A. \(\left( { - \frac{{15}}{4}; - \frac{{19}}{6}; - \frac{{43}}{{12}}} \right)\)
B. \(\left( {\frac{{15}}{4};\frac{{19}}{6};\frac{{43}}{{12}}} \right)\)
C. (45;28;43)
D. (-45;-28;-43)
A. M(0;2;0)
B. M(0;-1;0)
C. \(M\left( {0;\frac{5}{3};0} \right)\)
D. M(0;1;0)
A. (-1;1;1)
B. (1;-1;1)
C. (1;1;3)
D. (1;-2;-3)
A. 6x + 3y + 2z - 6 = 0
B. x - y + z - 2 = 0
C. x + 2y - 3z + 16 = 0
D. x - y + 2z = 0
A. \(m = \frac{3}{2};n = 4\)
B. \(m = - \frac{3}{2};n = 4\)
C. \(m = - \frac{3}{2};n = - 4\)
D. \(m = - 4;n = \frac{3}{2}\)
A. \(\frac{{x + 2}}{1} = \frac{{y - 1}}{2} = \frac{{z - 3}}{{ - 2}}\)
B. \(\frac{{x - 2}}{1} = \frac{{y + 1}}{2} = \frac{{z + 3}}{{ - 2}}\)
C. \(\frac{{x - 1}}{{ - 2}} = \frac{{y - 2}}{1} = \frac{{z + 2}}{3}\)
D. \(\frac{{x + 1}}{{ - 2}} = \frac{{y + 2}}{1} = \frac{{z - 2}}{3}\)
A. N(0;0;3)
B. N(0;0;4)
C. N(2;3;0)
D. Không tồn tại N
A. \(\left\{ \begin{array}{l} x = - 1 + t\\ y = 2\\ z = - 3 - t \end{array} \right.\left( {t \in R} \right)\)
B. \(\left\{ \begin{array}{l} x = 1\\ y = - 2\\ z = 3 - 2t \end{array} \right.\left( {t \in R} \right)\)
C. \(\left\{ \begin{array}{l} x = 1 + 2t\\ y = - 2\\ z = 3 + 2t \end{array} \right.\left( {t \in R} \right)\)
D. \(\left\{ \begin{array}{l} x = 1 + t\\ y = - 2\\ z = 3 - t \end{array} \right.\left( {t \in R} \right)\)
A. \(I\left( {\frac{7}{2};3; - \frac{5}{2}} \right)\)
B. I(4;2;3)
C. \(I\left( {2;\frac{3}{2}; - 1} \right)\)
D. \(I\left( { - 1; - \frac{1}{2}:\frac{5}{2}} \right)\)
A. 2x - z - 3 = 0
B. 2x + y + z - 3 = 0
C. 4x - y - 5z + 13 = 0
D. 9x - y + z - 16 = 0
A. \(d:\frac{{x - 2}}{1} = \frac{{y - 2}}{{ - 3}} = \frac{{z - 1}}{{ - 5}}\)
B. \(d:\frac{{x - 1}}{2} = \frac{y}{3} = \frac{{z - 2}}{{ - 4}}\)
C. \(d:\left\{ \begin{array}{l} x = 2 + t\\ y = 2\\ z = 1 - t \end{array} \right.\left( {t \in R} \right)\)
D. \(d:\frac{{x - 2}}{{ - 1}} = \frac{{y - 2}}{2} = \frac{{z - 1}}{{ - 3}}\)
A. \(d:\left\{ \begin{array}{l} x = - 3 + t\\ y = 1 - 2t\\ z = 1 - t \end{array} \right.\left( {t \in R} \right)\)
B. \(d:\left\{ \begin{array}{l} x = 3t\\ y = 2 + t\\ z = 2 + 2t \end{array} \right.\left( {t \in R} \right)\)
C. \(d:\left\{ \begin{array}{l} x = - 2 - 4t\\ y = - 1 + 3t\\ z = 4 - t \end{array} \right.\left( {t \in R} \right)\)
D. \(d:\left\{ \begin{array}{l} x = - 1 - t\\ y = 3 - 3t\\ z = 3 - 2t \end{array} \right.\left( {t \in R} \right)\)
A. \(\left[ \begin{array}{l} x + 2y + z + 2 = 0\\ x + 2y + z - 2 = 0 \end{array} \right.\)
B. \(\left[ \begin{array}{l} x + 2y - z - 10 = 0\\ x + 2y + z - 2 = 0 \end{array} \right.\)
C. \(\left[ \begin{array}{l} x + 2y + z + 2 = 0\\ - x - 2y - z - 10 = 0 \end{array} \right.\)
D. \(\left[ \begin{array}{l} x + 2y + z + 2 = 0\\ x + 2y + z - 10 = 0 \end{array} \right.\)
A. 2x + 3z - 11 = 0
B. y - 2z - 1 = 0
C. - 2y + 3z - 11 = 0
D. 2x + 3y - 11 = 0
A. \(\left[ \begin{array}{l} D\left( {0;0;0} \right)\\ D\left( {6;0;0} \right) \end{array} \right.\)
B. \(\left[ \begin{array}{l} D\left( {0;0;2} \right)\\ D\left( {8;0;0} \right) \end{array} \right.\)
C. \(\left[ \begin{array}{l} D\left( {2;0;0} \right)\\ D\left( {6;0;0} \right) \end{array} \right.\)
D. \(\left[ \begin{array}{l} D\left( {0;0;0} \right)\\ D\left( { - 6;0;0} \right) \end{array} \right.\)
A. \(M\left( { - \frac{3}{2}; - \frac{3}{4};\frac{1}{2}} \right);M\left( { - \frac{{15}}{2};\frac{9}{4};\frac{{ - 11}}{2}} \right)\)
B. \(M\left( { - \frac{3}{5}; - \frac{3}{4};\frac{1}{2}} \right);M\left( { - \frac{{15}}{2};\frac{9}{4};\frac{{11}}{2}} \right)\)
C. \(M\left( {\frac{3}{2}; - \frac{3}{4};\frac{1}{2}} \right);M\left( {\frac{{15}}{2};\frac{9}{4};\frac{{11}}{2}} \right)\)
D. \(M\left( {\frac{3}{5}; - \frac{3}{4};\frac{1}{2}} \right);M\left( {\frac{{15}}{2};\frac{9}{4};\frac{{11}}{2}} \right)\)
A. \(\left[ \begin{array}{l} 2x - 3y + 6z - 12 = 0\\ 2x - 3y - 6z = 0 \end{array} \right.\)
B. \(\left[ \begin{array}{l} 2x + 3y + 6z + 12 = 0\\ 2x + 3y - 6z - 1 = 0 \end{array} \right.\)
C. \(\left[ \begin{array}{l} 2x + 3y + 6z - 12 = 0\\ 2x + 3y - 6z = 0 \end{array} \right.\)
D. \(\left[ \begin{array}{l} 2x - 3y + 6z - 12 = 0\\ 2x - 3y - 6z + 1 = 0 \end{array} \right.\)
A. \(\left( Q \right):2x + 2y + 3z - 7 = 0\)
B. \(\left( Q \right):2x - 2y + 3z - 7 = 0\)
C. \(\left( Q \right):2x + 2y + 3z - 9 = 0\)
D. \(\left( Q \right):x + 2y + 3z - 7 = 0\)
A. \(\left\{ \begin{array}{l} x = - 1 - t\\ y = 1 + t\\ z = 1 + 3t \end{array} \right.\left( {t \in R} \right)\)
B. \(\left\{ \begin{array}{l} x = - t\\ y = 1 + t\\ z = 3 + t \end{array} \right.\left( {t \in R} \right)\)
C. \(\left\{ \begin{array}{l} x = - 1 - t\\ y = 1 - t\\ z = 3 + t \end{array} \right.\left( {t \in R} \right)\)
D. \(\left\{ \begin{array}{l} x = - 1 - t\\ y = 1 + t\\ z = 3 + t \end{array} \right.\left( {t \in R} \right)\)
A. \(\frac{{x - 1}}{{ - 1}} = \frac{{y - 2}}{{ - 3}} = \frac{{z - 3}}{{ - 5}}\)
B. \(\frac{x}{2} = \frac{{y + 1}}{1} = \frac{{z - 1}}{1}\)
C. \(\frac{{x - 1}}{1} = \frac{{y - 2}}{3} = \frac{{z - 3}}{5}\)
D. \(\frac{{x - 1}}{1} = \frac{{y - 2}}{{ - 3}} = \frac{{z - 3}}{{ - 5}}\)
A. 2x - y + 2z + 22 = 0
B. 2x - y + 2z + 13 = 0
C. 2x - y + 2z - 13 = 0
D. 2x + y + 2z - 22 = 0
A. \(\left( {0; - \frac{4}{3};\frac{2}{3}} \right)\)
B. \(\left( {0; - \frac{2}{3};\frac{4}{3}} \right)\)
C. \(\left( {0; - \frac{2}{3};\frac{8}{3}} \right)\)
D. \(\left( {0;\frac{2}{3}; - \frac{8}{3}} \right)\)
A. x + y - z + 1 = 0
B. x - y - z + 1 = 0
C. x + y - 2z - 3 = 0
D. x + y + z - 3 = 0
A. \(S = \sqrt 3 \)
B. \(S = \sqrt 2 \)
C. \(S = \frac{1}{2}\)
D. S = 1
A. x + 2y + 3z - 8 = 0
B. x + y + z - 4 = 0
C. x + 2y + z - 6 = 0
D. \(\frac{x}{1} + \frac{y}{2} + \frac{z}{1} = 1\)
A. \(\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1\)
B. \(x + \frac{y}{2} + \frac{z}{3} = 3\)
C. x + y + z - 6 = 0
D. x + 2y + 3z - 14 = 0
A. M(3;0;-1)
B. M(1;0;0)
C. M(1;0;3)
D. M(1;1;3)
A. \(\frac{9}{{14}}\)
B. \(\frac{3}{{14}}\)
C. \(\frac{9}{{\sqrt {14} }}\)
D. \(\frac{3}{{\sqrt {14} }}\)
A. 2x + y - 2z - 10 = 0
B. 2x + y - 2z - 12 = 0
C. x - 2y - z - 1 = 0
D. x - 4y + z - 3 = 0
A. 0
B. 1
C. 2
D. 3
A. D(0;-7;0) hoặc D(0;-8;0).
B. D(0;9;0) hoặc D(0;8;0).
C. D(0;7;0) hoặc D(0;8;0).
D. D(0;-7;0) hoặc D(0;8;0).
A. \(\frac{\sqrt{31}}{5}.\)
B. \(\frac{\sqrt{30}}{5}.\)
C. \(\frac{\sqrt{32}}{5}.\)
D. \(\frac{\sqrt{33}}{5}.\)
A. \(\frac{\sqrt{3}}{2}\)
B. \(\frac{\sqrt{3}}{3}\)
C. \(\frac{\sqrt{3}}{4}\)
D. \(\frac{\sqrt{3}}{5}\)
A. \(\frac{\sqrt{3}}{3}.\)
B. \(\frac{\sqrt{2}}{2}.\)
C. \(\frac{\sqrt{3}}{2}.\)
D. \(\frac{\sqrt{3}}{4}.\)
A. \(\frac{1}{3}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{5}\)
D. \(\frac{1}{6}\)
A. \(\overrightarrow{a}=5.\overrightarrow{u}+7.\overrightarrow{v}-\overrightarrow{w}.\)
B. \(\overrightarrow{a}=-5.\overrightarrow{u}+7.\overrightarrow{v}-\overrightarrow{w}.\)
C. \(\overrightarrow{a}=-5.\overrightarrow{u}+7.\overrightarrow{v}+\overrightarrow{w}.\)
D. \(\overrightarrow{a}=-5.\overrightarrow{u}-7.\overrightarrow{v}-\overrightarrow{w}.\)
A. \(\frac{\sqrt{3}}{13}.\)
B. \(\frac{\sqrt{3}}{3}.\)
C. \(\frac{\sqrt{13}}{3}.\)
D. \(\frac{\sqrt{13}}{13}.\)
Lời giải có ở chi tiết câu hỏi nhé! (click chuột vào câu hỏi).
Copyright © 2021 HOCTAPSGK