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

The vector equation of the line passing through $P(1, 2, 3)$ and $Q(2, 3, 4)$ is

If the line $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$,then $4\lambda+9\mu$ is equal to :

Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3 \hat{i} + 2 \hat{j} - 2 \hat{k}$.

Lines $\frac{2x-5}{k} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other. Then,the value of $k$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module