Angle between two planes

Subject: Maths • Chapter: 3D Geometry • Subtopic: Angle between two planes

1. What is a normal vector? (first principles)

WHY are the coefficients the normal? Take two points $P=(x_1,y_1,z_1)$ and $Q=(x_2,y_2,z_2)$ both on the plane. Subtract their equations: $a(x_1-x_2)+b(y_1-y_2)+c(z_1-z_2)=0.$ The left side is exactly $\vec n \cdot \vec{QP}$ . So $\vec n \cdot (\text{any vector lying in the plane}) = 0$ — meaning $\vec n$ is perpendicular to every direction inside the plane. That is the definition of being normal to the plane.

2. Deriving the angle formula

HOW — from the dot product. For any two vectors, $\vec n_1 \cdot \vec n_2 = |\vec n_1|\,|\vec n_2|\cos\theta.$ This is the definition of dot product (projection of one vector onto another). Rearranging:

Formula Angle between two planes

Planes $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$ with normals $\vec n_1=\langle a_1,b_1,c_1\rangle$ , $\vec n_2=\langle a_2,b_2,c_2\rangle$ : $\boxed{\cos\theta = \frac{|\vec n_1 \cdot \vec n_2|}{|\vec n_1|\,|\vec n_2|} = \frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}}$ The absolute value guarantees the acute angle $0\le\theta\le 90^\circ$ .

WHY the modulus? Flipping a plane's equation by $-1$ (e.g. multiplying through) flips its normal, changing $\cos\theta$ to $-\cos\theta$ — but the plane is the same. The $|\cdot|$ removes this sign ambiguity and always returns the smaller wedge angle.

3. Worked examples

Worked example Example 1 — basic

Find the angle between $2x - y + 2z = 5$ and $x + 2y + 2z = 3$ .

$\vec n_1=\langle 2,-1,2\rangle$ , $\vec n_2=\langle 1,2,2\rangle$ . Why? Coefficients are the normals.
Dot: $2(1)+(-1)(2)+2(2)=2-2+4=4$ . Why? Numerator of formula.
Magnitudes: $|\vec n_1|=\sqrt{4+1+4}=3$ , $|\vec n_2|=\sqrt{1+4+4}=3$ . Why? Length normalises projection.
$\cos\theta=\dfrac{|4|}{3\cdot3}=\dfrac{4}{9}$ , so $\theta=\cos^{-1}\!\frac49\approx 63.6^\circ$ .

4. Forecast-then-Verify

Recall Forecast first, then check

Q: Two planes have normals $\langle1,0,0\rangle$ and $\langle0,0,1\rangle$ . Forecast the angle before computing. Verify: Dot $=0$ , so $\theta=90^\circ$ . The planes are the $yz$ -plane and $xy$ -plane — they meet at a right angle along the $y$ -axis. ✔

5. Common mistakes (Steel-man)

6. Flashcards

What vector represents a plane's orientation?

Its normal

\vec n=\langle a,b,c\rangle

, the coefficients of

x,y,z

Why are the coefficients

(a,b,c)

perpendicular to the plane?

Subtracting the plane equation at two points gives

\vec n\cdot(\text{in-plane vector})=0

, so

\vec n\perp

every in-plane direction.

Formula for angle between two planes

\cos\theta=\dfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}

Why the absolute value in the formula?

It removes the sign ambiguity from flipping a normal and always gives the acute angle.

Condition for two planes to be perpendicular

\vec n_1\cdot\vec n_2=0

, i.e.

a_1a_2+b_1b_2+c_1c_2=0

Condition for two planes to be parallel

\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

(normals are scalar multiples).

Angle between line and plane uses which ratio?

\sin\theta=\dfrac{|\vec b\cdot\vec n|}{|\vec b||\vec n|}

(note: sine, not cosine).

Angle between

x+y+z=1

and

-x-y+z=2

\cos^{-1}(1/3)\approx70.5^\circ

Recall Feynman: explain to a 12-year-old

Imagine two flat cardboard sheets leaning against each other, making a "V". To say how open the V is, you don't touch the cardboard — you stick a pencil straight up out of each sheet and measure the angle between the two pencils. Those pencils are the "normals". For a plane written like $2x-y+2z=5$ , the pencil is just the three numbers in front: $(2,-1,2)$ . Multiply matching numbers, add them, divide by the lengths — and the cosine button on your calculator tells you the angle. We always keep the answer positive so we get the small, sensible angle.

Connections

Equation of a plane — where $\langle a,b,c\rangle$ comes from.
Normal vector and dot product — the engine behind the formula.
Angle between two lines — same dot-product idea, different objects.
Angle between a line and a plane — uses $\sin$ , the complement.
Distance from a point to a plane — also built from the normal.
Scalar (dot) product — $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, do planes ke beech ka angle nikalna bahut simple hai — bas ek trick samajh lo: har plane ko uska normal vector define karta hai, jo plane ke bilkul perpendicular ek arrow hota hai. Agar plane ka equation $ax+by+cz+d=0$ hai, to uska normal seedha $\langle a,b,c\rangle$ — yaani $x,y,z$ ke aage wale numbers. Bas itna hi.

Ab do planes ke beech ka angle = un dono normals ke beech ka angle. Aur do vectors ka angle nikalne ke liye humara purana dost dot product use hota hai: $\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}$ . Numerator mein matching components multiply karke add karo, denominator mein dono ki lengths. Modulus (absolute value) isliye lagate hain taaki hamesha acute angle mile — kyunki normal ka direction ulta karne se plane same rehta hai, sirf sign badalta hai.

Do important cases yaad rakho: agar dot product zero ho jaye to planes perpendicular hain ( $90^\circ$ ). Agar normals ek dusre ke scalar multiple ho (ratios equal), to planes parallel hain ( $0^\circ$ ). Ek common galti — line aur plane ka angle nikalte waqt $\sin$ lagta hai, lekin do planes ke beech $\cos$ . Mnemonic: "Cosine for planes, Sine for the line." Bas isse exam mein confusion khatam.

1. What is a normal vector? (first principles)

2. Deriving the angle formula

3. Worked examples

4. Forecast-then-Verify

5. Common mistakes (Steel-man)

6. Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — 3D Geometry

Connections