Find the cartesian equation of the line which | vedclass

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

Question

(A) The given line is $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.The direction ratios of this line are $a = 3, b = 5, c = 6$.Since the required l

Vedclass · Accepted Answer

$\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}$

Vedclass · Answer

(A) The given line is $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.The direction ratios of this line are $a = 3, b = 5, c = 6$.Since the required line is parallel to the given line,its direction ratios will also be proportional to $3, 5, 6$.The line passes through the point $(x_1, y_1, z_1) = (-2, 4, -5)$.The cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Substituting the values,we get $\frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6}$.Thus,the required equation is $\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}$.

Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and is parallel to the line given by $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

Similar Questions

The distance between the lines $\frac{x}{2}=\frac{y}{-1}=\frac{z}{2}$ and $\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z-2}{2}$ is

The vector equation of the line joining the points $i - 2j + k$ and $-2j + 3k$ is

Show that the points $A(1, 2, 7)$,$B(2, 6, 3)$,and $C(3, 10, -1)$ are collinear.

If $P$ is a point on the line parallel to the vector $2 \hat{i}-3 \hat{j}-6 \hat{k}$ and passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-2 \hat{k}$ and $AP=21$,then the position vector of $P$ can be

The equation of a line passing through the point $(2, -1, 1)$ and parallel to the line joining the points $\hat{i} + 2\hat{j} + 2\hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module