ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Phương trình đường thẳng

A.\(d:\left\{ {\begin{array}{*{20}{c}}{x = {x_0} + at}\\{y = {y_0} + bt}\\{z = {z_0} + ct}\end{array}} \right.\left( {t \in \mathbb{Z}} \right)\)

B. \(d:\left\{ {\begin{array}{*{20}{c}}{x = {x_0} + at}\\{y = {y_0} + bt}\\{z = {z_0} + ct}\end{array}} \right.\left( {t \in \mathbb{R}} \right)\)

C. \(d:\left\{ {\begin{array}{*{20}{c}}{x = a + {x_0}t}\\{y = b + {y_0}t}\\{z = c + {z_0}t}\end{array}} \right.\left( {t \in \mathbb{R}} \right)\)

D. \(d:\left\{ {\begin{array}{*{20}{c}}{x = a + {x_0}t}\\{y = b + {y_0}t}\\{z = c + {z_0}t}\end{array}} \right.\left( {t \in \mathbb{Z}} \right)\)

A.\[\left( {a;b;c} \right)\]

B. \[\left( {a;b;c} \right)\]

C. \[\left( {{x_0};{y_0};{z_0}} \right)\]

D. \[\left( { - {x_0}; - {y_0}; - {z_0}} \right)\]

A.\[\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\]

B. \[\frac{{x - {x_0}}}{{ - a}} = \frac{{y - {y_0}}}{{ - b}} = \frac{{z - {z_0}}}{{ - c}}\]

C. \[\frac{{x + {x_0}}}{a} = \frac{{y + {y_0}}}{b} = \frac{{z + {z_0}}}{c}\]

D. \[\frac{{x + {x_0}}}{a} = \frac{{y + {y_0}}}{{ - b}} = \frac{{z + {z_0}}}{c}\]

A.(−1;−1;1)     

B.(−1;1;1)        

C.(0;1;1)

D.(0;1;0)

A.(0;1;2)           

B.(1;0;1)

C.(2;−2;1)        

D.(3;−4;1) 

A.A và B đều thuộc d  

B.B và C đều thuộc d

C.A và C đều thuộc d

D.chỉ có A thuộc d  

A.Phương trình chính tắc của \[(d):\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\]

B.Phương trình tham số của \(d:\left\{ {\begin{array}{*{20}{c}}{x = {x_0} + at}\\{y = {y_0} + bt}\\{z = {z_0} + ct}\end{array}} \right.\left( {t \in \mathbb{R}} \right)\)

C.Nếu \[k \ne 0\;\] thì \[\vec v = k.\vec u\]là một vecto chỉ phương của đường thẳng (d).

D.Phương trình chính tắc của\[(d):\frac{{x + {x_0}}}{a} = \frac{{y + {y_0}}}{b} = \frac{{z + {z_0}}}{c}\]

A.\(\left\{ {\begin{array}{*{20}{c}}{x = t}\\{y = t}\\{z = t}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{x = t}\\{y = 0}\\{z = 0}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{x = t}\\{y = t}\\{z = 0}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{x = 0}\\{y = 0}\\{z = t}\end{array}} \right.\)

A.M(0,0,3)

B.N(0,1,0)

C.P(−2,0,0)

D.Q(1,0,1)  

A.\(\Delta :\left\{ {\begin{array}{*{20}{c}}{x = 1 - 4t}\\{y = 2 + 3t}\\{z = - 1 - 2t}\end{array}} \right.\)

B. \(\Delta :\left\{ {\begin{array}{*{20}{c}}{x = - 4 + t}\\{y = 3 + 2t}\\{z = - 2 - t}\end{array}} \right.\)

C. \(\Delta :\left\{ {\begin{array}{*{20}{c}}{x = 4 + t}\\{y = - 3 + 2t}\\{z = 2 - t}\end{array}} \right.\)

D. \(\Delta :\left\{ {\begin{array}{*{20}{c}}{x = 1 + 4t}\\{y = 2 - 3t}\\{z = - 1 + 2t}\end{array}} \right.\)

A. \(\left\{ {\begin{array}{*{20}{c}}{x = 2 + 2t}\\{y = - 3t}\\{z = - 1 + t}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{x = - 2 + 2t}\\{y = - 3t}\\{z = 1 + t}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{x = - 2 + 4t}\\{y = - 6t}\\{z = 1 + 2t}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{x = 4 + 2t}\\{y = - 3t}\\{z = 2 + t}\end{array}} \right.\)

A.\[\frac{{x + 1}}{2} = \frac{{y + 2}}{{ - 3}} = \frac{{z - 3}}{4}\]

B. \[\frac{{x - 1}}{3} = \frac{{y - 2}}{{ - 1}} = \frac{{z + 3}}{1}\]

C. \[\frac{{x - 3}}{1} = \frac{{y + 1}}{2} = \frac{{z - 1}}{{ - 3}}\]

D. \[\frac{{x - 1}}{2} = \frac{{y - 2}}{{ - 3}} = \frac{{z + 3}}{4}\]

A.\[\frac{x}{2} = \frac{y}{{ - 1}} = \frac{z}{1}\]

B. \[\frac{x}{2} = \frac{y}{1} = \frac{z}{1}\]

C. \[\frac{x}{2} = \frac{y}{{ - 1}} = \frac{z}{{ - 1}}\]

D. \[\frac{x}{{ - 2}} = \frac{y}{1} = \frac{z}{1}\]

A.\(\left\{ {\begin{array}{*{20}{c}}{x = - 2 + t}\\{y = 3 - t}\\{z = 1}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{x = 2 - t}\\{y = - 1 + t}\\{z = 1}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{x = 2 - 2t}\\{y = - 1 + 2t}\\{z = 1}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{x = - 2 + t}\\{y = 3 + t}\\{z = 1}\end{array}} \right.\)

A. \(\left\{ {\begin{array}{*{20}{c}}{x = 2 + 3t}\\{y = 1 + t}\\{z = 3 + t}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{x = - 1 + 3t}\\{y = t}\\{z = 2 + t}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{x = 5 - 3t}\\{y = 2 - t}\\{z = 4 - t}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{x = - 4 + 3t}\\{y = - 1 + t}\\{z = 2 + t}\end{array}} \right.\)

A. \(\left\{ {\begin{array}{*{20}{c}}{x = 1 + t}\\{y = 2}\\{z = - 3}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{x = 1}\\{y = 2 + t}\\{z = - 3}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{x = 1}\\{y = 2}\\{z = 3 + t}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{x = 1 + t}\\{y = 2 + t}\\{z = - 3}\end{array}} \right.\)

A.\[d:\frac{{x - 1}}{4} = \frac{{y - 2}}{{ - 7}} = \frac{{z - 3}}{{ - 1}}\]

B. \[d:\frac{{x - 1}}{4} = \frac{{y - 2}}{7} = \frac{{z - 3}}{1}\]

C. \[d:\frac{{x - 1}}{{ - 4}} = \frac{{y - 2}}{{ - 7}} = \frac{{z - 3}}{1}\]

D. \[d:\frac{{x - 1}}{4} = \frac{{y - 2}}{{ - 7}} = \frac{{z - 3}}{1}\]

A.\(\left\{ {\begin{array}{*{20}{c}}{x = 6t}\\{y = - 4t}\\{z = - 3t}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{x = 6t}\\{y = 2 + 4t}\\{z = - 3t}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{x = 6t}\\{y = 4t}\\{z = - 3t}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{x = 6t}\\{y = 4t}\\{z = 1 - 3t}\end{array}} \right.\)

A.\[\frac{5}{6}\]

B. \[\frac{5}{3}\]

C. \[\frac{7}{3}\]

D. \[\frac{3}{5}\]

A.\[A \notin d,\,\,B \in d\]

B. \[A \in d,\,\,B \in d\]

C. \[A \in d,\,\,B \notin d\]

D. \[A \notin d,\,\,B \notin d\]