The parametric equations of the line passing | vedclass

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

Question

(C) The vector equation of a line passing through points $A(\vec{a})$ and $B(\vec{b})$ is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.He

Vedclass · Accepted Answer

$x=3-2 \lambda, \quad y=4-5 \lambda, \quad z=-7+13 \lambda$

Vedclass · Answer

(C) The vector equation of a line passing through points $A(\vec{a})$ and $B(\vec{b})$ is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.Here,$\vec{a} = 3\hat{i} + 4\hat{j} - 7\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 6\hat{k}$.The direction vector $\vec{v} = \vec{b} - \vec{a} = (1-3)\hat{i} + (-1-4)\hat{j} + (6 - (-7))\hat{k} = -2\hat{i} - 5\hat{j} + 13\hat{k}$.Thus,the parametric equations are $x = x_1 + v_x \lambda$,$y = y_1 + v_y \lambda$,$z = z_1 + v_z \lambda$.Substituting the values: $x = 3 - 2\lambda$,$y = 4 - 5\lambda$,$z = -7 + 13\lambda$.

The parametric equations of the line passing through the points $A(3,4,-7)$ and $B(1,-1,6)$ are

Similar Questions

The Cartesian equation of a line is $3x + 1 = 6y - 2 = -z + 1$. Find its vector equation.

The equation of the line of the shortest distance between the lines $\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$ and $\frac{x - 1}{0} = \frac{y + 1}{-2} = \frac{z}{1}$ is

$ABC$ 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:

The foot of the perpendicular from $(0,2,3)$ to the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module