Plane — Equation, Normal, and Construction

A plane in space has three equivalent forms:

• Cartesian form: $ax + by + cz = d$. The coefficients give the normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$.
• Vector form: $\vec{r}\cdot\vec{n} = d$, where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is the position vector of a general point.
• Point-normal form: through a point $A(x_0,y_0,z_0)$ with normal $\vec{n} = (a,b,c)$: $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.

Two planes are parallel when their normals are parallel (proportional coefficients). Two planes are perpendicular when their normals are perpendicular: $\vec{n_1}\cdot\vec{n_2} = 0$.

If a normal $\vec{n}$ makes angles $\alpha, \beta, \gamma$ with the $X, Y, Z$ axes, its direction cosines $\cos\alpha, \cos\beta, \cos\gamma$ satisfy:

Parallel to a coordinate plane (normal along one axis):

• Parallel to XY-plane: $z = k$. Parallel to YZ-plane: $x = k$. Parallel to ZX-plane: $y = k$.

Parallel to a given plane $ax+by+cz = d$: the required plane is $ax+by+cz = d'$ with the same normal $(a,b,c)$; substitute the given point to find $d'$.

Let $M(x_0,y_0,z_0)$ be the foot of perpendicular from the origin to the plane. Then:

• The normal is $\vec{n} = \overrightarrow{OM} = (x_0,y_0,z_0)$.
• The plane passes through $M$, so the constant is $\vec{n}\cdot\overrightarrow{OM} = x_0^2 + y_0^2 + z_0^2$.

Cartesian: $x_0 x + y_0 y + z_0 z = x_0^2 + y_0^2 + z_0^2$. Vector: $\vec{r}\cdot(x_0\hat{i} + y_0\hat{j} + z_0\hat{k}) = x_0^2 + y_0^2 + z_0^2$.

Given the angles a normal makes with the axes, recover its direction ratio via $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, then write the plane through the point $A$:

To build a plane through a point $A$ perpendicular to planes with normals $\vec{n_1}$ and $\vec{n_2}$:

• The required normal is $\vec{n} = \vec{n_1}\times\vec{n_2}$ (perpendicular to both, so both given normals lie IN the required plane).
• Then write the plane through $A$: $\vec{n}\cdot(\vec{r} - \vec{a}) = 0$.

This is the cross-product-normal engine — one of the two HARD workhorses of this subtopic.

For a plane through a point $A$ parallel to two lines with direction vectors $\vec{d_1}, \vec{d_2}$:

• The normal is $\vec{n} = \vec{d_1}\times\vec{d_2}$.
• Plane: $\vec{n}\cdot(\vec{r} - \vec{a}) = 0$.

Read each line's direction straight off its symmetric form $\frac{x - x_1}{p} = \frac{y - y_1}{q} = \frac{z - z_1}{s}$: the direction is $(p, q, s)$.

For points $A, B, C$:

• Form two in-plane vectors $\overrightarrow{AB} = \vec{b} - \vec{a}$ and $\overrightarrow{AC} = \vec{c} - \vec{a}$.
• Normal $\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}$; plane through $A$.

Determinant form (equivalent):

For the plane perpendicular to segment $PQ$ and passing through its midpoint:

• Midpoint $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.
• Normal $\vec{n} = \overrightarrow{PQ} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
• Plane: $\vec{n}\cdot(\vec{r} - \vec{m}) = 0$.

The family of planes through the line of intersection of $P_1: a_1x+b_1y+c_1z+d_1 = 0$ and $P_2: a_2x+b_2y+c_2z+d_2 = 0$ is:

• Perpendicular to XY-plane → $z$-coefficient $= 0$: $c_1 + \lambda c_2 = 0$.
• Parallel to X / Y / Z-axis → the matching coefficient $= 0$ (e.g. parallel to Y-axis → $b_1 + \lambda b_2 = 0$).
• Perpendicular to a third plane with normal $\vec{m}$ → family-normal $\cdot\, \vec{m} = 0$.
• Parallel to a line with direction $\vec{d}$ → family-normal $\cdot\, \vec{d} = 0$.
• Through a point → substitute the point.

Intercept form: $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$, with axis points $A(a,0,0), B(0,b,0), C(0,0,c)$.

• Centroid of $\triangle ABC$: $\left(\dfrac{a}{3}, \dfrac{b}{3}, \dfrac{c}{3}\right)$.
• Area of $\triangle ABC$: $\dfrac{1}{2}\sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2}$.

Given a point $P$ and its mirror image $P'$ in an unknown plane:

• The plane passes through the midpoint $M = \tfrac12(P + P')$.
• Its normal is $\overrightarrow{PP'} = P' - P$ (the segment is perpendicular to the plane).

Build the plane, then test which option-point satisfies it. (The fuller treatment of finding an image or foot of perpendicular lives on the Foot of Perpendicular, Image, and Projection page; here we only need this reverse construction.)