设一平面垂直于平面$z=0$，并通过从点$(1, - 1,1)$到直线$\left\{ \matrix{ y - z + 1 = 0\cr x = 0\cr} \right.$的垂线，则此平面方程为( ).
A: $x + 2y + 1 = 0$
B: $x + 2y = 0$
C: $x + 2y + 1 +z= 0$
D: $x + 2y + 2 = 0$
A: $x + 2y + 1 = 0$
B: $x + 2y = 0$
C: $x + 2y + 1 +z= 0$
D: $x + 2y + 2 = 0$
举一反三
- 曲线\( \left\{ {\matrix{ { { x^2} + {y^2} = {z^2}} \cr { { z^2} = y} \cr } } \right. \)在坐标面\( yoz \) 上的投影曲线方程为（ ） A: \( \left\{ {\matrix{ { { x^2} + { { \left( {y - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \) B: \( \left\{ {\matrix{ { { z^2} = y} \cr {x = 0} \cr } } \right. \) C: \( \left\{ {\matrix{ {z = {y^2}} \cr {x = 0} \cr } } \right. \) D: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)
- 过点(1, -2, -2)且与平面x -2 y + 3z = 2平行的平面方程为 A: x -2 y + z = 6; B: x -2y + 3z = 0; C: x -2y + 3z = 0; D: 2x - y + 3z = 9.
- 求解常微分方程初值问题[img=224x61]1803072f6b2a05a.png[/img]应用的语句是 A: DSolve[2y[x]y"[x]==1+(y'[x])^2,y[0]==1,y'[0]==0,y[x],x B: DSolve[{2y[x]y" [x]==1+(y'[x])^2,y[0]==1,y'[0]==0},y[x],x] C: DSolve[{2y[x]y" [x]==1+(y^' [x])^2;y[0]==1;y'[0]==0},y[x],x] D: DSolve[{2yy"==1+(y^' )^2&&y[0]==1&&y'[0]==0},y[x],x]
- 设\(f\left( {x,y,z} \right) = x{y^2} + y{z^2} + z{x^2}\)，则\({f_{yz}}\left( {0,-1,0} \right) = \)( ) A: 1 B: 0 C: -1 D: 2
- 已知直线的一般方程\( \left\{ {\matrix{ {x - 2y - z + 4 = 0} \cr {5x + y - 2z + 8 = 0} \cr } } \right. \), 则其点向式方程为（ ） A: \( { { x - 2} \over 2} = {y \over { - 3}} = { { z - 4} \over {11}} \) B: \( {x \over 5} = {y \over { - 3}} = { { z - 4} \over {11}} \) C: \( { { x - 2} \over 5} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \) D: \( { { x - 2} \over 2} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \)