20 câu Trắc nghiệm Toán 12 Kết nối tri thức Bài 15. Phương trình đường thẳng trong không gian có đáp án

A. \[\overrightarrow u = \left( { - 1;2;1} \right).\]

B. \[\overrightarrow u = \left( {2;1;0} \right).\]

C. \[\overrightarrow u = \left( {2;1;1} \right).\]

D. \[\overrightarrow u = \left( { - 1;2;0} \right).\]

A. \[\overrightarrow u = \left( {5; - 2; - 1} \right).\]

B. \[\overrightarrow u = \left( {10; - 4;1} \right).\]

C. \[\overrightarrow u = \left( {0;2; - 11} \right).\]

D. \[\overrightarrow u = \left( {0;2;11} \right).\]

A. \[M\left( {4;1;3} \right).\]

B. \[N\left( {2;1;5} \right).\]

C. \[P\left( {3; - 2; - 1} \right).\]

D. \[Q\left( {1; - 3;4} \right).\]

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

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

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

D. \[\frac{{x + 1}}{3} = \frac{{y - 3}}{{ - 3}} = \frac{{z + 5}}{2}.\]

A. \[\frac{{x - 5}}{2} = \frac{{y - 4}}{1} = \frac{{z + 1}}{2}.\]

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

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

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

A. \[\left\{ \begin{array}{l}x = 2 + 2t\\y = - t\\z = - 1 + t\end{array} \right.{\rm{ }}\left( {t \in \mathbb{R}} \right).\]

B. \[\left\{ \begin{array}{l}x = 2 + 2t\\y = - t\\z = - 1 - t\end{array} \right.{\rm{ }}\left( {t \in \mathbb{R}} \right).\]

C. \[\left\{ \begin{array}{l}x = 2 + 2t\\y = - 1\\z = - 1 + t\end{array} \right.{\rm{ }}\left( {t \in \mathbb{R}} \right).\]

D. \[\left\{ \begin{array}{l}x = 2 + 2t\\y = - t\\z = 1 - t\end{array} \right.{\rm{ }}\left( {t \in \mathbb{R}} \right).\]

A. \[Q\left( {5; - 2;3} \right).\]

B. \[N\left( { - 1;1;0} \right).\]

C. \[M\left( {3; - 1;2} \right).\]

D. \[P\left( { - 3;2;1} \right).\]

A. Chéo nhau.

B. Trùng nhau.

C. Song song.

D. Cắt nhau.

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

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

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

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

A. \[\frac{{\sqrt {21} }}{6}.\]

B. \[\frac{{\sqrt 6 }}{2}.\]

C. \[2\sqrt 2 .\]

D. \[\frac{{\sqrt {14} }}{2}.\]

A. \[d \subset \left( P \right).\]

B. \[d\parallel \left( P \right).\]

C. \[d \bot \left( P \right).\]

D. \[d\] cắt \[\left( P \right).\]

A. \[m = - \frac{1}{2}.\]

B. \[m = 0,5.\]

C. \[m = 1.\]

D. \[m = \frac{1}{2}.\]

A. \[m \in \left\{ { - 1;3} \right\}.\]

B. \[m = - 1.\]

C. \[m = 3.\]

D. Không có giá trị \[m\] thỏa mãn.

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

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

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

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

A. \[\left\{ \begin{array}{l}x = 1 + 3t\\y = 3t\\z = 1 - t.\end{array} \right.\]

B. \[\left\{ \begin{array}{l}x = 1 + t\\y = 3t\\z = 1 - t.\end{array} \right.\]

C. \[\left\{ \begin{array}{l}x = 1 + 3t\\y = 1 + 3t\\z = 1 - t.\end{array} \right.\]

D. \[\left\{ \begin{array}{l}x = 1 + 3t\\y = 3t\\z = 1 + t.\end{array} \right.\]

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

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

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

D. \[d:\frac{x}{2} = \frac{{y + 2}}{1} = \frac{{z + 2}}{1}.\]

A. \[\frac{x}{2} = \frac{y}{3} = \frac{{z + 3}}{{ - 5}}.\]

B. \[\frac{x}{{ - 2}} = \frac{y}{3} = \frac{{z + 3}}{5}.\]

C. \[\frac{x}{{ - 2}} = \frac{y}{3} = \frac{{z - 3}}{5}.\]

D. \[\frac{x}{2} = \frac{y}{3} = \frac{{z - 3}}{{ - 5}}.\]

A. \[\frac{{x - 4}}{8} = \frac{{y - 3}}{1} = \frac{{z - 2}}{{ - 4}}.\]

B. \[\frac{{x - 4}}{9} = \frac{{y - 3}}{2} = \frac{{z - 2}}{{ - 1}}.\]

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

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

A. \[\left\{ \begin{array}{l}x = 1\\y = 2 + t\\z = 3 + 4t.\end{array} \right.\]

B. \[\left\{ \begin{array}{l}x = 1 + t\\y = 1 + 3t\\z = 3 + 2t.\end{array} \right.\]

C. \[\left\{ \begin{array}{l}x = 0\\y = - 1 + t\\z = 3 + 4t.\end{array} \right.\]

D. \[\left\{ \begin{array}{l}x = 2 + t\\y = 3 + 3t\\z = 3 + 2t.\end{array} \right.\]

A. 1.

B. 0.

C. 2.

D. −1.