Which of the following lines in {R^3} is coin | vedclass

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(B) The given line is $\vec{r} = (1, 1, 1) + k(2, 3, 4)$. This line passes through the point $A(1, 1, 1)$ and has a direction vector $\vec{v} = (2, 3,

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$\frac{3 - x}{6} = \frac{4 - y}{9} = \frac{5 - z}{12}$

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(B) The given line is $\vec{r} = (1, 1, 1) + k(2, 3, 4)$. This line passes through the point $A(1, 1, 1)$ and has a direction vector $\vec{v} = (2, 3, 4)$.For two lines to be coincident,they must have the same direction vector (or a scalar multiple) and the same point must satisfy both equations.Consider option $B$: $\frac{3 - x}{6} = \frac{4 - y}{9} = \frac{5 - z}{12}$.This can be rewritten as $\frac{x - 3}{-6} = \frac{y - 4}{-9} = \frac{z - 5}{-12}$.The direction vector is $(-6, -9, -12) = -3(2, 3, 4)$,which is parallel to the given line.Now,check if the point $(1, 1, 1)$ lies on this line:$\frac{3 - 1}{6} = \frac{2}{6} = \frac{1}{3}$$\frac{4 - 1}{9} = \frac{3}{9} = \frac{1}{3}$$\frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3}$Since all ratios are equal to $\frac{1}{3}$,the point $(1, 1, 1)$ lies on the line.Thus,the line in option $B$ is coincident with the given line.

Which of the following lines in ${R^3}$ is coincident with the line $\vec{r} = (1, 1, 1) + k(2, 3, 4), k \in R$?

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