The equation of the x-axis is: | vedclass

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

Question

(C) The $x$-axis passes through the origin $(0, 0, 0)$.Its direction ratios are $(1, 0, 0)$ because it is parallel to the $x$-axis.The symmetric form

Vedclass · Accepted Answer

$\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$

Vedclass · Answer

(C) The $x$-axis passes through the origin $(0, 0, 0)$.Its direction ratios are $(1, 0, 0)$ because it is parallel to the $x$-axis.The symmetric form 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 $(x_1, y_1, z_1) = (0, 0, 0)$ and $(a, b, c) = (1, 0, 0)$,we get $\frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{0}$.This simplifies to $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$.

The equation of the $x$-axis is:

Similar Questions

When are the two lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ perpendicular to each other?

If for some $\alpha \in R$,the lines $L_1: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $L_2: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}$ are coplanar,then the line $L_2$ passes through the point

The point of intersection of the lines $\vec{r}=2 \vec{b}+t(6 \vec{c}-\vec{a})$ and $\vec{r}=\vec{a}+s(\vec{b}-3 \vec{c})$ is

Find the coordinates of the foot of the perpendicular drawn from the point $A(1, 8, 4)$ to the line joining the points $B(0, -1, 3)$ and $C(2, -3, -1)$.

The parametric equations of the line passing through $A(3, 4, -7)$ and $B(1, -1, 6)$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module