84 câu trắc nghiệm Toán 12 Chân trời sáng tạo Bài 1: Phương trình mặt phẳng có đáp án - Đề 1

A. \(x - {y^2} - 2 = 0\).

B. \(x + {z^2} - 3 = 0\).

C. \(x - z - 4 = 0\).

D. \({x^2} + {y^2} + {z^2} - 1 = 0\).

A. \({\vec n_1} = ( - 1;2;3)\).

B. \({\vec n_2} = (1;2;3)\).

C. \({\vec n_3} = ( - 1;2;0)\).

D. \({\vec n_4} = ( - x;2y;3)\).

A. \({\vec n_1} = (3;6;12)\).

B. \({\vec n_2} = (3x;6y;12z)\).

C. \({\vec n_3} = (3x; - 6y;12z)\).

D. \({\vec n_4} = ( - 1;2; - 4)\).

A. \({\vec n_1} = (3;4;1)\).

B. \({\vec n_2} = (3;4; - 1)\).

C. \({\vec n_3} = (3;4;5)\).D. \({\vec n_4} = (3;4; - 5)\).

A. \(x - {x_0} = 0\).

B. \(y - {y_0} = 0\).

C. \(z - {z_0} = 0\).

D. \(x + y + z - {x_0} - {y_0} - {z_0} = 0\).

A. \(\left| {{x_0}} \right|\).

B. \(\left| {{y_0}} \right|\).

C. \(\left| {{z_0}} \right|\).

D. \(\left| {{x_0} + {y_0} + {z_0}} \right|\).

A. \(\frac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}\).

B. \(\frac{{\left| {a{x_0} + b{y_0} + c{z_0}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\).

C. \(\frac{{\left| {a{y_0} + b{z_0} + c} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\).

D. \(\frac{{\left| {a{y_0} + b{z_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}\).

A. \(\left( {2;3; - 1} \right)\).

B. \(\left( {3;5; - 2} \right)\).

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

D. \(\left( {3; - 5; - 1} \right)\).

A. \[\overrightarrow c = \left( {2;6; - 1} \right)\].

B. \[\overrightarrow c = \left( {4;6; - 1} \right)\].

C. \[\overrightarrow c = \left( {4; - 6; - 1} \right)\].

D. \[\overrightarrow c = \left( {2; - 6; - 1} \right)\].

A. \(\vec n = (8;4; - 3)\).

B. \(\vec n = ( - 18;0; - 3)\).

C. \(\vec n = ( - 18;4; - 3)\).

D. \(\vec n = (1;4; - 3)\).

A. \(x - 3{y^2} + z - 1 = 0\).

B. \({x^2} + 2y + 4z - 2 = 0\).

C. \(2x - 3y + 4z - 2024 = 0\).

D. \(2x - 3y + 4{z^2} - 2025 = 0\).

A. \(\overrightarrow n = ( - 3;1; - 2)\).

B. \(\overrightarrow n = (3;1;2)\)

C. \(\overrightarrow n = (3; - 1;2)\)

D. \(\overrightarrow n = (6; - 2;4)\)

A. \(I\left( {1;1;1} \right)\)

B. \(R = 5\)

C. \(h = d\left( {I,(P)} \right) = \frac{{\left| {1 + 2 + 2 + 7} \right|}}{{\sqrt {{1^2} + {2^2} + {2^2}} }} = 4\)

D. \(r = \sqrt {{R^2} - {h^2}} = \sqrt {25 - 16} = 3\)

A. \(\left( P \right):x + y - z - 13 = 0\)

B. \(\left( S \right)\)

C. \(\left( P \right)\)

D. \(\overrightarrow d = \left( {3;\;2;\;0} \right)\)

A. \(2\)

B. \(1\)

C. \(\left( S \right)\)

D. \(A\left( {1\,;\,2\,;\, - 1} \right)\)

A. \(I\left( {0;5;0} \right)\).

B. \(MN\).

C. \(\left( P \right)\).

D. \(\left( P \right):3x - y + 4z + 5 = 0\).

A. \[M\left( {3;\,2;\,1} \right)\]

B. \[3\sqrt 3 \]

C. \[X\left( {a;\,b;\,c} \right)\]

D. \[a + b + c < - 2\]

A. \[\left( P \right)\]B. \[M\left( {3;\,2;\,1} \right)\]C. \[3\sqrt 3 \]D. \[ \Leftrightarrow \frac{{\left| {6 + d} \right|}}{{\sqrt 3 }} = 3\sqrt 3 \]

A. \(\left\{ \begin{array}{l}\left( {{P_m}} \right) \bot \left( \alpha \right)\\\left( {{Q_m}} \right) \bot \left( \alpha \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\overrightarrow {{n_1}} \bot \overrightarrow {{n_\alpha }} \\\overrightarrow {{n_2}} \bot \overrightarrow {{n_\alpha }} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\overrightarrow {{n_1}} .\overrightarrow {{n_\alpha }} = 0\\\overrightarrow {{n_2}} .\overrightarrow {{n_\alpha }} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}4m - 2 - 6n = 0\\4 + m - 6n = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = 2\\n = 1\end{array} \right..\).

B. \(m + n = 3\).

C. \[Oxyz\].

D. \[\left( P \right)\].

A. \[OM = ON\].

B. \[\left| d \right| = \left| {\frac{d}{b}} \right|\].

C. \[d = 0\].

D. \[O\].

A.

B.

C.

D.

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D. .

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B. .

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D. .

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A. \(z = 0\).

B. \(x = 0\).

C. \(x + y + z = 0\).

D. \(y = 0\).

A. \(z = 0\).

B. \(x = 0\).

C. \(y = 0\).

D. \(x + y = 0\).

A.

B.

C.

D.

A.

B.

C.

D.

A. \(3x - 5y - z - 6 = 0\).

B. \(3x - 5y - z + 6 = 0\).

C. \(3x + 5y - z + 6 = 0\).

D. \(3x - 5y + z - 6 = 0\).

A. \(2x + 6y - z - 7 = 0\).

B. \(2x - 6y - z + 5 = 0\).

C. \(2x + 6y + z + 5 = 0\).

D. \(2x - 6y - z + 7 = 0\).