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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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,

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $\vec{r} = (1, 1, 1) + k(2, 3, 4), k \in R$?

Similar Questions

Let $L$ be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$, whose distance from the point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

The foot of the perpendicular from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$ is

The shortest distance between the lines $\vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(\hat{i} + 4\hat{j} - 5\hat{k})$ is:

The shortest distance between the skew-lines $\frac{x-3}{-1}=\frac{y-4}{2}=\frac{z+2}{1}$ and $\frac{x-1}{1}=\frac{y+7}{3}=\frac{z+2}{2}$ is

Find the equation of a line parallel to the $x$-axis and passing through the origin.

Vedclass Test Series

Exam Paper Generator

Online Exam Module