Trắc nghiệm Phương trình đường thẳng trong không gian có đáp án (Nhận biết)

A. $d:\left(\begin{array}{c}x=x_0+\mathrm{at}\\ y=y_0+\mathrm{bt}\\ z=z_0+\mathrm{ct}\end{array}\right)\left(t\in Z\right)$

B. $d:\left(\begin{array}{c}x=x_0+\mathrm{at}\\ y=y_0+\mathrm{bt}\\ z=z_0+\mathrm{ct}\end{array}\right)\left(t\in R\right)$

C. $d:\left(\begin{array}{c}x=a+x_{0}t\\ y=b+y_{0}t\\ z=c+z_{0}t\end{array}\right)\left(t\in R\right)$

D. $d:\left(\begin{array}{c}x=a+x_{0}t\\ y=b+y_{0}t\\ z=c+z_{0}t\end{array}\right)\left(t\in Z\right)$

A. $\frac{x-a}{x_0}=\frac{y-b}{y_0}=\frac{z-c}{z_0}$

B. $\frac{x-x_0}{x_0}=\frac{y-y_0}{y_0}=\frac{z-z_0}{z_0}$

C. $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

D. $d:\left(\begin{array}{c}x=x_0+\mathrm{at}\\ y=y_0+\mathrm{bt}\\ z=z_0+\mathrm{ct}\end{array}\right)\left(t\in R\right)$

A. $\left(a;b;c\right)$

B. $\left(a;-b;c\right)$

C. $\left(x_0;y_0;z_0\right)$

D. $\left(-x_0;-y_0;-z_0\right)$

A. $\left(a;b;c\right)$

B. $\left(a;-b;c\right)$

C. $\left(x_0;y_0;z_0\right)$

D. $\left(-x_0;-y_0;-z_0\right)$

A. $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

B. $\frac{x-x_0}{-a}=\frac{y-y_0}{-b}=\frac{z-z_0}{-c}$

C. $\frac{x+x_0}{a}=\frac{y+y_0}{b}=\frac{z+z_0}{c}$

D. $\frac{x+x_0}{a}=\frac{y+y_0}{-b}=\frac{z+z_0}{c}$

A. $\left(-1;-1;1\right)$

B. $\left(-1;1;1\right)$

C. $\left(0;1;1\right)$

D. $\left(0;1;0\right)$

A. Phương trình chính tắc của $\left(d\right):\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

B. Phương trình tham số của $d:\left(\begin{array}{c}x=x_0+\mathrm{at}\\ y=y_0+\mathrm{bt}\\ z=z_0+\mathrm{ct}\end{array}\right)\left(t\in R\right)$

C. Nếu $k\ne 0$ thì $\stackrel{\to }{v}=k.\stackrel{\to }{u}$ là một vectơ chỉ phương của đường thẳng (d)

D. Phương trình chính tắc của $\left(d\right):\frac{x+x_0}{a}=\frac{y+y_0}{b}=\frac{z+z_0}{c}$

A. Phương trình chính tắc của $\left(d\right):\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

B. Phương trình tham số của $d:\left(\begin{array}{c}x=x_0+\mathrm{at}\\ y=y_0-\mathrm{bt}\\ z=z_0-\mathrm{ct}\end{array}\right)\left(t\in R\right)$

C. Đường thẳng d chỉ có duy nhất một vec tơ chỉ phương là $\stackrel{\to }{u}=\left(a;b;c\right)$

D. Phương trình chính tắc của $\left(d\right):\frac{x+x_0}{a}=\frac{y+y_0}{b}=\frac{z+z_0}{c}$

A. $\stackrel{\to }{u_1}=\left(0;3;-1\right)$

B. $\stackrel{\to }{u_1}=\left(1;3;-1\right)$

C. $\stackrel{\to }{u_1}=\left(1;-3;-1\right)$

D. $\stackrel{\to }{u_1}=\left(1;2;5\right)$

A. $\left(\begin{array}{c}x=t\\ y=t\\ z=t\end{array}\right)$

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

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

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

A. $\left(\begin{array}{c}x=t\\ y=t\\ z=t\end{array}\right)\left(t\in R\right)$

B. $\left(\begin{array}{c}x=t\\ y=0\\ z=0\end{array}\right)\left(t\in R\right)$

C. $\left(\begin{array}{c}x=t\\ y=t\\ z=0\end{array}\right)\left(t\in R\right)$

D. $\left(\begin{array}{c}x=0\\ y=0\\ z=t\end{array}\right)\left(t\in R\right)$

A. $M\left(0;0;3\right)$

B. $N\left(0;1;0\right)$

C. $P\left(-2;0;0\right)$

D. $Q\left(1;0;1\right)$

A. $\left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right)=\stackrel{\to }{0}$

B. $\left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right)=\left(\stackrel{\to }{u},\stackrel{\to }{\mathrm{MM}\text{'}}\right)$

C. $\left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right)=\left(\stackrel{\to }{u},\stackrel{\to }{\mathrm{MM}\text{'}}\right)=\stackrel{\to }{0}$

D. $\left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right)\ne \left(\stackrel{\to }{u},\stackrel{\to }{\mathrm{MM}\text{'}}\right)$

A. $d//d\text{'}$

B. $d\equiv d\text{'}$

C. d cắt d’

D. A hoặc B đúng

A. $\left(\begin{array}{c}\left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right)\ne \stackrel{\to }{0}\\ \left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right).\stackrel{\to }{\mathrm{MM}\text{'}}=0\end{array}\right)$

B. $\left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right)\ne \stackrel{\to }{0}$

C. $\left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right).\stackrel{\to }{\mathrm{MM}\text{'}}=0$

D. $\left(\stackrel{\to }{u},\stackrel{\to }{u\text{'}}\right)=\stackrel{\to }{0}$

A. $d//d\text{'}$

B. $d\perp d\text{'}$

C. $d\equiv d\text{'}$

D. d cắt d'

A. d // d'

B. d $\equiv$ d'

C. d cắt d'

D. d chéo d'

A. Cắt nhau

B. Song song

C. Chéo nhau

D. Trùng nhau

A. $d\left(A;d\text{'}\right)=\frac{\left(\left(\stackrel{\to }{\mathrm{AM}\text{'}},\stackrel{\to }{u\text{'}}\right)\right)}{\left(\stackrel{\to }{u\text{'}}\right)}$

B. $d\left(A;d\text{'}\right)=\frac{\left(\left(\stackrel{\to }{\mathrm{AM}\text{'}},\stackrel{\to }{u\text{'}}\right)\right)}{\stackrel{\to }{u\text{'}}}$

C. $d\left(A;d\text{'}\right)=\frac{\left(\stackrel{\to }{\mathrm{AM}\text{'}},\stackrel{\to }{u\text{'}}\right)}{\stackrel{\to }{u\text{'}}}$

D. $d\left(A;d\text{'}\right)=\frac{\left(\stackrel{\to }{\mathrm{AM}\text{'}},\stackrel{\to }{u\text{'}}\right)}{\left(\stackrel{\to }{u\text{'}}\right)}$

A. $\mathrm{cos\phi }=\frac{\left(\stackrel{\to }{u}.\stackrel{\to }{u\text{'}}\right)}{\left(\stackrel{\to }{u}\right).\left(\stackrel{\to }{u\text{'}}\right)}$

B. $\mathrm{cos\phi }=\frac{\stackrel{\to }{u}.\stackrel{\to }{u\text{'}}}{\left(\stackrel{\to }{u}\right).\left(\stackrel{\to }{u\text{'}}\right)}$

C. $\mathrm{cos\phi }=-\frac{\stackrel{\to }{u}.\stackrel{\to }{u\text{'}}}{\left(\stackrel{\to }{u}\right).\left(\stackrel{\to }{u\text{'}}\right)}$

D. $\mathrm{cos\phi }=-\frac{\left(\stackrel{\to }{u}.\stackrel{\to }{u\text{'}}\right)}{\left(\stackrel{\to }{u}\right).\left(\stackrel{\to }{u\text{'}}\right)}$