Multivariable-Calculus · Unit 1 · Video 5 · Interactive Practice

Dot Product Zero Means Perpendicular: Why x + 2y + 3z = 0 Is a Plane

IKey Formulas

Formula	Name	What you need
$\vec{A}\cdot\vec{B} = \|\vec{A}\|\,\|\vec{B}\|\cos\theta$	Geometric form of the dot product	Two vectors and the angle between them
$\vec{A}\perp\vec{B} \iff \vec{A}\cdot\vec{B} = 0$	Orthogonality test, for nonzero $\vec{A},\vec{B}$	Components only — no angle is ever computed
$ax + by + cz = \langle a, b, c\rangle\cdot\langle x, y, z\rangle$	A linear expression is a dot product	The coefficients, read in order
$ax + by + cz = 0$ is the plane through $O$ with normal $\vec{A} = \langle a, b, c\rangle$	Normal vector of a plane through the origin	$\vec{A} \neq \vec{0}$

Key Insight: The normal vector is never hidden — it is the coefficient list, so $x + 2y + 3z = 0$ is the plane through the origin perpendicular to $\langle 1, 2, 3\rangle$ .

IIZero Is the Right Angle

One number decides the angle's kind: positive is acute, negative is obtuse, zero is exactly $90^\circ$ .

B: b₁ 3 B: b₂ 1

💡 Challenge: exactly four nonzero integer positions of $\vec{B}$ in this window give $\vec{A}\cdot\vec{B} = 0$ , and all four lie on one line through the origin.

IIIA Linear Equation Is a Dot Product

The left-hand side of $x + 2y + 3z = 0$ is a sum of products — that is, a dot product in disguise.

Step 1

Step 1 — A point of the solution set

Let

O = (0,0,0)

and let

P = (x, y, z)

satisfy the equation, so that

\overrightarrow{OP} = \langle x, y, z\rangle

The point drawn is

P = (-1, 2, -1)

, and indeed

-1 + 4 - 3 = 0

Step 2 — The left side is a sum of products

x + 2y + 3z = x\cdot 1 + y\cdot 2 + z\cdot 3

Multiply matching entries, then add — that is precisely what a dot product does.

Step 3 — Name the other vector

\vec{A} = \langle 1, 2, 3\rangle \quad\Longrightarrow\quad \vec{A}\cdot\overrightarrow{OP} = x + 2y + 3z

So the equation is a geometric statement:

\ \vec{A}\cdot\overrightarrow{OP} = 0

Step 4 — Read it with the geometric formula

|\vec{A}|\,|\overrightarrow{OP}|\cos\theta = 0, \qquad |\vec{A}| = \sqrt{14} \neq 0

So either

\overrightarrow{OP} = \vec{0}

— that is

P = O

, which does solve the equation — or

\cos\theta = 0

, forcing

\theta = \pi/2

Step 5 — All those directions at once

x + 2y + 3z = 0 \quad\Longleftrightarrow\quad \overrightarrow{OP} \perp \vec{A}

Every direction perpendicular to one fixed vector, anchored at

O

, sweeps out one flat sheet: the plane through the origin with normal vector

\vec{A} = \langle 1, 2, 3\rangle

IVThe Coefficients Are the Normal Vector

Retilt the normal vector and the whole sheet follows, because the equation's coefficients are its components.

a₁ (of x) 1 a₂ (of y) 2 a₃ (of z) 3

💡 Challenge: flatten the sheet into the horizontal floor — the coefficients that do it spell an equation you have known since before this course.

VQuiz Questions

Problem 1 · Apply the Test

Given: $\vec{A} = \langle 2, -3, 1\rangle$ and $\vec{B} = \langle 3, 1, -3\rangle$ — compute $\vec{A}\cdot\vec{B}$ and decide whether the two vectors are perpendicular.

✅ Correct!

6 + (-3) + (-3) = 0

, and both vectors are nonzero, so

\cos\theta = 0

and

\theta = 90^\circ

— reached without computing a single angle.

❌ Not quite. Two of the three products are negative:

(-3)(1) = -3

and

(1)(-3) = -3

. Carry both minus signs into the sum.

❌ Not quite. That total comes from adding matching components,

5 + (-2) + (-2) = 1

. The dot product multiplies them first:

(2)(3) + (-3)(1) + (1)(-3)

❌ Not quite. Multiply matching components, then add:

(2)(3) + (-3)(1) + (1)(-3)

Show solution

Multiply the first components, the second components, the third components, and add:

\vec{A}\cdot\vec{B} = (2)(3) + (-3)(1) + (1)(-3) = 6 - 3 - 3 = 0

Now read the geometric formula backwards. Since $\vec{A}\neq\vec{0}$ and $\vec{B}\neq\vec{0}$ , both lengths are strictly positive, so

|\vec{A}|\,|\vec{B}|\cos\theta = 0 \quad\Longrightarrow\quad \cos\theta = 0 \quad\Longrightarrow\quad \theta = 90^\circ

The vectors are perpendicular. Note that the lengths $|\vec{A}| = \sqrt{14}$ and $|\vec{B}| = \sqrt{19}$ were never needed — the sign of the dot product carries the whole verdict.

Problem 2 · Solve for the Missing Component

Given: $\vec{A} = \langle 2, c, -1\rangle$ and $\vec{B} = \langle 3, 1, 4\rangle$ — find every value of $c$ that makes $\vec{A}\perp\vec{B}$ .

✅ Correct!

6 + c - 4 = 0

gives

c = -2

, and the check confirms it:

\langle 2, -2, -1\rangle\cdot\langle 3, 1, 4\rangle = 6 - 2 - 4 = 0

❌ Close, but check the sign.

c + 2 = 0

solves to

c = -2

; with

c = 2

the dot product is

6 + 2 - 4 = 4 \neq 0

, an acute angle.

❌ Not quite. The third product is

(-1)(4) = -4

, not

+4

. That one sign turns the correct

c + 2 = 0

into

c + 10 = 0

❌ Not quite. The condition

\vec{A}\cdot\vec{B} = 0

is a linear equation in

c

whose coefficient is

1

, so it has exactly one solution.

❌ Not quite. Write out

\vec{A}\cdot\vec{B}

with

c

carried along, set it equal to

0

, and solve.

Show solution

Perpendicular means the dot product vanishes, so build it componentwise and keep $c$ as an unknown:

\vec{A}\cdot\vec{B} = (2)(3) + (c)(1) + (-1)(4) = 6 + c - 4 = c + 2

Setting that equal to zero:

c + 2 = 0 \quad\Longrightarrow\quad c = -2

Check: $\langle 2, -2, -1\rangle\cdot\langle 3, 1, 4\rangle = 6 - 2 - 4 = 0$ , so $\theta = 90^\circ$ .

Because the coefficient of $c$ is $1 \neq 0$ , the equation is genuinely linear and one value of $c$ is the whole answer.

Problem 3 · From Two Points to a Plane

Given: $P = (1, 0, 2)$ and $Q = (4, -2, 3)$ — find the plane through the origin perpendicular to $\overrightarrow{PQ}$ .

What is $\overrightarrow{PQ}$ ?

Which equation is that plane?

✅ Correct!

\overrightarrow{PQ} = \langle 3, -2, 1\rangle

, and the coefficients of

3x - 2y + z = 0

are exactly its components.

❌ Check the vector.

\overrightarrow{PQ}

runs from

P

Q

, so subtract the tail from the head, component by component:

\langle 4 - 1,\; -2 - 0,\; 3 - 2\rangle

❌ Check the equation. Its coefficients are the normal's components in order, and "through the origin" forces the right-hand side to be

0

, since

(0,0,0)

must satisfy it.

Show solution

Step 1 — the normal vector. Head minus tail:

\overrightarrow{PQ} = \langle 4 - 1,\; -2 - 0,\; 3 - 2\rangle = \langle 3, -2, 1\rangle

Step 2 — the membership condition. A point $(x,y,z)$ lies on the plane exactly when its position vector is perpendicular to $\overrightarrow{PQ}$ :

\langle 3, -2, 1\rangle\cdot\langle x, y, z\rangle = 0 \quad\Longleftrightarrow\quad 3x - 2y + z = 0

Check: the origin gives $3(0) - 2(0) + 0 = 0$ ✓, and $\langle 3,-2,1\rangle$ is read straight off the coefficients.

The near miss $3x - 2y + z = 5$ carries the right coefficients but the wrong constant: substituting the origin gives $3(0) - 2(0) + 0 = 0 \neq 5$ , so $(0,0,0)$ fails it. The $5$ is simply what $P$ itself returns, $3(1) - 2(0) + 2 = 5$ .

Problem 4 · Two Planes at a Right Angle

Given: two planes through the origin meet at a right angle exactly when their normal vectors are perpendicular — decide whether $x + 2y + 3z = 0$ and $2x + 2y - 2z = 0$ meet at a right angle.

✅ Correct!

\langle 1, 2, 3\rangle\cdot\langle 2, 2, -2\rangle = 2 + 4 - 6 = 0

, so the normals are perpendicular and therefore so are the planes.

❌ Backwards. Parallel planes have parallel normals, where

\cos\theta = \pm 1

; a vanishing dot product is the perpendicular case,

\cos\theta = 0

❌ Watch the third coefficient. In

2x + 2y - 2z = 0

it is

-2

, so the third product is

(3)(-2) = -6

and the total is

2 + 4 - 6 = 0

❌ They do. Each equation hands you its normal directly, and whether that pair's dot product is zero settles the question without any inverse cosine.

❌ Not quite. Read a normal off each equation's coefficients, then apply the orthogonality test to those two vectors.

Show solution

Each equation is a dot product set to zero, so its coefficients are a normal vector:

x + 2y + 3z = 0 \;\longrightarrow\; \vec{A} = \langle 1, 2, 3\rangle, \qquad 2x + 2y - 2z = 0 \;\longrightarrow\; \vec{B} = \langle 2, 2, -2\rangle

Now test that pair:

\vec{A}\cdot\vec{B} = (1)(2) + (2)(2) + (3)(-2) = 2 + 4 - 6 = 0

The normals are perpendicular, so the planes meet at a right angle — yes.

Both vectors are nonzero ( $|\vec{A}| = \sqrt{14}$ , $|\vec{B}| = \sqrt{12}$ ), which is what licenses the step from $|\vec{A}|\,|\vec{B}|\cos\theta = 0$ to $\cos\theta = 0$ .

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