The Cartesian equation of the line passing th | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{(x-1)}{1}=\frac{(y+2)}{1}=\frac{(z-1)}{3}$

Vedclass · Answer

(C) The equation of a line passing through a point $\vec{a} = x_1 \bar{i} + y_1 \bar{j} + z_1 \bar{k}$ and parallel to a vector $\vec{b} = a \bar{i} + b \bar{j} + c \bar{k}$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Given point is $(1, -2, 1)$,so $x_1 = 1, y_1 = -2, z_1 = 1$.Given parallel vector is $\bar{i} + \bar{j} + 3 \bar{k}$,so $a = 1, b = 1, c = 3$.Substituting these values into the formula,we get $\frac{x-1}{1} = \frac{y-(-2)}{1} = \frac{z-1}{3}$.This simplifies to $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-1}{3}$.

The Cartesian equation of the line passing through the point $\bar{i}-2 \bar{j}+\bar{k}$ and parallel to the vector $\bar{i}+\bar{j}+3 \bar{k}$ is

Similar Questions

The shortest distance between the lines $r = (3t - 4)\hat{i} - 2\hat{j} - (1 + 2t)\hat{k}$ and $r = (6 + s)\hat{i} + (2 - 2s)\hat{j} + 2(1 + s)\hat{k}$ is

The perpendicular distance of the point $(2, 4, -1)$ from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9}$ is

The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}$:

Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{-1}$ and $\frac{x + 5}{7} = \frac{y - 2}{-6} = \frac{z - 3}{4}$ intersect at the point $R$. The reflection of $R$ in the $xy$-plane has coordinates

Vedclass Test Series

Exam Paper Generator

Online Exam Module