The parametric equations of the line passing | vedclass

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

Question

(A) The line passes through points $A(3, 4, -7)$ and $B(1, -1, 6)$.First,find the direction vector $\vec{v} = B - A = (1 - 3, -1 - 4, 6 - (-7)) = (-2,

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The line passes through points $A(3, 4, -7)$ and $B(1, -1, 6)$.First,find the direction vector $\vec{v} = B - A = (1 - 3, -1 - 4, 6 - (-7)) = (-2, -5, 13)$.The parametric equations of a line passing through $(x_1, y_1, z_1)$ with direction vector $(a, b, c)$ are given by $x = x_1 + a\lambda, y = y_1 + b\lambda, z = z_1 + c\lambda$.Using point $A(3, 4, -7)$ and direction vector $(-2, -5, 13)$,we get:$x = 3 - 2\lambda$$y = 4 - 5\lambda$$z = -7 + 13\lambda$Thus,the correct option is $A$.

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

Similar Questions

$P, Q, R$ and $S$ are four points with the position vectors $3i-4j+5k, 0i+0j+4k, -4i+5j+1k$ and $-3i+4j+3k$,respectively. Then,the line $PQ$ meets the line $RS$ at the point

The point of intersection of the lines $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$ is

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

The square of the distance of the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + \hat{k})$ from the origin is:

The vector equation of the line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module