The Cartesian equation of the line which is p | vedclass

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

Question

(D) The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and parallel to a vector $\vec{b} = a \hat{i} + b \hat{j} + c \hat{k}$

Vedclass · Accepted Answer

$\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$

Vedclass · Answer

(D) The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and parallel to a vector $\vec{b} = a \hat{i} + b \hat{j} + c \hat{k}$ is given by the formula: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ Here,the point is $(x_1, y_1, z_1) = (5, 2, -4)$ and the direction vector is $\vec{b} = 3 \hat{i} + 2 \hat{j} - 8 \hat{k}$,so $a=3, b=2, c=-8$. Substituting these values into the formula,we get: $\frac{x-5}{3} = \frac{y-2}{2} = \frac{z-(-4)}{-8}$ $\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$ Thus,the correct option is $D$.

The Cartesian equation of the line which is parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and passes through the point $(5, 2, -4)$ is . . . . . . .

Similar Questions

If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$,then the largest possible value of $|\lambda|$ is equal to ..........

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4},$ $l \neq 0$ is $(3,5,7),$ then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to:

The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to $............$

The angle between two lines $\frac{x + 1}{2} = \frac{y + 3}{2} = \frac{z - 4}{-1}$ and $\frac{x - 4}{1} = \frac{y + 4}{2} = \frac{z + 1}{2}$ is

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module