Exámenes Selectividad UNED resueltos

8/12/2019 Exámenes Selectividad UNED resueltos

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A =1 2 30 −1 21 0 3

B = 1 35 −2

x

3 −3 x1 −x x + 1 −1

2 0 x

x

3 −3 x1

−x x + 1

−1

2 0 x= 3 x(x + 1) + 6

−2x(x + 1) + 3 x(1

−x) =

= 3x2 + 3 x + 6 −2x2 −2x + 3 x − 3x2 = −2x2 + 4 x + 6

−2x2 + 4 x + 6 > 0



−2x2 + 4 x + 6 = 0 =⇒x2 −2x −3 = 0

x = 2 ±√ 4 + 12

2 =

2 ±4

2

x = 3

x = −1

(−∞, −1) (−1, 3) (3, + ∞)

−2x2 + 4 x + 6 − + − (−1, 3)

X AX + B = X

A = 1 11 1

B = 1 −20 −1

X

AX + B = X =⇒AX −X = −B =⇒(A −I ) X = −B =⇒X = ( A −I )− 1

·(−B )

A − I

A −I = 1 11 1 −

1 00 1

= 0 11 0

(A −I )− 1 = x yz t

0 11 0 · x y

z t= 1 0

0 1=⇒(A −I )− 1 = 0 1

1 0

X = ( A −I )− 1 ·(−B ) = 0 11 0 · −1 2

0 1=

0 1

−1 2

X AX = B

A = 1 20 1

B = 1 23 4



A |A| = 1 = 0 X

A

A

·X =

·B =

⇒

A− 1

·A

·X = A− 1

·B =

⇒

X = A− 1

·B

x yz t ·

1 20 1

= 1 00 1

=⇒

x = 12x + y = 0 =⇒y = −2z = 02z + t = 1 =⇒t = 1

X = A− 1

·B =

1

−2

0 1 · 1 2

3 4 = −5

−6

3 4

a, b

A

2 2 b0 5 a3 1 b

Rg(A) ≥2

2 20 5

= 10 = 0 |A|

|A| =2 2 b0 5 a3 1 b

= 10 b + 6 a −15b−2a = −5b + 4 a

−5b + 4 a = 0 =

⇒

4a = 5b =

⇒

b = 4a

5 a = 5 b = 4 .

A, B

A ·B B ·A

A = 1 32 5

B = 2 31 1



A ·B = 1 32 5 ·

2 31 1

= 5 69 11

B ·A = 2 31 1 · 1 3

2 5= 8 21

3 8

X A ·X = X ·A

A = 1 11 1

X = a bc d

A ·X = 1 11 1

a bc d

= a + c b+ da + c b+ d

X ·A = a bc d

1 11 1

= a + b a + bc + d c + d

A ·X = X ·A

a + c b+ da + c b+ d

= a + b a + bc + d c + d

a + c = a + b =

⇒

c = b

b + d = a + b =⇒a = d

a + c = c + d =⇒a = d

b + d = c + d =⇒c = b

X

X = a bb a



X =0 1 11 0 11 1 0

X =1 3 2 0

−1 1 0 31 −1 3 + 1 2 + 0 0 + 3

=1 3 2 0

−1 1 0 30 4 2 3

A =1 2 33 1 24 3 5

A

A =1 2 33 1 2

4 3 0

=⇒ |A | =1 2 33 1 2

4 3 0

= 16 + 27 −12 −6 = 0

A =0 1 11 0 11 1 0

|A| = 2 = 0

A = 0 1 01 0 11 1 0

|A | = 1 = 0



A A2 = A + I I

A

B = A −I A

A ·B = A ·(A −I ) = A2 −A = A + I − A = I

B ·A = ( A −I ) ·A = A2 −A = A + I − A = I

B A A

A2 = A + I

A2 −A = I =⇒A(A −I ) = I

|A(A −I )| = |I | =⇒ |A| |A −I | = |I | = 1

|A| = 0 =⇒A

1 2 3 45 6 7 86 8 10 12

1 2 3 a2 4 6 83 6 9 12



F 3 = 32

F 2

1 a2 8

= 8 −2a = 0 =⇒a = 4

a = 4 =⇒RgA = 1

a = 4 =⇒RgA = 2

A

2A |2A| = 8

A

x |2A| = 8 A

A =x 1 1

x + 1 2 2x 2 −x 1

2A

|2A| = 2 3 · |A| |A| = 1

|A| = 1

x 1 1x + 1 2 2

x 2−x 1= 2 x + 2 x + ( x + 1)(2 −x) −2x −(x + 1) −2x(2 −x) =

= 2 x + 2 x + 2 x + 2 −x2 −x −2x −x −1 −4x + 2 x2 = x2 −2x + 1

|A| = x2 −2x + 1 = 1 =⇒x2 −2x = 0

x(x −2) = 0 =⇒x = 0x −2 = 0 =⇒x = 2



X A2X = A

A = 1 21 1

A2

A2 = A ·A = 1 21 1 ·

1 21 1

= 3 42 3

3 42 3 ·X =

1 21 1

X = 3 42 3

− 1

· 1 21 1

3 4

2 3 −→

3 2

4 3 −→

3 −2

−4 3 −→

3 −4

−2 3

A− 1 =

3 −4

−2 31

= 3 −4

−2 3

X = 3 −4

−2 3 · 1 21 1

= −1 21 −1

A =−1 2 bb b−3 −10 2 1

−1 2 bb b−3 −10 2 1

= −(b−3) + 2 b2 −2b−2 = −b + 3 + 2 b2 −2b−2 = 2 b2 −3b + 1 = 0



b = 3±√ 9 −8

4 =

3 ±14

=

3 + 14

= 1

3 −1

4 =

2

4 =

1

2

b = 1 , 12

b = 1

−1 2 11 −2 −10 2 1

b = 12

−1 2 1

212 −

52 −1

0 2 1

A =1 1 11 2 −12 1 1

A =1 1 11 2

−1

2 1 1

(2, 1, 1) = a(1, 1, 1) + b(1, 2, −1)



a + b = 2a + 2b = 1a

− b = 1

a + b = 2a − b = 1

=⇒2a = 3 =⇒a = 32

b = 2 − 32

= 12

32 + 22 = 52 = 1

A A |A| = 2

|2A|

A− 1

|A

·At

| At

A

A

A

2A A

A

|2A| = 2 3 · |A| = 8 ·2 = 16

|A ·B | = |A| · |B | A ·A− 1 = I

A

·A− 1 =

|I

|⇒A− 1

· |A

|= 1

⇒

A− 1 = 1

|A| =

1

2 |At | = |A|

A ·At = |A| · At = |A|2 = 4



−2

A

b

A =0 b b1 0 1b −2 0

|A| =0 b b1 0 1b −2 0

= b2 −2b = 0⇒b(b−2) = 0 ⇒b = 0b = 2

b = 0 , 2⇒RgA = 3

b = 0

A =0 0 01 0 10 −2 0

RgA = 2

1 00 −2

= −2 = 0

b = 2

A =0 2 21 0 1

2 −2 0

RgA = 2

0 21 0

= −2 = 0

A =1

−2

−1, B = 1 −2 2 , X =

xyz

, O =000

A ·B

A ·B ·X = O



A ·B A B

A ·B

A ·B =1

−2

−1· 1 −2 2 =

1 −2 2

−2 4 −4

−1 2 −2

A ·B ·X = 0⇒

1 −2 2

−2 4 −4−1 2 −2 ·

x

yz =

0

00

x − 2y + 2z = 0

−2x + 4y − 4z = 0

−x + 2y − 2z = 0

y z

x −2y + 2 z = 0y = αz = β

⇒

x = 2α − 2β y = αz = β

A =1 1 1a b ca2 b2 c2

A

|A| = ( b−a)(c −a)(c −b)

a b c A



|A| =1 1 1a b c

a2 b2 c2

F 2 = F 2 − aF 1

F 3 = F 3 − a 2 f 1

=1 1 10 b−a c −a

0 b2 −a2 c2 −a2

= 1 ·A11 =

=b−a c −a

b2 −a2 c2 −a2 = ( b−a) c2 −a2 −(c −a) b2 −a2

= ( b−a)(c −a)(c + a) −(c −a)(b−a)(b + a) =

= ( b−a)(c −a) [(c + a) −(b + a)] = ( b−a)(c −a)(c −b)

RgA = 1

a, b, c

A b

A =1 2 1

b + 1 1 11 b b−1

1 2 1b + 1 1 1

1 b b−1= b−1 + 2 + b(b + 1) −1 −2(b + 1)( b−1) −b =

= b−1 + 2 + b2 + b−1 −2b2 + 2 −b = −b2 + b + 2 = 0 =⇒

=⇒b = 2b = −1

b = −1, 2 =⇒RgA = 3

b = −1

A =1 2 10 1 11 −1 −2

1 20 1

= 1 = 0



b = 2

A =1 2 13 1 11 2 1

1 23 1

= 1 −6 = −5 = 0

A =−1 1 11 2 −1

−2 1 2

A

RgA 2 −1 1

1 2=

−3 = 0

|A| =−1 1 11 2 −1

−2 1 2= −4 + 2 + 1 + 4 −2 −1 = 0

RgA = 2

B C C

B

A =1 1 01 0 10 1 1

A− 1



1 1 01 0 10 1 1

= −1 −1 = −2 = 0

1 1 01 0 10 1 1

Menores

−−−−−−→−1 1 11 1 11 1 −1

Adjuntos

−−−−−−→−1 −1 1

−1 1 −11 −1 −1

T raspuesta

−−−−−−−→

T raspuesta

−−−−−−−→−1 −1 1

−1 1 −11 −1 −1

A− 1 = 1

−2−1 −1 1

−1 1 −11 −1 −1

=

12

12 −1

212 −1

212

−12

12

12

A a

A =0 1 2

a + 1 −1 a −2

−1 a + 1 2

|A| =0 1 2

a + 1 −1 a −2

−1 a + 1 2= −(a −2) + 2( a + 1) 2 −2 −2(a + 1) =

= −a + 2 + 2 a2 + 4 a + 2 −2 −2a −2 = 2 a2 + a = 0 =⇒

a = 0

a = −12

a = 0 , −12

=⇒RgA = 3

a = 0

A =0 1 21 −1 −2

−1 1 2



RgA = 2

0 11 −1

= −1 = 0

a = −12

A =

0 1 212 −1 −

52

−1 1

2 2

RgA = 2

0 112 −1

= −12 = 0

X = x 1y 0

X ·X t = 1 00 1

X t X

X

X t

=

x y

1 0

X ·X t = x 1y 0 ·

x y1 0

= x2 + 1 xy

xy y2

x2 + 1 = 1xy = 0y2 = 1

x = 0

y = ±1

X 1 = 0 11 0

y X 2 = 0 1

−1 0

A

A =1 2 −3 00 1 −1 1

−1 0 1 −1.



A

c3 = −c1 −c2

A A

A =1 2 −3 00 1 −1 1

−1 0 1 −1

RgA 2

1 20 1

= 1 = 0

A

c1 , c2 , c3 c1 , c2 , c4

1 2 00 1 1

−1 0 −1= −1 −2 = −3 = 0

RgA = 3



M

M =−1 0 33 1 10 2 1

−1 0 33 1 10 2 1

F 3 = F 3 − F 2

−−−−−−−→−1 0 33 1 1

−3 1 0

F 2 = F 2 − F 3

−−−−−−−→−1 0 36 0 1

−3 1 0

F 1 =3 F 1 − F 3

−−−−−−−−→0 −1 96 0 1

−3 1 0

M =0 −1 96 0 1

−3 1 0

x + y + z = 1x + y + z = 2

a

(a −3)x + 4z = 2x − 2z = −1

−x + ay + 2z = a

a −3 0 4 21 0 −2 −1

−1 a 2 a



a −3 0 41 0 −2

−1 a 2

= 4 a + 2 a(a −3) = 4 a + 2 a2 −6a = 2a2 −2a

2a2 −2a = 0 =⇒(2a −2) a = 0 =⇒ a = 0a = 1

a = 0 , 1 =⇒RgA = RgA = 3 = no =⇒

a = 0

−3 0 4 21 0 −2 −1

−1 0 2 0

a = 1

−2 0 4 21 0 −2 −1

−1 1 2 1

A A

2

1 0

−1 1= 1 = 0

RgA = 2 = RgA < 3 = no

a

ax − ay + az = a(3 −2a)z = 1

x + ( a −1)y = 0



A =a −a a a0 0 3−2a 1

1 a −1 0 0

a −a a0 0 3−2a1 a −1 0

= −a(3 −2a) −a(a −1)(3 −2a) = −3a + 2 a2 −(3a2 −2a3 −3a + 2 a2) =

= − 3a + 2 a2 −3a2 + 2 a3 + 3a −2a2 = 2 a3 −3a2 = 0 =⇒a2 = 0 =⇒a = 0

2a −3 = 0 =⇒a = 32

a = 0 , 32

=⇒RgA = RgA = 3 = no =⇒

a = 0

A =0 0 0 00 0 3 11 −1 0 0

RgA = 2 = RgA < no

=⇒

a = 32

A =

32 −

32

32

32

0 0 0 1

1 1

2 0 0

a

a y + ( a + 1) z = aa x + z = a

x + a z = a

0 a a + 1 aa 0 1 a1 0 a a



0 a a + 1a 0 11 0 a

= a −a3 = 0 =⇒a(1 −a2) = 0 =⇒ a = 01 −a2 = 0 =⇒a = ±1

a = 0 , 1, −1 =⇒RgA = 3 = RgA = n ◦ =⇒

a = 0

0 0 1 00 0 1 0

1 0 0 0

RgA = 2

RgA = 2 = RgA < 3 = n ◦ =⇒

a = 10 1 2 11 0 1 11 0 1 1

RgA = 2

1 2 10 1 10 1 1

= 1 −1 = 0

RgA = 2

RgA = 2 = RgA < 3 = n ◦ =⇒

a = −1

0 −1 0 −1

−1 0 1 −1

1 0 −1 −1



M =

0 1

−1

−1 0 20 4 4

0 1 −1

−1 0 2

0 4 4

F 1 ←→ F 2

−−−−−−→−1 0 20 1

−1

0 4 4

F 3 = F 3 − 4F 2

−−−−−−−−→−1 0 20 1

−1

0 0 8

F 1 =4 F 1 − F 3

−−−−−−−−→F 2 =8 F 2 + F 3

−4 0 00 8 00 0 8

a

ay + az = 0x + z = 0

4x − 2y + az = a

0 a a 01 0 1 04 −2 a a

|A| =0 a a1 0 14 −2 a

= 4 a −2a −a2 = −a2 + 2 a

−a2 + 2 a = 0 =

⇒

a = 0

a = 2

a = 0 , 2



a = 00 0 0 01 0 1 04 −2 0 0

1 04 −2

= −2 = 0

RgA = RgA = 2 < 3 = no

a = 20 2 2 01 0 1 04 −2 2 2

0 21 0 = −2 = 0

0 2 01 0 04 −2 2

= −4 = 0 =⇒RgA = 2 = 3 = RgA

a

x + y + z = 13x + 2y + z = 1

y + 2z = a

1 1 1 13 2 1 10 1 2 a

RgA = RgA < n o A

RgA ≥2

1 13 2

= −1 = 0

RgA = 21 1 13 2 10 1 2

= 4 + 3 −6 −1 = 0



RgA = 2

0 =1 1 13 2 10 1 a

= 2 a + 3 −3a −1 = −a + 2 = 0 =⇒a = 2

x + y + z = 12x − y = 33x + z = 4

a

x + y + z = ax −y + z = 1x −3y + z = 0

A =1 1 1 a1 −1 1 11 −3 1 0

RgA ≥2

1 11 −1

= −1 −1 = −2 = 0

RgA = 2

1 1 a1 −1 11 −3 0

= 1 −3a + a + 3 = −2a + 4 = 0 =⇒a = 2



x + y + z = 3x + y + z = 5

2x −y + z = 3

x + y + z = 32x −y + z = 33x +2 z = 8



y −x = zx −z = yy + z = x

A =−1 1 −1 01 −1 −1 0

−1 1 1 0

RgA

RgA ≥2

1 −1

−1 −1= −1 −1 = −1 = 0

|A|

|A| =−1 1 −11 −1 −1

−1 1 1

= 1 + 1 −1 + 1 −1 −1 = 0

RgA = 2 =⇒

x = λ

y −z = λ −y −z = −λ

−2z = 0z = 0

x = λ z = 0 =⇒y = λ (λ,λ, 0)

RgA = 2

RgA = 1



RgA = 2

2x + 3y − 2z = 3x

− y + z = 4

3x + 2y − z = 7

RgA = 12x + 3y − 2z = 34x + 6y − 4z = 6

−2x − 3y + 2z = −3

x + 2y − z = 2x + (1 + b)y − bz = 2bx + by + (1 + b)z = 1

1 2 −1 21 1 + b −b 2b1 b 1 + b 1

|A| =1 2 −11 1 + b −b1 b 1 + b

= (1 + b)2 −2b−b + (1 + b) −2(1 + b) + b2 = 1 + b2 + 2b− 2b− b + 1+

+ b− 2 −2b + b2 = 2 b2 −2b = 0 =⇒b = 0 b = 1

b = 0 , 1 =⇒

b = 01 2 −1 21 1 0 01 0 1 1

RgA = 2

1 21 1

= 1 −2 = −1 = 0



RgA

1 2 21 1 01 0 1

= 1 −2 −2 = −3 = 0

RgA = 2 = 3 = RgA =⇒

b = 11 2 −1 21 2 −1 21 1 2 1

RgA = 2

1 21 1

= 1 −2 = −1 = 0

RgA RgA = 2 = RgA < 3 = no

=⇒

x +2 y −z = 1x + y −z = 1x −z = 1

1 2 −1 11 1 −1 11 0 −1 1

1 2 −1 11 1 −1 11 0 −1 1

F 2 = F 2 − F 1

−−−−−−−→F 3 = F 3 − F 1

1 2 −1 10 −1 0 00 −2 0 0

y = 0 y = 0

z = λx −λ = 1 =⇒x = 1 + λ

x = 1 + λy = 0z = λ



x + y + z = ax + y + az = 1x + ay + z = 1

1 1 1 a1 1 a 11 a 1 1

1 1 11 1 a

1 a 1

= 1 + a + a −1 −a2 −1 = −a2 + 2 a −1

−a2 + 2 a −1 = 0 =⇒a2 −2a + 1 = 0 =⇒a = 1

• a = 1

• a = 1

1 1 1 11 1 1 11 1 1 1

RgA = 1 = RgA < n o =⇒

−x + 2y + z = 1

ax − y + 2z = 22x + ( a −1)z = 2



−1 2 1 1a

−1 2 2

2 0 a −1 2

−1 2 1a −1 22 0 a −1

= a −1 + 8 + 2 −2a(a −1) = a −1 + 8 + 2 −2a2 + 2 a = −2a2 + 3 a + 9 = 0

a = −3 ±√ 9 + 724

= −3 ±94

=

−3 + 9

4 =

6

4 =

2

3

−3 −94

= −124

= −3

a = 23

, −3 =⇒RgA = RgA = 3 =⇒

a = 23

−1 2 1 123 −1 2 2

2 0 −13

2

RgA = 2

23 −1

2 0= 2 = 0

−1 2 123 −1 2

2 0 2

= 2 + 8 + 2 − 83

= 36

3 − 83

= 28

3 = 0

RgA = 3

RgA = 2 = 3 = RgA =⇒

a = −3

−1 2 1 1

−3 −1 2 22 0 −4 2



RgA = 2

−3 −12 0

= 2 = 0

−1 2 1

−3 −1 22 0 2

= 2 + 8 + 2 + 12 = 24 = 0

RgA = 3

RgA = 2 = 3 = RgA =⇒

ax + ay = 0x + z = a

−2y + az = a

a a 0 01 0 1 a0 −2 a a

a a 01 0 10 −2 a

= −a2 + 2 a = 0 =⇒a(−a + 2) = 0 =⇒ a = 0a = 2

a = 0 , 2 =⇒RgA = RgA = 3 =⇒

a = 0

0 0 0 01 0 1 00 −2 0 0

RgA = RgA = 2 < 3 =

=⇒



a = 2

2 2 0 01 0 1 20

−2 2 2

RgA = 2

2 21 0

= −2 = 0

2 2 01 0 20 −2 2

= −4 + 8 = 4 = 0

RgA = 3

RgA = 2 = 3 = RgA =⇒

2x

− y + z = 1

−x + y − z = 0y − z = 1

2 −1 1 1

−1 1 −1 00 1 −1 1

2 −1

−1 1= 2 −1 = 1 = 0 =⇒RgA 2

2 −1 1

−1 1 −10 1 −1

= −2 −1 + 1 + 2 = 0 =⇒RgA = 2



RgA 2

RgA

2 −1 1

−1 1 0

0 1 1

= 2

−1

−1 = 0

RgA = RgA = 2 < 3 =

2x − y + λ = 1

−x + y

− λ = 0

=⇒ 2x − y = 1 −λ

−x + y = λ

2x − y = 1 −λ

−x + y = λ

x = 1

y = λ + 1

x = 1y = 1 + λz = λ

b

bx + by = 13x + bz = b−2

− y + z = b−3

b b 0 13 0 b b−20 −1 1 b−3

b b 03 0 b0 −1 1

= 3 b−b2 = 0 =⇒ b = 0b = 3



b = 0 , 3 =⇒RgA = RgA = 3 =

=⇒

b = 0

0 0 0 13 0 0 −20 −1 1 −3

RgA = 2

3 00 −1

= −3 = 0

0 0 1

3 0 −20 −1 −3

= −3 = 0

RgA = 2 = 3 = RgA =⇒

b = 3

3 3 0 13 0 3 10 −1 1 0

RgA = 2

3 33 0

= −9 = 0

3 3 13 0 10 −1 0

= −3 + 3 = 0

RgA = 2 = RgA = 3 =

=⇒

y − z = 1

−x + 4z = 02y − z = 1



0 1 −1

−1 0 4

0 2 −1

= 2

−1 = 1 = 0

x =

1 1 −10 0 41 2 −1

|A| =

4−81

= −4

y =

0 1

−1

−1 0 40 1 −1

|A| =

1 −11

= 0

z =

0 1 1

−1 0 00 2 1

|A| = −2 + 1

1 = −1

a

x + y = a + 1

−2x − y + az = −2(a + 1) x + y − z = 2

1 1 0 a + 1

−2 −1 a −2a + 1 1 −1 2

1 1 0

−2 −1 aa + 1 1 −1

= 1 + a(a + 1) −2 −a = 1 + a2 + a −2 −a = a2 −1 = 0

= a2

−1 = 0 =⇒a = ±1

a = ±1 =⇒RgA = RgA = 3 =

=⇒



a = 1

1 1 0 2

−2 −1 1 −22 1 −1 2

RgA = 2

1 1

−2 −1= 1 = 0

1 1 2

−2 −1 −22 1 2

= 0 c3 = 2c2

RgA = RgA = 2 = 3 =

=

⇒ a = −1

1 1 0 0

−2 −1 −1 −20 1 −1 2

RgA = 2

1 1

−2 −1= 1 = 0

1 1 0

−2 −1 −20 1 2

= −2 + 4 + 2 = 4 = 0

RgA = 2 = 3 = RgA =⇒

a

−x + 2y + z = ax + ( a −1)y + az = 0

ax + 2y + z = −1

−1 2 1 a1 a −1 a 0a 2 1 −1



−1 2 11 a −1 aa 2 1

= −(a −1) + 2 a2 + 2 −a(a −1) −2 + 2 a = −a + 1 + 2 a2 + 2 −

− a2 + a −2 + 2 a = a2 + 2 a + 1 = 0 =⇒a = −1

a = −1 =⇒RgA = 3 = RgA =

=⇒

a = −1

−1 2 1 −11 −2 −1 0

−1 2 1 −1

RgA = 1 RgA = 2

1 −1

−1 0 = −1 = 0

RgA = 1 = 2 = RgA =⇒

b

y + bz = 1 + b

x + z = 3 −bbx − by = 1 −b

0 1 b 1 + b1 0 1 3−bb −b 0 1−b

0 1 b1 0 1b −b 0

= b−b2 = 0 =⇒b = 0b = 1

b = 0 , 1 =⇒RgA = 3 = RgA =

=⇒

b = 00 1 0 11 0 1 30 0 0 1



RgA = 2

0 11 0

= −1 = 0 RgA

0 1 11 0 30 0 1

=

−1 = 0

RgA = 2 = 3 = RgA =⇒ b = 1

0 1 1 21 0 1 21 −1 0 0

RgA = 2

0 11 0

= −1 = 0 RgA

0 1 21 0 21 −1 0

= 2 −2 = 0

RgA = 2 = RgA = 3 =

=⇒



y = √ 3 x (3, 0) tg60o = √ 3

tg30o =√ 33

y = √ 3 x 60o

30o

tg 30o = r

3 =

⇒

r = 3

·tg 30o = 3

√ 33

= √ 3 C 3, √ 3

(x −3)2 + y −√ 3 2= 3



x + y + z = 3

x = 0 z = 0

z = 1 y = 0

x + y+ z = 1

x = 0 z = 0 =⇒A(0, 1, 0)

z = 1 y = 0 =⇒B (0, 0, 1)

A B A(0, 1, 0)

−−→AB = (0 , −1, 1)

x = 0y = 1 −λz = + λ

x −2y + z = 1

(2, 0, 1), (0, 2, 1) (1, −1, 0)

A(2, 0, 1), B (0, 2, 1), C (1, −1, 0) −−→AB −→AC

−−→AB = (−2, 2, 0) −→AC = (−1, −1, −1)

A, −−→AB, −→AC

x −2 −2 −1y 2 −1

z −1 0 −1= −2(x −2) + 2( z −1) + 2( z −1) −2y =

= −2x + 4 + 2 z −2 + 2 z −2 −2y = −2x −2y + 4 z −2 = 0

x + y −2z + 1 = 0

x − 2y + z − 1 = 0x + y − 2z + 1 = 0



x − 2y + z − 3 = 0x + y − 2z + 8 = 0

(1, 0, 2) (2, 1, 0)

u∧v = i j k

1 0 22 1 0

= 4 j + k −2 i

w = (−2, 4, 1) | w| = √ 4 + 16 + 1 = √ 21

o

o = w

| w| = −2

√ 21,

4√ 21

, 1√ 21

x + z = 2

x + y = 0 , y + z = 2

x + z = 0

α(x + y) + y + z −2 = 0

x + z = 0α(x + y) + y + z −2 = 0

=⇒ x + z = 0αx + ( α + 1) y + z = 2

α

x + z = 0αx + ( α + 1) y + z = 2

z = λ =⇒x = −λ



α(−λ) + ( α + 1) y + λ = 2 =⇒(α + 1) y = 2 −λ + αλ =⇒y = 2α + 1

+ α −1α + 1

λ

α

x = − λ

y = 2α + 1

+ α −1

α + 1λ

z = λ

=⇒v = −1, α −1α + 1

, 1

x + y = 0y + z = 2

y = λ =⇒x = −λ z = 2 −λ

x = − λy = λz = 2 − λ

=⇒u = (−1, 1, −1)

α u v

u · v = 0

u · v = (−1, 1, −1) · −1, α −1α + 1

, 1 = 1 + α −1α + 1 −1 = 0 =⇒

α −1α + 1

= 0 =⇒α = 1

α = 1

x + z = 0x + 2y + z = 2

e v

e−v

|e

|=

|v

|=

|e

−v

|

cos( u, w) = u · w

|u| · | w| e−v e−v

cos(

e−v, e−v) = (e−v) ·( e−v)

|e−v| · |e−v| =

e· e+ v · v−2e · v

|e−v| · |e−v| =

= |e|2 + |v|

2−2 |e| · |v|cos(

e, v)

|e

−v

| · |e

−v

| cos(

e−v, e−v) = 1

1 = |e|2 + |e|

2−2 |e|

2 cos( e, v)

|e|2 =⇒2 |e|

2−2 |e|

2 cos( e, v) = |e|2 =⇒



=⇒2 −2 cos( e, v) = 1 =⇒−2 cos( e, v) = −1 =⇒cos( e, v) = 12

(

e, v) =

π3

e (1, 1, 3)

u v

u · e = v· e = 0

u = (2 , 1, −1) =⇒u · e = (2 , 1, −1) ·(1, 1, 3) = 2 + 1 −3 = 0v = (1 , 2, −1) =⇒v · e = (1 , 2, −1) ·(1, 1, 3) = 1 + 2 −3 = 0

ABCD

V

ABV BCV

A(2, 0, 0); B (2, 2, 0); C (0, 2, 0) V (1, 1, 3)

π ≡ABV A

−−→AB −→AV A(2, 0, 0); −−→AB = (0 , 2, 0) −→AV = (−1, 1, 3)

x −2 0 −1y 2 1z 0 3

= 6( x −2) + 2 z = 6x −12 + 2z = 0

π ≡3x + z = 6



π ≡BCV B −−→BC −−→BV B(2, 2, 0)−−→BC = ( −2, 0, 0) −−→BV = (−1, −1, 3)

x −2 −2 −1y

−2 0

−1

z 0 3= 2 z + 6( y

−2) = 2 z + 6 y

−12 = 0

π ≡3y + z = 6

π π

n = (3 , 0, 1) n = (0 , 3, 1)

cosα =n · n

|n| · | n |=

1√ 10 √ 10

= 110

=⇒α = 84o 15 39

3x

−2y + z = 1

−x = 3y + 3 z, y + 2 z = −1

(1, −1, 1) (−1, −1, 0)

u

x + 3y + 3z = 0y + 2z = −1

=

⇒

−→n1 = (1 , 3, 3)=⇒−→n2 = (0 , 1, 2)

u = −→n1∧−→n2 =

i j k1 3 30 1 2

= 6 i + k −2 j −3 i = 3 i −2 j + k =⇒u = (3 , −2, 1)

P (1, −1, 1) Q(−1, −1, 0)

−−→PQ = (−2, 0, −1) n −−→P Q

n ·−−→P Q = (3 , −2, 1).(−2, 0, −1) = −6 −1 = −7 = 0



ABCD

−−→CB −−→AD

−−→CB −−→BA

D(0, 0, 0) A(3, 0, 0) B(3, 1, 0) C (0, 1, 0).

−−→CB = (3 , 0, 0)−−→AD = (−3, 0, 0)

=⇒−−→CB ·−−→AD = (3 , 0, 0) ·(−3, 0, 0) = −9

−−→BA −−→CB

−−→BA = (0 , −1, 0)

−−→CB ∧−−→BA =

i j k3 0 00 −1 0

= −3 k

−−→CB ∧−−→BA = 3

x + y = 1 , y + z = 2

(1, 1, 0) (0, 1, 1)



x + y = 1

y + z = 2

=⇒−→n1 = (1 , 1, 0)

=⇒−→n2 = (0 , 1, 1)

=

⇒

u = −→n1

∧

−→n2 =

i j k1 1 0

0 1 1

= i + k

− j

u = (1 , −1, 1)

v = (0 , 1, 1) −(1, 1, 0) = ( −1, 0, 1)

0 =x y z1 −1 1

−1 0 1= −x −y −z −y = −x −2y −z

x + 2 y + z = 0

ax + by + cz = d a x + b y + c z = d

aa

= bb

= cc

= dd

(1, 1, 0); (1, 0, 1) (0, 1, 1)

x + 2 y + 3 z = 0

A(1, 1, 0); B (1, 0, 1) C (0, 1, 1)

−−→AB −→AC



−−→AB = (0 , −1, 1) −→AC = ( −1, 0, 1)

π ≡x −1 0 −1y −1 −1 0

z 1 1= −(x −1) −(y −1) −z = −x + 1 −y + 1 −z = 0 =⇒

=⇒x + y + z −2 = 0

−→x + y + z −2 = 0 −→x + y + z + 3 = 0

−→x + 2 y + 3 z = 0 −→x + 2 y + 3 z −1 = 0

x + y + z + 3 = 0x + 2 y + 3 z −1 = 0

(0, 1, 1) (2, 1, 0)

u = (0 , 1, 1) v = (2 , 1, 0)

w = u∧v =

i j k0 1 12 1 0

= 2 j −2 k − i =⇒ w = (−1, 2, −2)

w

w

| w| =

(−1, 2, −2)√ 1 + 4 + 4

= −13

, 23

, −23

a b

(1, 0, 0), (a,b, 0), (a, 0, b) (0,a ,b)

−−→AB, −→AC −−→AD

0

−−→AB = ( a −1, b, 0); −→AC = ( a −1, 0, b) −−→AD = (−1,a ,b)

a −1 b 0a −1 0 b

−1 a b= −b2 −b2(a −1) −ab(a −1) = 0 =⇒

b2 −ab2 + b2 −a2b + ab = 0 =⇒



=⇒ab(−b−a + 1) = 0

(1, 2, 3)

x + y = 1, y + z = 1

A

u

A, −−→AB, u y = λ

x + λ = 1y = λ

λ + z = 1

=⇒

x = 1 −λy = λ

z = 1 −λ

=⇒ B(1, 0, 1)u = (

−1, 1,

−1)

A(1, 2, 3)

−−→AB = (0 , −2, −2)

u = (−1, 1, −1)

x −1 0 −1y −2 −2 1z −3 −2 −1

= 2( x −1) + 2( y −2) −2(z −3) + 2( x −1) =

= 2 x −2 + 2 y −4 −2z + 6 + 2 x −2 = 4 x + 2 y −2z −2 = 0

2x + y −z −1 = 0

(1, 0, 0) (0, 1, 0)

x + y + z = 2x −y + z = 2

A(1, 0, 0) B(0, 1, 0)

−−→AB u −−→AB u

−−→AB = (−1, 1, 0)

u



x + y + z = 2x −y + z = 2

=⇒−→n1 = (1 , 1, 1)

−→n2 = (1 , −1, 1)=⇒u = −→n1∧−→n2

u = −→n1∧−→n2 =

i j k1 1 11 −1 1

= i + j − k − k − j + i = 2 i −2 k =⇒u = (2 , 0, −2)

x −1 −1 2y 1 0z 0

−2

= −2(x −1) −2z −2y = −2x + 2 −2z −2y = 0

x + y + z −1 = 0

P =

(0,a ,b) A = (1 , 0, 0), B = (1 , 1, 1)

C = (0 , 2, 1)

−−→AB (0, 1, 1) −→AC = ( −1, 2, 1)

0 =x −1 y z

0 1 1

−1 2 1= ( x −1) −y + z −2(x −1) = x −1 −y + z −2x + 2 = −x −y + z + 1

π ≡x + y −z −1 = 0 P ∈π

a −b−1 = 0

(1, 2, 1)

u(1, 2, 1) v

u · v = 0 v(1, 1, 3) u

|v| = √ 1 + 1 + 9 = √ 11



w 1√ 11

, 1√ 11

, −3√ 11

x −y + z = 2

x2

= y + 1

1 =

z + 2

−1

x −y + z = 2 =⇒n = (1 , −1, 1) A = (0 , 0, 2)

x2

= y + 1

1 =

z + 2

−1 =⇒u = (2 , 1, −1) B = (0 , −1, −2)

u · n = 2 −1 −1 = 0 =⇒

B ∈π

0 + 1 −2 = 2

d(r, π ) = d(B, π ) = |0 + 1 −2 −2|√ 1 + 1 + 1=

3√ 3 = √ 3 u

u v

u + v u−v

v v |u| = 4 |v| = 3

u + v

| u + v

|

(1)=

( u + v)

·( u + v)

(2)=

|u

|

2 +

|v

|

2 + 2 u

·v

(3)=

|u

|

2 +

|v

|

2 = √ 16 + 9 = 5

u −v

| u −v|(1)= ( u −v) ·( u −v)

(2)= |u|

2 + |v|2−2u · v

(3)= |u|

2 + |v|2 = √ 16 + 9 = 5



u · u = |u|2

a b

a b

a

b

a∧ b = |a| · b ·sen a, b

(1, 1, 1)

x + y = 1

x + y = 1

π : x + y = 1 =⇒n(1, 1, 0)

x = 1 + λy = 1 + λz = 1

x −11

= y −1

1 =

z −10

x −1 = y −1z = 1

=⇒ x −y = 0

z = 1



1 + λ + 1 + λ = 1 =⇒2λ = −1 =⇒λ = −12

x = 1 − 12

y = 1 − 12

z = 1

=⇒Q12

, 12

, 1

(1, 1, 1)

x −12

= y1

= z + 1

1

n(2, 1, 1)

2x + y + z + D = 0

(1, 1, 1)

D

2 + 1 + 1 + D = 0 =⇒D = −4

2x + y + z −4 = 0

r ≡x = 1 + 2 λy = λz = −1 + λ

λ

2(1 + 2 λ) + λ −1 + λ −4 = 0 =⇒2 + 4 λ + λ −1 + λ −4 = 0 =⇒6λ = 3 =⇒λ = 12



x = 1 + 1 = 2

y = 1

2 =

12

z = −1 + 1

2 = −

12

=⇒Q 2, 12

, −12

r : x + y + z = 0x −y + z = 1

, r : x + y + z = 0ax + + bz = 0

a b

r r

r r

r r

−→n1 = (1 , 1, 1)

−→n2 = (1 , −1, 1)

d = −→n1∧−→n2 =

i j k1 1 11 −1 1

= i + j − k − k − j + i = 2 i −2 k⇒ d = (2 , 0, −2)

r −→n1 = (1 , 1, 1)

−→n2 = ( a, 0, b)

−→d = −→n1∧−→n2 =

i j k1 1 1a 0 b

= b i + a j −a k −b j = b i + ( a −b) j −a k⇒−→d = ( b, a −b,−a)

d −→d

2b

= 0a −b

= −2

−a ⇒ −2a = −2b⇒a = b

d ·−→d = 0 ⇒(2, 0, −2) ·(b, a −b,−a) = 2 b + 2 a = 0⇒2b = −2a⇒b = −a



Π : x + y = 0

r :

x = λ

y = −2z = 1 + λ

s Π

r

r (λ, −2, 1 + λ)

λ

λ −2 = 0⇒λ = 2

P (2, −2, 3)

s P

Π r

Π

Π

n = (1 , 1, 0)

r

d = (1 , 0, 1) s

u = d∧n =

i j k1 0 11 1 0

= j + k − i⇒u = (−1, 1, 1)

s

s :x = 2 − λy =

−2 + λ

z = 3 + λ

r :x = λy = − λz = 1

s : x + y = 0x −z = 1

r s

r s



r s

r

r :x = λy = − λz = 1

⇒ P (0, 0, 1)u = (1 , −1, 0)

s x = α

s : x + y = 0x

− z = 1 ⇒

x = αy = − α

z = −1 + α⇒

Q(0, 0, −1)v = (1 ,

−1, 1)

P Q

−−→P Q = (0 , 0, −2)

det u, v,−−→P Q =1 −1 01 −1 10 0 −2

= 2 −2 = 0

• r s

• u∧v

r

R(λ,λ, 1)

s

λ − λ = 0λ − 1 = 1 ⇒λ = 2

R(2, −2, 1)

d = u∧v =

i j k1 −1 01 −1 1

= − i − k + k − j ⇒ d = (−1, −1, 0)

r 1 :x = 2 − λy = −2 − λz = 1



P = (0 , 0, −1) Q = (1 , 0, 0)

x + y = 1 x −z = 0

Π r P

P −−→P Q P

−−→P Q = (1 , 0, 1)

z = λ ; x + y = 1

x −z = 0=⇒

y = 1 −λx = λ

x = λy = 1 − λz = λ

u = (1 , −1, 1)

x = λ + µy = − µz = −1 + λ + µ

A = (1 , 1, 1) B = (1 , 0, 0) C = (0 , 2, 1) r

A B Π C r

P 0 r Π

−−→AB = (0 , −1, −1)

x = 1y = − λz = − λ

−−→AB

−y −z + D = 0

−2 −1 + D = 0 =⇒D = 3

−y −z + 3 = 0

P (1, −λ, −λ)

λ + λ + 3 = 0 =⇒λ = 23



P 1, −23

, −23

θ u = ( λ, 1, 0) v = (1 , µ, 0) λ

µ

λ µ cosθ = 0

λ µ senθ = 0

cos( u, v) = u · v

|u| · |v|

cosθ = λ ·1 + 1 ·µ + 0 ·0√ λ2 + 1 + 1 + µ2

= 0 =⇒ λ + µ

√ λ2 + 1 + 1 + µ2= 0 =⇒

=⇒λ + µ = 0 =⇒µ = −λ

senθ = 0 =

⇒

cosθ = 1

cosθ = λ + µ

√ λ2 + 1 + 1 + µ2= 1 =⇒λ + µ = λ 2 + 1 + 1 + µ2

r : x = 1y = z

s : y = 0x = z

P r Q s −−→PQ

(−1, 0, 1)

r :x = 1y = λz = λ

s :x = µy = 0z = µ

P (1,λ ,λ ) Q(µ, 0, µ) −−→P Q = ( µ −1, −λ, µ −λ) λ

µ−−→P Q = 1 =⇒ (µ −1)2 + λ2 + ( µ −λ)2 = 1 =⇒(µ −1)2 + λ2 + ( µ −λ)2 = 1

−−→P Q⊥(−1, 0, −1) =⇒(µ −1, −λ, µ −λ) ·(−1, 0, 1) = 0 =⇒−µ + 1 + µ −λ = 0



λ = 1

(µ −1)2 + 1 + ( µ −1)2 = 1 =⇒ µ2 −2µ + 1 + 1 + µ2 −2µ + 1 = 1 =⇒

=⇒ 2µ2

−4µ + 2 = 0 =⇒µ2

−2µ + 1 = 0 =⇒

=⇒ µ = 1

P (1, 1, 1) Q(1, 0, 1)

Π (1, 0, 1)

x + y + z = 0 x −z = 1

r Π

x + y + z = 0x

− z = 0

=⇒ z = λx = λ y =

−2λ

P (1, 0, 1) n = (1 , −2, 1)

x −2y + z + D = 0

1 + 1 + D = 0 =⇒D = −2

x −2y + z −2 = 0

Q(λ, −2λ, λ )

λλ + 4 λ + λ −2 = 0 =⇒6λ −2 = 0 =⇒λ =

26

= 13

Q13

, −23

, 13

a b r : x = 0y = 0

Π ≡x + y + az = b



x = 0 , y = 0 , x + y + az = b

O(0, 0, 0) k = (0 , 0, 1)

n = (1 , 1, a ) P (b,0, 0) y = z = 0

k

·n = a

a = 0 r Π a = 0

O

b = 0

a = 0 =⇒ r Π

a = 0 b = 0 =⇒ r

a = 0 b = 0 =⇒ r

x = 0y = 0

x + y + az = b

1 0 0 00 1 0 01 1 a b

1 00 1

= 1 = 0

1 0 00 1 01 1 a

= a

a = 0

a = 0



1 0 00 1 01 1 b

= b

a = 0 b = 0 =⇒RgA = RgA = 2 =⇒ r⊂Π

a = 0 b = 0 =⇒RgA = 2 = 3 = RgA =⇒ r Π

r : x + y = 0x −z = 1

r :x = 1y = λz = λ

Π r

s

Π

r x = µ

r :x = µy =

− µ

z = −1 + µ

r⊥s Π⊥s

r⊥s Π r r Π

A(0, 0, −1) r

Π : y + z = bA(0, 0, −1)∈Π

=⇒b = −1 =⇒Π : y + z = −1

s Π s P (1,λ ,λ ) Π

λ + λ = −1 =⇒λ = −12

=⇒P 1, −12

, −12



(1, 1, 2)

(1, 1, 0); (1, 0, 1) (0, 1, 1)

A(1, 1, 2) B(1, 1, 0) C (1, 0, 1) D(0, 1, 1) B, C D B, −−→BC −−→BD

−−→BC (0, −1, 1) −−→BD (−1, 0, 1)

x −1 y −1 z0 −1 1

−1 0 1

= −(x −1) −(y −1) −z = −x + 1 −y + 1 −z = −x −y −z + 2

π ≡x + y + z −2 = 0

d(A, π ) = |1 + 1 + 2 −2|√ 1 + 1 + 1=

2√ 3 =

2√ 33

u

(3, 5, 0)

(0, 1, 2) (0, 1, 1)

A(3, 5, 0), B (0, 1, 2) C (0, 1, 1)

B C B(0, 1, 2) −−→BC = (0 , 0, −1)

d(A, r ) =−−→AB∧

−−→BC

−−→BC

−−→AB = (−3, −4, 2)

−−→AB ∧−−→BC =

i j k

−3 −4 20 0 −1

= 4 i −3 j =⇒−−→AB ∧

−−→BC = (4 , −3, 0)

d(A, r ) =√ 16 + 9

√ 1 = 5



(1, 0, 1), (2, 0, 2), (3, 1, 3) (1, 2, 1)

A(1, 0, 1); B (2, 0, 2) C (3, 1, 3) D(1, 2, 1) −−→AB = (1 , 0, 1); −→AC = (2 , 1, 2) −−→AD = (0 , 2, 0)

1 0 12 1 20 2 0

= 4 −4 = 0

ABC ACD

ABC AABC =

12

−−→AB ∧−→AC

−−→AB∧−→AC =

i j k

1 0 12 1 2

= 2 j + k − 2 j − i =⇒−−→AB∧−→AC = (−1, 0, 1)

AABC = 12 |(−1, 0, 1)| =

12√ 2 =

√ 22

2

ACDAACD =

12

−→AC ∧−−→AD

−→AC

∧−−→AD =

i j k

2 1 20 2 0= 4 k

−4 i =

⇒−→AC

∧−−→AD = (

−4, 0, 4)

AACD = 12 |(−4, 0, 4)| =

12√ 32 = 2√ 2

2



A = AABD + AACD =√ 22

+ 2√ 2 = 5√ 2

2

2

a

ax + y + z = 2

π/ 3 z = 0

cosα = |−→n1 ·−→n2||−→n1| · |−→n2|

ax + y + z = 2 =⇒ −→n1 = ( a, 1, 1)z = 0 =⇒ −→n2 = (0 , 0, 1)

cosα = |(a, 1, 1) ·(0, 0, 1)||(a, 1, 1)| · |(0, 0, 1)|

= 1

√ a2 + 2 ·1 α = π/ 3

1√ a2 + 2

= cosπ3

= 12

a2 + 2 = 2 =⇒a2 + 2 = 4 =⇒a2 = 2 =⇒a = ±√ 2

(1, 0, 0)

(0, 1, 1) (1, 2, 0) (2, 1, 1)

A(1, 0, 0); B (0, 1, 1); C (1, 2, 0)

−−→AB −→AC A(1, 0, 0), −−→AB = (−1, 1, 1) −→AC = (0 , 2, 0)

π =x −1 −1 0

y 1 2z 1 0

= −2z −2(x −1) = −2x −2z + 2 = 0 ⇒x + z −1 = 0

P (2, 1, 1)

d(P, π ) = |2 + 1 −1|√ 12 + 1 2 =

2√ 2 =

2√ 22

= √ 2



x −11

= y1

= z −1

−1

u A

A(1, 0, 1)

u = (1 , 1, −1) u v = (0 , 1, 1)

u⊥v (0, 1, 1).(1, 1, −1) = 0

v

w = (0 ,√

2,√

2)

P = A + w = (1 , 0, 1) + (0 , √ 2, √ 2) = (1 , √ 2, 1 + √ 2)

A(0, 0, 0); B (0, 0, 1); C (0, 4, 1); D (0, 4, 0); P (0, 2, 1).



−→AP = (0 , 2, 1) −→AC = (0 , 4, 1)

cosα =−→AP ·−→AC

−→AP

·−→AC

= (0, 2, 1) ·(0, 4, 1)

√ 02 + 2 2 + 1 2 ·√ 02 + 4 2 + 1 2 =

0 + 8 + 1√ 4 + 1 ·√ 16 + 1

= 9√ 85

α = arccos 9√ 85

= 12 o 31 44

(1, 0, 0);(0 , 1, 0) (0, 0, 1)

−−→AB −→AC

AT = 12

−−→AB∧−→AC

−−→AB = (−1, 1, 0) −→AC = (−1, 0, 1)

−−→AB∧−→AC =

i j k

−1 1 0

−1 0 1= i + k + j =⇒−−→AB ∧−→AC = (1 , 1, 1)

AT = |(1, 1, 1)|2

=√ 32

u2

cos(

−−→AB, −→AC ) =−−→AB ·−→AC

−−→AB · −→AC = |1 + 0 + 0 |

√ 2 ·√ 2=

1

2

α = arccos12

= 60o

x + y + z = 0

x + y = 1 y + z = 1

cos(90o −α) = |u · n||u| · |n|

u n



x + y + z = 0 n = (1 , 1, 1)

−→n1 −→n2

x + y = 1 =⇒−→n1 = (1 , 1, 0)

y + z = 1 =⇒−→n2 = (0 , 1, 1)

u = −→n1∧−→n2 =

i j k1 1 00 1 1

= i + k − j =⇒u = (1 , −1, 1)

cos(90o −α) = |(1, 1, 1) ·(1, −1, 1)|√ 1 + 1 + 1 ·√ 1 + 1 + 1= |1 −1 + 1 |√ 3 ·√ 3 =

13

90o −α = arccos13

=⇒90o −α = 70o 31 42 =⇒α = 90 o −70o 31 42 = 19 o 28 18

x + y + z = 1

=⇒(y = 0; z = 0) =⇒x = 1 −→A(1, 0, 0)

=⇒(x = 0; z = 0) =⇒y = 1 −→B (0, 1, 0)

=⇒(x = 0; y = 0) =⇒z = 1 −→C (0, 0, 1)

AT = 12 −−→AB∧−→AC

−−→AB = (−1, 1, 0) −→AC = (−1, 0, 1)

−−→AB ∧−→AC =

i j k

−1 1 0

−1 0 1= i + k + j

−−→AB

∧

−→AC = (1 , 1, 1)

AT = 12

−−→AB ∧−→AC =

12√ 1 + 1 + 1 =

√ 32

u2



r :x = 1 + λy = λz = λ

Π : x + y + z = 1

r Π √ 3

r

d = (1 , 1, 1)

Π

n = (1 , 1, 1)

r

Π r

P (1 + λ,λ,λ ) Π

√ 3d(P, Π) = |1 + λ + λ + λ −1|√ 3 = √ 3⇒ |3λ| = 3

|3λ| = 3 ⇒

3λ = 3

⇒

λ = 1

⇒

P 1(2, 1, 1)3λ = −3⇒λ = −1⇒P 2(0, −1, −1)

√ 3x −z = 3

x + y = 1 y −x = −1

senα = |v · n||v| · |n| n = √ 3, 0, −1

z = λ

2y = 0 =⇒y = 0

x = 1

r ≡x = 1y = 0z = λ

v = (0 , 0, 1)



senα =√ 3 ·0 + 0 ·0 −1 ·1

√ 3 + 1 ·√ 1 = 12

α = arcsen 12

= π6

rad

λ µ

P = ( λ, 1, µ) A = (1 , 1, 1) B = (1 , 0, 0)

C = (0 , 2, 1)

A

−−→AB −→AC A(1, 1, 1) −−→AB = (0 , −1, −1) −→AC = (−1, 1, 0)

x −1 y −1 z −10 −1 −1

−1 1 0= y −1 −(z −1) + x −1 = x + y −z −1 = 0

x + y −z −1 = 0

d(P, Π) = |λ + 1 −µ −1|√ 1 + 1 + 1= |λ −µ|√ 3

|λ −µ|√ 3 =⇒ |λ −µ| = √ 3

A = (1 , 0, 0) B = (0 , 1, 0)

λ µ P = ( λ,µ, 0)

ABP

−−→AB −→AP −−→AB = (−1, 1, 0) −→AP = ( λ −1, µ, 0)

A = 12

−−→AB ∧−→AP

−−→AB

∧−→AP =

i j k

−1 1 0λ −1 µ 0=

−µ k

−(λ

−1) k = ( λ

−µ

−1) k

−−→AB∧−→AP = λ −µ −1



A = λ −µ −1

2 = 1 =⇒λ −µ −1 = 2 =⇒λ −µ = 1

r A = (1 , 0, 0) B = (1 ,

−1, 0) s

C = (0 , 1, 1) D = (1 , 0, −1)

Π s r

r s

r s

r A −−→AB −−→AB = (0 ,

−1, 0)

r :x = 1y = − λz = 0

s C −−→CD −−→CD = (1 , −1, −2)

s :x = µy = 1 − µz = 1 − 2µ

C −−→CD s⊂Π −−→AB r Π

Π :x = + µy = 1 − λ − µz = 1 − 2µ

x y −1 z −10 −1 01 −1 −2

= 2 x + ( z −1) = 0 =⇒2x + z = 1

r s

d(r, s ) =−→AC, u, v

|u∧v|

−→AC, u, v =−1 1 10 −1 01 −1 −2

= −2 + 1 = −1



u∧v =

i j k0 −1 01 −1 −2

= 2 i + k =⇒u∧v = (2 , 0, 1)

d(r, s ) = | −1||(2, 0, 1)| = 1√ 4 + 1 = 1√ 5 =√

55



r A = (1 , 0, 0) B (−1, 0, −1)

r

C = (0 , −1, 0)

A −−→AB−−→AB = (−2, 0, −1) −−→AB = (2 , 0, 1)

x −12

= y0

= z1

y = 0x −1 = 2z

=⇒ y = 0

x −2z = 1

r r

r Π ≡ax + by + cz = d A∈Π =⇒a = d =⇒Π ≡ax + by + cz = a

−−→AB ⊥n =⇒(2, 0, 1) ·(a,b,c) = 0 =⇒2a + c = 0 =⇒c = −2a

Π ≡ax + by −2az = a

d(C, Π) = | −b|√ a2 + b2 + 4 a2 = 1 =⇒

| −b|√ 5a2 + b2 = 1 =⇒ 5a2 + b2 = | −b|

=⇒ 5a2 + b2 = | −b|2 =⇒5a2 + b2 = b2 =⇒5a2 = 0 =⇒a = 0

by = 0 =⇒y = 0

Exámenes Selectividad UNED resueltos

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