If the two lines frac{x-1}{2}=frac{y+1}{3}=fr | vedclass

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

Question

(C) Let the first line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$. Any point on this line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$. Since

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(C) Let the first line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$. Any point on this line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$. Since the lines have a point in common,this point must satisfy the second line equation $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$. Substituting the coordinates into the second line equation: $\frac{2\lambda+1-3}{1} = \frac{3\lambda-1-k}{2} = \frac{4\lambda+1}{1}$. Equating the first and third parts: $2\lambda-2 = 4\lambda+1 \Rightarrow -3 = 2\lambda \Rightarrow \lambda = -\frac{3}{2}$. Now,equate the first and second parts: $\frac{2\lambda-2}{1} = \frac{3\lambda-1-k}{2}$. Substituting $\lambda = -\frac{3}{2}$: $2(-\frac{3}{2})-2 = \frac{3(-\frac{3}{2})-1-k}{2} \Rightarrow -3-2 = \frac{-\frac{9}{2}-1-k}{2} \Rightarrow -5 = \frac{-\frac{11}{2}-k}{2} \Rightarrow -10 = -\frac{11}{2}-k \Rightarrow k = -\frac{11}{2} + 10 = \frac{9}{2}$.

If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ have a point in common,then $k=$

Similar Questions

The equation of the line of shortest distance between the lines $\frac{x - 6}{3} = \frac{y - 7}{-1} = \frac{z - 4}{1}$ and $\frac{x}{-3} = \frac{y + 9}{2} = \frac{z - 2}{4}$ is:

Show that the straight lines whose direction cosines are given by $2l + 2m - n = 0$ and $mn + nl + lm = 0$ are at right angles.

Show that the line through the points $(1, -1, 2)$ and $(3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.

The graph of the equation $y^2 + z^2 = 0$ in three-dimensional space is

If the direction cosines of two lines are given by $l+m+n=0$ and $mn-2lm-2nl=0$,then the acute angle between those lines is

Vedclass Test Series

Exam Paper Generator

Online Exam Module