Exercícios Resolvidos-2023 1

9. Calcule as distâncias: (a) Entre as retas r : (x, y, z) = (0, 1, 1) + t(1, 3, 1) e s : x = 1 + 2t y = 2 + 3t z = 3t (c) Entre os planos π₁ : 2x - y + 2z + 9 = 0 e π₂ : 4x - 2y + 4z - 21 = 0 10. Calcule a distância do ponto P = (1, 2, 3) e a reta x = 1 - 2t r : y = 2t z = 2 - t 9 a) Retas r : (x,y,z) = (0,1,1) + t(1,3,1) s : (x,y,z) = (1,2,0) + t(2,3,3) A₁(0,1,1) \overrightarrow{V₁ₐ} = (1,3,1) A₂₋₁ = (-1,1,1) V₂ₐ = (2,3,3) \overrightarrow{V₁ₐ} x \overrightarrow{V₂ₐ} = |\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 1 \\ 2 & 3 & 3 \end{vmatrix}| = 6\hat{i} - 3\hat{k} \frac{\overrightarrow{V₁ₐ} x \overrightarrow{V₂ₐ}}{\left|\overrightarrow{V₁ₐ} x \overrightarrow{V₂ₐ} \right|} = ⁶/\sqrt{46} \overrightarrow{V₁ₐ}, \overrightarrow{V₂ₐ}, A₁A₂) = |\begin{vmatrix} 3 & 1 & 1 \\ 2 & 3 & 3 \\ -1 & 1 & 1 \end{vmatrix}| = -10 \overrightarrow{V₁ₐ} x \overrightarrow{V₂ₐ} d(r,s) = \frac{|\overrightarrow{V₁ₐ}, \overrightarrow{V₂ₐ}, A₁A₂)|}{\left|\overrightarrow{V₁ₐ} x \overrightarrow{V₂ₐ} \right|} = \frac{1 - 101}{\sqrt{46}} = \frac{10}{\sqrt{46}} = \frac{10\sqrt{46}}{46} = \frac{5\sqrt{46}}{23} u. C. \quad \L = 3 \cdot C c) \pi₁ : 2x - y + 2z + 9 = 0 \pi₂ : 4x - 2y + 4z - 21 = 0 \overrightarrow{Vπ₁} = (2,-1,2) \overrightarrow{Vπ2} x:0 e z:0 0 - y + 0 + 9 = 0 \not y = 9 A₁(0,9,0) d(\pi₁, \pi₂) = d(A₁, \overrightarrow{V₂}) = \frac{|x₀+y₀-z₀+d|}{|\overrightarrow{Vₙ₂|}} = \frac{|0+9+0+9}{\sqrt{4+(-2)²+1²}} = \frac{18}{\sqrt{36}} = \frac{18}{6} = 3 \mu. C, \text{Mu. C.} 10. (x,y,z) = (1,0,2) + t(-2,-2,-1) P (1,2,3) |\overrightarrow{Vₙ₂|} = \sqrt{(-2)²+2²+1} \overrightarrow{AP}= P-A_=(0,2,1) \overrightarrow{AP} x \overrightarrow{Vₙ₂|} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 & 1 \\ -2 & 2 & 1 \end{vmatrix} = -4k + 2 \overrightarrow{AP} \times \overrightarrow{Vₙ₂} = \dot{-8} & 4k = (-20, 4) d(P,r) = \frac{|\overrightarrow{AP} x \overrightarrow{Vₙ₂}|}{|\overrightarrow{Vₙ₂|}} = \frac{\sqrt{|0+4(-2)+1²|}}{3} = \frac{\sqrt{20}}{3} = \frac{2\sqrt{2} \cdot \mu. C, \frac{2\sqrt{15}}{3}}