Home › JEE Main › Mathematics › Three Dimensional Geometry › The equation of a plane in normal form is $\vec{…
The equation of a plane in normal form is $\vec{r} \cdot \hat{n} = d$, where $\hat{n}$ is a unit vector normal to the plane and $d$ is:
AThe negative of the $z$-coordinate
BThe number of intercepts
CThe distance of the plane from the origin
DThe angle the plane makes with the $x$-axis
Answer & Solution
Correct answer: C. The distance of the plane from the origin
In the normal form $\vec{r} \cdot \hat{n} = d$, the constant $d$ is the *perpendicular distance* from the origin to the plane, measured along the unit normal $\hat{n}$.
Derivation: for any point $\vec{r}$ on the plane, the projection $\vec{r} \cdot \hat{n}$ is the perpendicular distance from origin to the plane (constant for all points on the plane). Call that constant $d$.
If $d > 0$, $\hat{n}$ points from origin toward the plane. If you flip $\hat{n}$, $d$ flips sign. Some textbooks require $d > 0$ by convention.
Related questions
The image of the point (1, 2, 3) in the plane x + y + z = 0 isThe foot of the perpendicular from origin to the plane 2x + 3y - 6z = 14 isThe line through the origin with direction ratios (1, 1, 1) meets the plane x + y + z = 6 A line has direction ratios (1, 1, 2) and meets the plane x - y + z = 0. The angle ϕ betweThe plane through the points (1,1,0), (1,0,1) and (0,1,1) has equationThe equation of a plane parallel to x + 2y - 3z = 5 and passing through (1, 1, 1) isThe shortest distance between r = (1,2,3) + λ(2,3,4) and r = (2,4,5) + μ(3,4,5) isThe line r = (2, -1, 4) + λ(3, 0, 2), expressed in symmetric Cartesian form, is