2; 3; 1)
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Bài 15: Chứng minh rằng hai đường thẳng sau đây cắt nhau:
 x = 2t
 x = t '+ 5

( d1 ) :  y = 3t − 2, t ∈ R; ( d 2 ) :  y = −4t '− 1, t ' ∈ R
 z = 4t + 6
 z = t '+ 20



ĐS: M(3; 7; 18)
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Bài 16: Chứng tỏ rằng hai đường thẳng sau đây không cắt nhau và vuông góc nhau:

( d1 ) :

3x + y − 5 z + 1 = 0
x y −1 z
=
= ; ( d2 ) : 
1
−2
3
2 x + 3 y − 8 z + 1 = 0

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Bài 17: Viết phương trình chính tắc của đường thẳng (d) qua M(1; 5; 0) và cắt cả hai đường
2 x − z − 1 = 0
3x + y − 2 = 0
; ( d2 ) : 
x + y − 4 = 0
y − z − 2 = 0
x −1 y − 5 z
=

=
ĐS:
1
3
0

thẳng: ( d1 ) : 

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Bài 18: Viết phương trình chính tắc của đường thẳng (d) qua A(0; 1; 1), vuông góc với

x −1 y + 2 z
=
= và cắt đường thẳng Viết phương trình chính tắc của đường thẳng
3
1
1
x + y − z + 2 = 0
(d) qua M(1; 5; 0) và cắt cả hai đường thẳng: ( d 2 ) : 

x +1 = 0
x −1 y −1 z −1
=
=
ĐS:
1
−1
−2

thẳng: ( d1 ) :

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Bài
19: Lập phương trình đường thẳng đi qua điểm A(-1, 2, -3), vuông góc với vectơ
r
a = (6; −2; −3) và cắt đường thẳng (d):

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( d) :
ĐS:

x −1 y +1 z − 3
=
=
2
−3
6

x −1 y +1 z − 3
=
=
3
2
−5

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Bài 20: Cho 2 đường thẳng
2 x − 3 y − 2 = 0
2 x − 3 y + 9 = 0
; ( d2 ) : 
 x + 3z + 2 = 0
 y + 2z +1 = 0

( d1 ) : 

a) Chứng minh (d1)//(d2). Viết phương mặt phẳng chứa (d1) và (d2).
b) Tìm tọa độ điểm N đối xứng với M(-2; 3; -4) qua (d1)
ĐS: x + 4y + 11z +10 = 0; N(4; -3; 2).
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x +1 = 0

Lập phương trình đường thẳng qua A, vuông góc (d1) và cắt (d2).
ĐS:

x
y −1 z −1
=
=
−1
1
2

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Bài 22: Trong không gian cho 2 đường thẳng

( d1 ) :

x −7 y −3 z −9
x − 3 y −1 z −1
=
=
; ( d2 ) :
=
=
1
2
−1
−7
2
3

Chứng tỏ rằng hai đường thẳng đó chéo nhau.
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Bài 23: Cho đường thẳng xác định bởi phương trình:

x+2 y+2 z
=
=
và điểm M(4; -3; 2). Tìm tọa độ điểm N là hình chiếu vuông góc của điểm
3
2
−1

M lên đường thẳng đã cho.
ĐS: N(1; 0; -1)
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Bài 26: Trong không gian cho 2 đường thẳng
 x = 2t + 1
x = 2 + s

( d1 ) :  y = t + 2 và ( d 2 ) :  y = −3 + 2s
 z = 3t − 2
 z = 1 + 3s



a) Chứng tỏ rằng (d1) và (d2) chéo nhau.
b) Tính khoảng cách giữa (d1) và (d2).
ĐS:

8 3

3

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Bài 27: Trong không gian cho 2 đường thẳng

( d1 ) :

x −7 y −3 z −9
x − 3 y − 1 z −1
=
=
và ( d 2 ) :

=
=
1
2
−1
−7
2
3

a) Chứng tỏ rằng (d1) và (d2) chéo nhau.
b) Tính khoảng cách giữa (d1) và (d2).
ĐS: 2 21
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- 01689583116

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4
 1 11 8 
; A ; ; − ÷
3
3 3 3

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- 01689583116

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