Cho hai điểm A(1;−2;0),B(0;1;1), độ dài đường cao OH của tam giác OAB là:

A.\(\left[ {\begin{array}{*{20}{c}}{m = 4}\\{m = - 2}\end{array}} \right.\)

B. \(\left[ {\begin{array}{*{20}{c}}{m = - 4}\\{m = 2}\end{array}} \right.\)

C. \(\left[ {\begin{array}{*{20}{c}}{m = 4}\\{m = 2}\end{array}} \right.\)

D. \(\left[ {\begin{array}{*{20}{c}}{m = - 4}\\{m = - 2}\end{array}} \right.\)

A.\(\left\{ {\begin{array}{*{20}{c}}{\left[ {\vec u,\overrightarrow {u'} } \right] \ne \vec 0}\\{\left[ {\vec u,\overrightarrow {u'} } \right]\overrightarrow {MM'} = 0}\end{array}} \right.\)

B. \[\left[ {\vec u,\overrightarrow {u'} } \right] \ne \vec 0\]

C. \[\left[ {\vec u,\overrightarrow {u'} } \right]\overrightarrow {MM'} = 0\]

D. \[\left[ {\vec u,\overrightarrow {u'} } \right] = \vec 0\]

A.d//d′

B.d≡d′

C.d cắt d′

D.d chéo d′

A.Song song.

B.Trùng nhau.

C.Cắt nhau.

D.Chéo nhau.

A.\({d_1}:\left\{ {\begin{array}{*{20}{c}}{x = 3t}\\{y = 1 + t}\\{z = 5t}\end{array}} \right.\)

B. \({d_2}:\left\{ {\begin{array}{*{20}{c}}{x = 2}\\{y = 2 + t}\\{z = 1 + t}\end{array}} \right.\)

C. \[{d_3}:\frac{{x - 2}}{3} = \frac{y}{2} = \frac{{z - 1}}{{ - 5}}\]

D. \[{d_4}:\frac{{x + 2}}{2} = \frac{y}{{ - 1}} = \frac{{z + 1}}{2}\]

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

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

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

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