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

(B) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b}

Vedclass · Accepted Answer

$\frac{x-1}{3}=\frac{y+3}{5}=\frac{z-5}{6}$

Vedclass · Answer

(B) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Since the required line is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$,it will have the same direction ratios,which are $(3, 5, 6)$.Given that the line passes through the point $(1, -3, 5)$,we substitute these values into the formula:$\frac{x-1}{3} = \frac{y-(-3)}{5} = \frac{z-5}{6}$This simplifies to $\frac{x-1}{3} = \frac{y+3}{5} = \frac{z-5}{6}$.Thus,the correct option is $B$.

The Cartesian equation of the line which is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ and passing through the point $(1, -3, 5)$ is:

Similar Questions

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then $k=$

If the shortest distance between the lines $r=(3 \hat{i}+4 \hat{j}-2 \hat{k})+t(-\hat{i}+2 \hat{j}+\hat{k})$ and $r=(\hat{i}-7 \hat{j}-2 \hat{k})+s(\hat{i}+3 \hat{j}+2 \hat{k})$ is equivalent to the projection of $P=-2 \hat{i}+11 \hat{j}$ on $Q$,then a possible vector $Q$ is

The distance of the point $B(1, 2, 3)$ from the line passing through $A(4, 2, 2)$ and parallel to the vector $\vec{c} = 2i + 3j + 6k$ is

The ratio in which the $YZ$-plane divides the line segment joining the points $(2, 4, 5)$ and $(3, 5, -4)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module