Equation of a plane — normal form, intercept form, general form: Question bank

Before we start, one picture to hold in your head for the whole page.

Figure — Equation of a plane — normal form, intercept form, general form

That red arrow is the normal $n$ — the one direction perpendicular to the flat sheet. The plane is everything the arrow points across, never along. Almost every trap below comes from confusing this arrow with something else (a point in the plane, an intercept, a distance). Keep looking back at it.

True or false — justify

Recall Answer with a

reason, never a bare yes/no The coefficients $(a, b, c)$ in $a x + b y + cz + d = 0$ are the coordinates of a point lying on the plane. ::: False. They are the components of the normal $n = (a, b, c)$ — a direction perpendicular to the plane, not a location on it. Two planes with the same normal vector are always the same plane. ::: False. Same normal means parallel; they can sit at different distances (different $d$ ). Only equal $d$ (after matching scale) makes them identical. Multiplying a plane equation through by $- 1$ gives a different plane. ::: False. $a x + b y + cz + d = 0$ and $- a x - b y - cz - d = 0$ have the same solution set; the normal just flips direction, which is still perpendicular to the same sheet. In normal form $l x + m y + n z = p$ the number $p$ can be negative. ::: False. $p$ is a perpendicular distance from the origin, so $p \geq 0$ by construction. If your algebra gives $p < 0$ , you divided by the wrong sign of $a^{2} + b^{2} + c^{2}$ . Every plane in 3D can be written in intercept form. ::: False. A plane through the origin ( $d = 0$ ) has intercept $0$ on some axis, and $\frac{x}{0}$ is undefined — no intercept form exists for it. If a plane is parallel to the $x$ -axis, its equation has no $x$ term. ::: True. Parallel to the $x$ -axis means the normal is perpendicular to $\hat{i}$ , so the $x$ -component of $n$ is $0$ , hence coefficient of $x$ is $0$ . The equation $x = 3$ describes a plane, not just a line. ::: True. In 3D it is $1 x + 0 y + 0 z = 3$ : all points with $x = 3$ , a whole plane with normal $(1, 0, 0)$ parallel to the $y z$ -plane. Dividing $a x + b y + cz + d = 0$ by $∣ n ∣$ automatically produces intercept form. ::: False. Dividing by $∣ n ∣ = a^{2} + b^{2} + c^{2}$ gives normal form (unit normal). Intercept form comes from making the right side equal 1.

Spot the error

Recall Each line contains one flawed statement — name the flaw

"Plane $2 x - y + 2 z = 6$ has intercepts $2, - 1, 2$ (the coefficients)." ::: Error: coefficients are normal components. Intercepts come from dividing by $6$ : $\frac{x}{3} + \frac{y}{- 6} + \frac{z}{3} = 1$ , giving intercepts $3, - 6, 3$ . "Normal form of $x + y + z = 3$ is just $x + y + z = 3$ since it's already tidy." ::: Error: the normal must be a unit vector. Here $∣ n ∣ = 3$ , so divide: $\frac{x + y + z}{3} = 3$ , and $p = 3$ . "Distance of the origin from $2 x - y + 2 z = - 6$ is $p = - 2$ ." ::: Error: $p$ can't be negative. Divide by $- 3$ (to flip the negative RHS positive): $- \frac{2}{3} x + \frac{1}{3} y - \frac{2}{3} z = 2$ , so $p = 2$ . "The plane through the origin $x + 2 y + 3 z = 0$ has intercepts $0, 0, 0$ , so all three axes are its intercepts." ::: Error: it meets each axis only at the origin, so there are no distinct finite intercepts and no valid intercept form ( $\frac{x}{0}$ is undefined). "To get the plane's normal from three points, add the three position vectors." ::: Error: adding gives a point-average (centroid direction), not a perpendicular. Use the cross product $A B \times A C$ , which is perpendicular to both in-plane vectors. " $4 x + 4 y + 4 z = 8$ and $x + y + z = 2$ are different planes because the coefficients differ." ::: Error: the second is the first divided by $4$ — identical solution set, same plane. Plane equations are defined up to a nonzero scalar.

Why questions

Recall Explain the

mechanism, not just the fact Why is a plane fixed by ONE perpendicular direction plus one distance, when there are infinitely many directions inside it? ::: The in-plane directions don't distinguish a plane (many planes share them). The perpendicular direction is unique up to sign, so it pins the tilt; the distance pins the position along that pencil. Why does $n \cdot (r - r_{0}) = 0$ capture "lies on the plane"? ::: $r - r_{0}$ is a vector inside the plane. A normal is perpendicular to every in-plane vector, and perpendicular ⇔ dot product $= 0$ — so the equation is true exactly for on-plane points. Why do we divide by $a^{2} + b^{2} + c^{2}$ specifically to reach normal form? ::: That square root is $∣ n ∣$ . Dividing rescales $n$ to length $1$ , turning the coefficients into direction cosines and turning the RHS into a genuine distance. Why must we divide by the RHS constant $k$ (not $∣ n ∣$ ) to get intercept form? ::: Intercept form needs the right side to read $1$ , so that $\frac{x}{a}$ etc. expose the intercepts directly. Dividing by $k$ (not $∣ n ∣$ ) is what forces RHS $= 1$ . Why is the cross product the right tool for "plane through three points"? ::: We need a vector perpendicular to two in-plane vectors at once. The cross product is defined to produce exactly such a mutually-perpendicular vector — precisely the normal. Why does a zero coefficient (say of $y$ ) mean the plane is parallel to that axis? ::: Zero $y$ -component in $n$ means $n ⊥ \hat{j}$ . If the normal is perpendicular to the $y$ -axis, the $y$ -axis lies along the plane's tilt — i.e. the plane runs parallel to it.

Edge cases

Recall Degenerate and boundary inputs — say what happens and why

What plane is $0 x + 0 y + 0 z = 5$ ? ::: None. The normal is the zero vector $(0, 0, 0)$ , which has no direction, so it defines no plane — the equation $0 = 5$ is simply false everywhere. What does $d = 0$ do to the plane geometrically, and to its forms? ::: The plane passes through the origin. Normal and general forms still work; intercept form breaks (each intercept is $0$ ). If $p = 0$ in normal form $r \cdot \overset{n}{^} = 0$ , where is the plane? ::: Distance from origin is $0$ , so the plane passes through the origin, containing $n$ 's perpendicular directions only. Can two of the intercepts be equal while the third differs — is that a contradiction? ::: No contradiction. E.g. $\frac{x}{3} + \frac{y}{3} + \frac{z}{- 6} = 1$ is a perfectly valid tilted plane; intercepts are independent numbers. What happens to intercept form as an intercept $\to \infty$ ? ::: $\frac{x}{a} \to 0$ , so that variable drops out — the plane becomes parallel to that axis (a finite tilt, infinitely far intercept). Is $(a, b, c) = (0, 0, 1)$ a legal normal, and what plane does $z = d$ give? ::: Yes, it's a unit vector along $\hat{k}$ . $z = d$ is a horizontal plane parallel to the $x y$ -plane at height $d$ . If someone writes $l x + m y + n z = p$ with $l^{2} + m^{2} + n^{2} = 4$ , what's wrong? ::: It's not true normal form — direction cosines must satisfy $l^{2} + m^{2} + n^{2} = 1$ . Here $∣ n ∣ = 2$ , so divide everything by $2$ first.

Question bank — Equation of a plane — normal form, intercept form, general form