Distance from a point to a plane

WHAT are we measuring?

The plane is written as $\pi:\ ax+by+cz+d=0,$ and its normal vector is $\vec n=(a,b,c)$ , because that's the direction the plane "faces."

HOW: derive the formula from scratch

Let $P=(x_1,y_1,z_1)$ be our point, and let $Q=(x_0,y_0,z_0)$ be any point lying on the plane (so $ax_0+by_0+cz_0+d=0$ ).

Why pick a point on the plane? Because the distance is the projection of the vector $\vec{QP}$ onto the normal — and to compute a projection I need a tail on the plane and a head at $P$ .

Step 1 — form the vector from plane to point. $\vec{QP}=(x_1-x_0,\ y_1-y_0,\ z_1-z_0)$ Why this step? This vector runs from the plane up to $P$ ; its shadow along $\vec n$ is exactly the perpendicular distance.

Step 2 — project onto the unit normal. The signed distance is the component of $\vec{QP}$ along $\hat n=\dfrac{\vec n}{|\vec n|}$ : $D=\vec{QP}\cdot\hat n=\frac{\vec n\cdot\vec{QP}}{|\vec n|}.$ Why this step? The projection of a vector onto a unit direction is a dot product — that gives the length of the part pointing along the normal, which is the only part that matters.

Step 3 — expand the dot product. $\vec n\cdot\vec{QP}=a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)$ $=ax_1+by_1+cz_1-(ax_0+by_0+cz_0).$

Step 4 — use that $Q$ is on the plane. Since $ax_0+by_0+cz_0=-d$ , substitute: $\vec n\cdot\vec{QP}=ax_1+by_1+cz_1+d.$ Why this step? The unknown point $Q$ magically disappears — the answer depends only on $P$ and the plane's coefficients.

Step 5 — divide by $|\vec n|$ and take magnitude (distance is non-negative):

The numerator = "plug the point into the plane equation." The denominator = "length of the normal," which converts that raw number into a true distance.

Figure — Distance from a point to a plane

The sign tells you which side

Worked examples

Worked example Example 3 — distance between two parallel planes

Find distance between $\pi_1:\ 2x-y+2z+3=0$ and $\pi_2:\ 2x-y+2z-6=0$ . Why we can do this: They're parallel (same normal). Pick any point on $\pi_2$ , get its distance to $\pi_1$ . Put $y=z=0$ in $\pi_2$ : $2x-6=0\Rightarrow x=3$ , so point $(3,0,0)$ . Distance to $\pi_1$ : $\dfrac{|2(3)-0+0+3|}{\sqrt{4+1+4}}=\dfrac{|9|}{3}=3$ . Shortcut (same normal coeffs): $D=\dfrac{|d_1-d_2|}{|\vec n|}=\dfrac{|3-(-6)|}{3}=3.$ ✓

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (hide and answer)

State the distance formula and name each part.
Why does the unknown point $Q$ vanish in the derivation?
What does the sign of the unsigned-stripped expression tell you?
How do you get the distance between two parallel planes fast?

Answers: 1) $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$ ; numerator = point in plane eqn, denominator = $|\vec n|$ . 2) Because $ax_0+by_0+cz_0=-d$ on the plane, eliminating $Q$ . 3) Which side of the plane $P$ lies on. 4) $\frac{|d_1-d_2|}{|\vec n|}$ after matching normals.

Recall Feynman: explain to a 12-year-old

Imagine a flat wall (the plane) and you standing somewhere in the room (the point). The shortest way to touch the wall is to walk straight at it — not sideways, straight forward, perpendicular. The "normal" is just the arrow showing "straight at the wall." To find how far you are, you take your position, feed it into the wall's secret recipe ( $ax+by+cz+d$ ), and then shrink that number by the size of the arrow ( $\sqrt{a^2+b^2+c^2}$ ). The result is exactly how many steps it takes to reach the wall.

Connections

Equation of a plane — where $(a,b,c,d)$ come from.
Normal vector to a plane — why $\vec n=(a,b,c)$ .
Projection of a vector — the dot-product step is a projection.
Distance between two parallel planes — direct corollary.
Distance from a point to a line in 3D — sister formula using cross products.
Foot of perpendicular from a point to a plane — find the actual landing point.

Distance from point

(x_1,y_1,z_1)

to plane

ax+by+cz+d=0

\dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}

Why divide by

\sqrt{a^2+b^2+c^2}

It is

|\vec n|

; it converts the scaled signed quantity into a true length (projection onto the unit normal).

In the derivation, why does the on-plane point

Q

disappear?

Because

ax_0+by_0+cz_0=-d

, substituting removes

Q

, leaving only

P

and

d

What does a positive signed value

\frac{ax_1+by_1+cz_1+d}{|\vec n|}

mean?

P

lies on the side the normal

\vec n

points toward.

Distance between parallel planes

ax+by+cz+d_1=0

and

ax+by+cz+d_2=0

\dfrac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}

(normals must be made identical first).

Distance from origin to

ax+by+cz+d=0

\dfrac{|d|}{\sqrt{a^2+b^2+c^2}}

Two points give the same sign in the plane equation — what does that mean?

They lie on the same side of the plane.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, idea bilkul simple hai. Ek plane space ko do hisson mein baant deta hai, aur tumhe ek point se us plane tak ka sabse chhota distance chahiye. Sabse chhota raasta hamesha plane ke perpendicular (yaani normal direction) ke along hota hai — tedha chaloge to lamba ho jayega. Isliye poora sawaal yeh ban jata hai: "normal ke along kitna slide karun taaki plane pe pahunch jaun?"

Formula nikalne ka tareeka: plane pe koi bhi point $Q$ lo (jiske coordinates $ax_0+by_0+cz_0=-d$ satisfy karte hain). Phir vector $\vec{QP}$ ka projection normal $\vec n=(a,b,c)$ par lo — yeh dot product se aata hai. Jab expand karoge to $Q$ apne aap gayab ho jata hai, aur bachta hai $ax_1+by_1+cz_1+d$ . Isko $|\vec n|=\sqrt{a^2+b^2+c^2}$ se divide karo, mod laga do, bas: $D=\dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$ .

Yaad rakhne ka mantra: "Plug, then Length" — upar point ko plane equation mein daal do, neeche normal ki length se divide kar do. Do galtiyaan sabse common hain: $|\vec n|$ se divide karna bhool jana, aur $+d$ chhod dena. Dono yaad rakho.

Bonus: agar mod hata do to sign batata hai point kis side hai. Do points ka same sign matlab same side — yeh geometry aur optimization mein bahut kaam aata hai. Aur do parallel planes ke beech distance? Bas $\dfrac{|d_1-d_2|}{|\vec n|}$ , pehle dono ke normals same bana lena.

WHAT are we measuring?

HOW: derive the formula from scratch

The sign tells you which side

Worked examples

Common mistakes (steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — 3D Geometry

Connections