The equation X = 0 represents:
YZ — Plane
In a three-dimensional coordinate system, points are represented by three coordinates: $(\text{x}, \text{y}, \text{z})$. Each coordinate describes the position of a point along a specific axis:
When we are given an equation like $\text{X} = 0$, it means that we are looking for all points $(\text{x}, \text{y}, \text{z})$ where the value of the x-coordinate is always zero. This condition restricts the position of the point. If $\text{x} = 0$, the point cannot move along the X-axis away from the origin (0,0,0) in the x-direction. The point will always be located on a surface where its distance from the YZ-plane along the X-axis is zero.
Consider a point in 3D space where the x-coordinate is fixed at zero. The y-coordinate and the z-coordinate, however, can take any real value. This defines a flat surface that extends infinitely in the Y and Z directions while always passing through the origin along the X-axis (since X is 0).
This specific flat surface is known as the YZ-plane. All points on the YZ-plane have an x-coordinate of 0. For example, points like $(\text{0}, \text{5}, \text{2})$, $(\text{0}, \text{-1}, \text{10})$, and $(\text{0}, \text{0}, \text{0})$ all lie on the YZ-plane because their x-coordinate is 0. The YZ-plane contains both the Y-axis and the Z-axis, as for any point on either of these axes, the x-coordinate is 0.
Therefore, the equation $\text{X} = 0$ uniquely represents the YZ-plane, where the x-coordinate of every point on the plane is zero.
If the foot of the perpendicular drawn from (-2, 1, 0) on a plane is (1, -2, 1), then the equation of the plane is
The equation of the plane which contain the points (0, 6, 0) and (-2, -3, 4) and which is parallel to the ray with direction ratios (2, 3, -2) is:
The image of the point (–3, 8, 4) in the plane 6x – 3y – 2z + 1 = 0, is -
The equation of the plane through the point (1, 2, –3) and normal to the straight line joining the points (–1, 3, 4) and (5, 2, –1) is-
Determine the vector equation of the plane passing through the intersection of the planes \(\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6\) and \(\vec{r}. (2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=-5\) , and the point (1, 1, 1)?