Bài tập Viết phương trình đường thẳng đi qua 1 điểm và có một vectơ chỉ phương lớp 12 (có lời giải)

A. \(\frac{{x - 1}}{4} = \frac{y}{2} = \frac{{ - z - 5}}{5}\);

B. \(\frac{{x + 1}}{4} = \frac{y}{2} = \frac{{z + 5}}{5}\);

C. \(\frac{{x - 1}}{4} = \frac{y}{2} = \frac{{z - 5}}{5}\);

D. \(\frac{{x - 1}}{5} = \frac{y}{2} = \frac{{z - 5}}{4}\).

A. \(\left\{ \begin{array}{l}x = - 3 + t\\y = 2 + 2t\\z = 1 - 3t\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = 1 - 3t\\y = 2 + 2t\\z = - 3 + t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = t\\y = 2t\\z = - 3t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = - 3t\\y = 2t\\z = t\end{array} \right.\).

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

B. \(\frac{{x - 1}}{1} = \frac{y}{3} = \frac{{z + 2}}{{ - 2}}\);

C. \(\frac{{x + 1}}{2} = \frac{y}{3} = \frac{{z - 2}}{{ - 2}}\);

D. \(\frac{{x - 1}}{2} = \frac{y}{3} = \frac{{z + 2}}{1}\).

A. z = 0;

B. \(\left\{ \begin{array}{l}x = 0\\y = t\\z = 0\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = t\\y = 0\\z = 0\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = 0\\y = 0\\z = t\end{array} \right.\).

A. x = 0;

B. y + z = 0;

C. \(\left\{ \begin{array}{l}x = 0\\y = 0\\z = t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = t\\y = 0\\z = 0\end{array} \right.\).

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

B. \(\frac{{x + 1}}{2} = \frac{{y + 3}}{{ - 1}} = \frac{{z - 2}}{1}\);

C. \(\frac{{x - 1}}{2} = \frac{{y - 3}}{{ - 1}} = \frac{{z + 2}}{1}\);

D. \(\frac{{x - 1}}{2} = \frac{{y + 3}}{{ - 1}} = \frac{{z + 2}}{1}\).

A. \(\frac{{x - 1}}{2} = \frac{{y + 3}}{{ - 3}} = \frac{{z + 6}}{8}\);

B. \(\frac{{x + 1}}{2} = \frac{{y - 3}}{{ - 3}} = \frac{{z - 6}}{8}\);

C. \(\frac{{x + 1}}{{ - 2}} = \frac{{y - 3}}{3} = \frac{{z - 6}}{8}\);

D. \(\frac{{x + 1}}{2} = \frac{{y - 3}}{3} = \frac{{z - 6}}{8}\).

A. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 - 4t\\z = 1 + 3t\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 - 2t\\z = 1 + 2t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = 2 + t\\y = - 1 - 2t\\z = 1 + t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 - t\\z = 1 + t\end{array} \right.\).

A. \(\left\{ \begin{array}{l}x = - 2t\\y = - 1 + t\\z = 3 + t\end{array} \right.\);

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

C. \(\frac{{x - 2}}{{ - 2}} = \frac{{y - 1}}{1} = \frac{{z - 3}}{1}\);

D. \(\frac{{x - 2}}{{ - 2}} = \frac{y}{1} = \frac{{z - 1}}{1}\).