Recenti prove d'esame di Geometria

GEOMETRIA - Settembre 2018 - Soluzioni

1)

v₁=(1,1,2) v₂=(2,m,1) v₃=(m,1,-2)

v₁∙v₂ = 2 - m + 2 = 4 - m = 0 → base ortogonale ↔ {4 - m = 0m - 3 = 02m - m - 2 = 0

m=4 m=3

Sia m_i = (u_i, m₃) B¹ = {v₁=(1,1,2), v₂=(2,4,1), v₃=(3,-1,-2)}base ortogonale

Cooa : (1 2 1) (1 2 -1) = -42 42 -16

(-4 8 -2) (4 -8 2) (-6 84)

v = (3,-1,-7)

A = ( -1 -2 1 ) ( 2 ) 0 1 2 -1 1 0 1 00 1 2

>=0

1 + (1) = 3 rango 3 - soluzioni

3){ x - y - z = 1 2x + 4y - 2 = 0 sono non paralleli

α // i + j + 3k

2(2k+1) + y(k-1) + z= -2k-1=0 2k+ 1+ k -1 + 3 (x + 3 ) 3k + 3=0

k=-1 ↷ α : x+2y-z-1=0 }

{ x=2-ty=2tz=tt∈R→ { x=2-2y=2zpiano contenente 1 x - 2 + 1 → ( 1 - 2t ) - 2 = 0

x2α) ↔ 0 ↔ x+2-1=0

[]

= [ x-1 ]

x+2y-z-1=0

{ x+z-2=0 }

(4)

A_β^f= [

• 1 1 -2
• -1 2 1
• 0 1 -2

dim Im f = r (A_β^f) = 2 dim Ker f = 1

• B_{Im f}= {(1,-1,0), (1,1,1)}
• B_{Ker f}= {(1,1,1)}

Tc A_β^f= 1 - 1 = 1 ≠ 2

  1 - λ          -2           0

 -1                 2 - λ           1

  0                    1      -2 - λ

autovalori λ₁ = 2,0,1

E_-2 = l₃ x+y -2z=0

            y+z=0

E₁ = l₁ x-2y+2z=0

            x-2y+2z=0

• E_-2= {(1, 1,-1)}
• E₁= {(t,-2,2,1)}
• B={[(-1,1,1), (1,1,0), (-2,2,1)]}

è una base di autovettori con A_β^f = [

• -2 0
• 0 0
• 0 1

(5)

v'=2i+j

τ: x-5y+4=0 x associati

vers τ= ¹/₂₆ (5i+j)

pr.ort. di v su τ (v" vers c) vers c) = ¹¹/₂₆ (5i+j) = ¹¹/₂₆ (5i+j)

• s: x-2 y+3

x-2y+z-4+6 s: x-2y-8=0

• 5(x-1)+1(y+2)
• t: 5x-y-3=0

(6)

Tema A

1. V
2. V
3. F
4. V
5. V
6. F

Tema B

1. F
2. F
3. V
4. F
5. F
6. F

3) A_B^t = ( 1 0 2

    0 1 -1

    -1 0 2 )

|λ - 1 0 2|

|0 1 -λ - 1| = 0

|1 0 2 - λ|

(1 - λ)(λ² -λ - 2λ + 2) = 0

λ = 0

E₀ = ker φ

 x + z = 0

 y - z = 0

 z = -t = y

 x = -2t

E₁ = {t(0, 1, 0), t ∈ ℝ}

E₃ = {t(-2, 1, 1 - 2), t ∈ ℝ}

β = {(-2, 1, 1), (0, 1, 0), (2, 1, -2)}

(5) y: kx + (z - k)y - kz + 3 - kx = 0

 k ∈ ℝ

α ∈ y per P(2, 0, 1 - 4)

 2k + k + 3 - k = 0  k = -3/2

 3x - 7y - 3z = 9 - 9 = 0

β ∈ y   2j - k

 4 - 2k + k = 0  k = 4

 β: 4x - 2y - 4z = -1 = 0

R = d(C₁, β) = |4 + 2 - 8 - 11|/√(16 + 4 + 16)

    = 3/6 = 1/2

(x - 1)² + (y + 1)² + (z - 2)² = 1/4

GEOMETRIA - Luglio 2019 - Soluzioni

1)

• V₁ = (1, 3)
• V₂ = ( - , , )
• V₃ = (1, -1, -1)
• a ∈ ℝ

V₁, V₂, V₃ sono L.T.

1. a 3
2. 1/4 a 2

⟺

2a + 3 + 3/2 a = 5/2

28/7 3a = 3 = 5a

a = -3 ± 4/4 a = 5/2 5 2

a_{V2 verso}

+ 1/4 16 + a2 = 1

a 7 5

a 7 16

(a 16 )

a (11 = 16 )

a =

2)

x₁ - x₂ + x₃ - x₄ = 1

x₁ + x₂ - x₃ + x₄ = 0

x₁ + x₃ - x₄ = 2

2 4

τ = 3 normali

x₁ = [t .]¹

-1 + t - t

= 1/2

x₂ = [t 1]

x₃ = [1 - 1]

= 3/2

sol. ( , 1, 1, 3, t), t ∈ ℝ

t (0, 0, 1, 1) + ( , 1, 3, 1)

3)

A(3,1) B(4, _{- , 1})

centro M (5,0)

Raggio: = d(A,M) = 4

= 5

≅ {(x-5)² + y² = 5

x² + y² - 10x + 20 = 0

rette da B verso A

retta per A⊥BA

4 2

cos x̂ C = -2 1 = cos ŷ C

2 (x - 3) - (y - 1) = 0

2 x - y - 5 = 0

5)

1. V
2. F
3. F
4. V
5. F
6. F

4

A(1,-1) B(3,2)

\overrightarrow{AB} = (2,3)

\gamma: \begin{cases} x = 1 + 2t \\ y = -1 + 3t \end{cases} t \in \mathbb{R}

\cos \widehat{x \gamma} = \frac{3}{\sqrt{13}}

\text{equazione retta} = \begin{cases} x = \frac{x - 1}{2} \\ y = \frac{y+1}{3} \end{cases}

\begin{cases} 3x - 3 = 2y + 2 \\ z: 3x = 2y - 5 = 0 \end{cases}

accomunare centro C(1,1,0) Tangente a \ r

(x - 4)^2 + (y - 1)^2 = \frac{25}{13}

5

S: x^2 + y^2 + z^2 - 2x + 2y - 2 = 0

C(1,1,0) \ \text{centro}

Raggio = \sqrt{1 + 1 + 2} = 2

\Pi \ \text{per C e} \ \parallel \alpha: x + y + z + 5 = 0

x + y + z + d = 0

1 - 1 + 0 + d = 0 \Rightarrow \Pi: x + y + z = 0

P \ \text{Tangente a} \ S \ \text{in} P(0,0,\sqrt{2})

\overrightarrow{CP} = (-1, 1, \sqrt{2})

(x-0) + (y-0) + \sqrt{2}(z-\sqrt{2}) = 0

-x + y + \sqrt{2}z - 2 = 0 \Rightarrow \sigma: x - y - \sqrt{2}z + 2 = 0

\text{vettori modulo 5 e } \perp \ \alpha

\text{generico vettore } 1\alpha \ e \ t(\hat{i} + \hat{j} + \hat{k})

\Rightarrow \sqrt{t^2 + t^2 + t^2} = 5

3t^2 = 25 \quad t^2 = \frac{5}{3}

= 0 \quad + \frac{5}{\sqrt{3}}\left(\hat{i} + \hat{j} + \hat{k}\right)\\right] = 0

6

1. V
2. F
3. F
4. F
5. V
6. V

GEOMETRIA – Febbraio 2018 – Solonari

1. \( v_1 = (1, -1, 2k) \quad v_2 = (2, 1, -1, 2 + k) \quad v_3 = (k, 1, -4k, 5+k, 1) \)

   Le domande equivalenti chiedete per quelli \( k \in \mathbb{R} \) che svolge 2 la matrice.

   \( A = \begin{pmatrix} 1 & -1 & k \\ 2 & 1 & 2 + k \\ k & 6 + k & 1 \\ \end{pmatrix} \)

   \( |1 \quad -1 \quad k| \)

   \( |\begin{pmatrix} 1 & -1 \\ 2 & 1 \\ k & 6 + k \end{pmatrix}| = \begin{vmatrix} -20 + 18 + 3k \\ 0 \\ 2k-2 \\ 0 \end{vmatrix} \)

   \( \rightarrow \text{rank} = 2 \)

2. \( 2x_1 + x_2 − x_3 + x_4 = 0 \)

   \( x_1 − x_3 + x_4 = 0 \)

   \( x_1 − x_2 − 2x_3 + x_4 = 0 \)

   \( A = \begin{pmatrix} 2 & 1 & −1 & 1 \\ 1 & 0 & −1 & 1 \\ 1 & −1 & −2 & 1 \\ 2 & 0 & 1 \\ \end{pmatrix} \)

   \( x_3 = T \quad x_4 = u \)

   \( 2x_1 + x_2 = T − u \)

3. \( A_{{\varphi} } = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 0 & 1 \end{pmatrix} \)

   \( rango = 2\)

   \( B_{\text{lmf}} = \{(1, 2, 0), (0, 1, 1)\} \)

   \( x + z = 0 \)

   \( x + z + 2 = 0 \)

   \( z = t \quad x = 1−t \quad y = 3 + t \)

   \( \lambda = 1, 0, 1, 2 \)

   \( E_0 = \ker \varphi \)

   \( E_{-1} = \{t (1, 0, -1, 2), t \in \mathbb{R}\} \)

   \( E_2 = \{t (1, 3, 1), t \in \mathbb{R}\} \)