The parametric equations of the line passing through $A(3, 4, -7)$ and $B(1, -1, 6)$ are:
- A$x = 3 - 2\lambda, \quad y = 4 - 5\lambda, \quad z = -7 + 13\lambda$
- B$x = -2 + 5\lambda, \quad y = -5 + 4\lambda, \quad z = 13 - 7\lambda$
- C$x = 1 + 3\lambda, \quad y = -1 + 4\lambda, \quad z = 6 - 7\lambda$
- D$x = 3 + \lambda, \quad y = -1 + 4\lambda, \quad z = -7 + 6\lambda$
Explore More
Similar Questions
Let $P$ be the point $(10, -2, -1)$ and $Q$ be the foot of the perpendicular drawn from the point $R(1, 7, 6)$ on the line passing through the points $(2, -5, 11)$ and $(-6, 7, -5)$. Then the length of the line segment $PQ$ is equal to ..........
$ABC$ is a triangle with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $(\lambda^3 + \mu^3 + 5)$ is equal to:
Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$. The distance of the point $A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $L$ along the line $\frac{3x-11}{2} = \frac{3y-11}{1} = \frac{3z-19}{2}$ is equal to:
DifficultJEE MAIN 2024
View SolutionThe equation of a line passing through the point $(2,1,3)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is
The direction cosines of two lines are connected by the relations $l+m-n=0$ and $lm-2mn+nl=0$. If $\theta$ is the acute angle between those lines,then $\cos \theta=$
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo