LINEAR-ALGEBRA · Unit 13 · Video 1 · Interactive Practice

Why Two Perpendicular Planes Aren't "Orthogonal"

IKey Formulas

Formula	Name	Description
$\mathbf{x}^T\mathbf{y} = x_1 y_1 + x_2 y_2 + \cdots + x_n y_n$	Dot Product	Sum of component-wise products
$\mathbf{x} \perp \mathbf{y} \iff \mathbf{x}^T\mathbf{y} = 0$	Orthogonal Vectors	Dot product equals zero
$\\|\mathbf{x}+\mathbf{y}\\|^2 = \\|\mathbf{x}\\|^2 + 2\mathbf{x}^T\mathbf{y} + \\|\mathbf{y}\\|^2$	Norm Expansion	Pythagorean theorem holds when cross term vanishes
$S \perp T$ : $\mathbf{x}^T\mathbf{y} = 0$ for all $\mathbf{x} \in S$ , all $\mathbf{y} \in T$	Orthogonal Subspaces	Every pair must be perpendicular

IIVisualization 1 — Dot Product and Right Angles

Two vectors are orthogonal exactly when their dot product is zero — the instant the Pythagorean theorem holds for the triangle they form.

IIIVisualization 2 — The All-Pairs Test

For two lines through the origin, one number — the dot product of their directions — decides whether every pair of vectors is orthogonal.

IVVisualization 3 — Two Planes Always Share a Line

In $\mathbb{R}^3$ , two planes through the origin always meet in at least a line, so a nonzero vector lives in both — which makes orthogonality impossible at any angle.

Dihedral angle 90°

VQuiz Questions

Question 1 · Computing a Dot Product

Let $\mathbf{x} = (3, -1, 2)$ and $\mathbf{y} = (1, 4, -1)$ .

What is $\mathbf{x}^T\mathbf{y}$ ?

✅ Correct! $(3)(1) + (-1)(4) + (2)(-1) = 3 - 4 - 2 = -3$ .

❌ Not quite. Carefully multiply each corresponding pair — watch the signs: $(3)(1) + (-1)(4) + (2)(-1)$ .

Show solution

Solution:

Multiply corresponding components and sum:

$\mathbf{x}^T\mathbf{y} = (3)(1) + (-1)(4) + (2)(-1) = 3 - 4 - 2 = -3$

Since $\mathbf{x}^T\mathbf{y} = -3 \neq 0$ , these vectors are not orthogonal.

Common mistakes:

Getting 3: forgetting the negative sign on $(-1)(4) = -4$
Getting -1: computing $3 - 4 = -1$ but forgetting the third term
Getting 7: using $|x_i y_i|$ (absolute values) instead of signed products

Question 2 · Perpendicular vs. Orthogonal Subspaces

True or False: Two planes in $\mathbb{R}^3$ that meet at a 90° dihedral angle are orthogonal subspaces.

✅ Correct! Two planes in R³ always share a line, so the all-pairs requirement can never be satisfied — they are never orthogonal subspaces.

❌ Not quite. Think about what the two planes must share. Any nonzero vector on that shared line would need to be orthogonal to itself...

Show solution

Solution: False.

Two planes through the origin in $\mathbb{R}^3$ always share at least a line, because:

$\dim(S \cap T) \geq \dim(S) + \dim(T) - n = 2 + 2 - 3 = 1$

Any nonzero vector $\mathbf{v}$ on that shared line is in both subspaces. The all-pairs orthogonality requirement demands $\mathbf{v}^T\mathbf{v} = 0$ , but:

$\mathbf{v}^T\mathbf{v} = \|\mathbf{v}\|^2 > 0 \quad \text{for any } \mathbf{v} \neq \mathbf{0}$

Contradiction! Two planes in $\mathbb{R}^3$ are never orthogonal subspaces, no matter the dihedral angle.

This is the key distinction: "perpendicular" (geometric appearance) $\neq$ "orthogonal subspaces" (the all-pairs linear algebra requirement).

Question 3 · Intersection of Orthogonal Subspaces

If subspaces $S$ and $T$ are orthogonal ( $S \perp T$ ), what must their intersection $S \cap T$ equal?

✅ Correct! The zero vector is the only vector orthogonal to itself ( $\mathbf{0}^T\mathbf{0} = 0$ ), so it's the only vector that can live in both subspaces.

❌ Close, but not quite. Every subspace must contain the zero vector, so $S \cap T$ always includes $\mathbf{0}$ . The intersection is never truly empty.

❌ Not quite. Think: if a nonzero vector $\mathbf{v}$ lived in both subspaces, what would $\mathbf{v}^T\mathbf{v}$ need to equal?

Show solution

Solution: Only the zero vector $\{\mathbf{0}\}$ .

Two key facts combine:

Every subspace contains $\mathbf{0}$ , so $S \cap T$ always includes $\mathbf{0}$ . (It's never the empty set!)
- And $\mathbf{0}^T\mathbf{0} = 0$ ✅ — no contradiction.
No nonzero vector can be shared. If $\mathbf{v} \neq \mathbf{0}$ were in both $S$ and $T$ , the all-pairs test would require: $\mathbf{v}^T\mathbf{v} = \|\mathbf{v}\|^2 > 0 = 0 \quad \text{Contradiction!}$

Therefore $S \cap T = \{\mathbf{0}\}$ — the zero vector is the only vector orthogonal to itself.

This doesn't depend on dimensions: it's a universal consequence of the all-pairs requirement.

Question 4 · Dimension Constraint

In $\mathbb{R}^4$ , which of the following pairs of subspaces cannot be orthogonal?

Hint: If $S \perp T$ , then $S \cap T = \{\mathbf{0}\}$ , which forces $\dim(S) + \dim(T) \leq n$ .

✅ Correct! Since $2 + 3 = 5 > 4$ , these subspaces must share at least a line, making orthogonality impossible — just like two planes in $\mathbb{R}^3$ !

❌ Not quite. Check whether $\dim(S) + \dim(T)$ exceeds 4. If it does, the subspaces must share a nonzero vector, and orthogonality is impossible.

Show solution

Solution: A plane (dim 2) and a 3-dimensional subspace cannot be orthogonal in $\mathbb{R}^4$ .

For orthogonal subspaces, $S \cap T = \{\mathbf{0}\}$ requires $\dim(S) + \dim(T) \leq n$ .

Check each option in $\mathbb{R}^4$ (where $n = 4$ ):

Pair	$\dim(S) + \dim(T)$	$\leq 4$ ?	Orthogonal possible?
Line + 3-dim	$1 + 3 = 4$	✅ Yes	Yes
Plane + Plane	$2 + 2 = 4$	✅ Yes	Yes
Plane + 3-dim	$\mathbf{2 + 3 = 5}$	❌ No	No
Line + Plane	$1 + 2 = 3$	✅ Yes	Yes

Since $2 + 3 = 5 > 4$ , these subspaces must share at least a line:

$\dim(S \cap T) \geq 2 + 3 - 4 = 1 > 0$

A shared nonzero vector makes orthogonality impossible — the same argument as two planes in $\mathbb{R}^3$ !

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