Chapter 2 Vectors: Exercises - Linear Algebra with Python

Structure of each exercise:

Problem + hint (MyST ::::{exercise} format)
An expandable complete solution (:::{solution} :class: dropdown)
Python numerical verification (with execution output)

Note. Some verification code blocks are shared by several exercises (for example, orthogonal bases and the generalized Pythagorean theorem).

The Two Kinds of Exercises in This Book

The exercises in this book follow two tracks, which play different roles:

Exercises embedded in the text (the {exercise} boxes scattered through the sections, most of which come without solutions) are meant for exploratory thinking: they encourage you to stop while reading, work things out by hand, and extend the ideas of the text. If you get stuck, you may set them aside for now; there is no need to solve them right away.
The exercises in this exercise book are meant as training for consolidation: every one of them comes with a complete solution (an expandable dropdown) and a Python numerical verification, which makes them suitable for systematic practice and self-assessment.

In addition, exercises marked ◇ are advanced: the concepts they need are often formally developed only in later chapters (for example, complex inner products belong to Chapter 9). You may skip them on a first reading and come back to them after reaching the corresponding chapter.

import numpy as np
from itertools import combinations
np.set_printoptions(precision=4, suppress=True, linewidth=100)

# ── Visual hierarchy helpers ─────────────────────────────────────────
def print_header(title):
    print("=" * 60)
    print(f"  {title}")
    print("=" * 60)

def print_step(step, desc):
    print(f"\n▶ Step {step}: {desc}")
    print("-" * 40)

def compare_print(label, actual, expected_desc):
    print(f"\n[{label}]")
    print(f"  Computed value: {actual}")
    print(f"  Expected value: {expected_desc}")

# ── Math utilities ─────────────────────────────────────────
def proj(u, v):
    """Projection vector of v onto the direction of u"""
    return np.dot(u, v) / np.dot(u, u) * u

def para_area(u, v):
    """Area of the parallelogram spanned by u, v"""
    return np.sqrt(np.dot(u,u)*np.dot(v,v) - np.dot(u,v)**2)

def triple(u, v, w):
    """Signed volume D(u,v,w) = u·(v×w)"""
    return np.dot(u, np.cross(v, w))

Basic Vector Operations¶

Exercise 1 (Computing a Linear Combination)

In $\mathbb{R}^4$ , let $\mathbf{a} = (2, 0, -1, 3)$ , $\mathbf{b} = (1, 4, 2, -1)$ , $\mathbf{c} = (-1, 2, 0, 1)$ .

Compute $2\mathbf{a} - 3\mathbf{b} + \mathbf{c}$ and $\|2\mathbf{a} - 3\mathbf{b} + \mathbf{c}\|$ .

Hint. Compute component by component, then use $\|\mathbf{v}\| = \sqrt{v_0^2 + v_1^2 + v_2^2 + v_3^2}$ .

# ============================================================
print_header("Exercise 1 | Computing a Linear Combination")
# ============================================================

print_step(1, "Define the vectors a, b, c ∈ ℝ⁴")
a = np.array([2, 0, -1, 3], float)
b = np.array([1, 4,  2, -1], float)
c = np.array([-1, 2, 0,  1], float)
print("a =", a)
print("b =", b)
print("c =", c)

print_step(2, "Compute 2a - 3b + c")
result = 2*a - 3*b + c
print("2a - 3b + c =", result)

compare_print("‖2a - 3b + c‖", f"{np.linalg.norm(result):.6f}", f"2√66 = {2*np.sqrt(66):.6f}")

Exercise 2 (The Average of Cyclically Shifted Vectors)

In $\mathbb{R}^n$ , define the vectors

\mathbf{a}_0 = (1,2,\ldots,n),\quad \mathbf{a}_1 = (2,3,\ldots,n,1),\quad \ldots,\quad \mathbf{a}_{n-1} = (n,1,\ldots,n-2,n-1)

(4)

( $\mathbf{a}_i$ is obtained by cyclically shifting $\mathbf{a}_0$ to the left $i$ times.)

Compute the average vector $\mathbf{A} = \dfrac{1}{n}(\mathbf{a}_0 + \mathbf{a}_1 + \cdots + \mathbf{a}_{n-1})$ , and find $\|\mathbf{A} - \mathbf{a}_i\|$ for $i = 0, 1, \ldots, n-1$ .

Hint. Observe that in each component position, the corresponding entries of the $n$ vectors are exactly some arrangement of $1, 2, \ldots, n$ , so all components have the same sum. Once you have found $\|\mathbf{A} - \mathbf{a}_i\|^2$ , note that $i$ only affects where the summation starts, so the result does not depend on $i$ .

# ============================================================
print_header("Exercise 2 | The Average of Cyclically Shifted Vectors (check with n=4)")
# ============================================================

print_step(1, "Generate the cyclically shifted vectors a_0, …, a_{n-1}")
n = 4
vecs = [np.roll(np.arange(1, n+1, dtype=float), -i) for i in range(n)]
for i, v in enumerate(vecs):
    print(f"  a_{i} =", v)

print_step(2, "Compute the average vector A")
A_avg = np.mean(vecs, axis=0)
compare_print("average vector A", A_avg, f"(n+1)/2 times the all-ones vector = {(n+1)/2} * [1,1,1,1]")

print_step(3, "Check that all the vectors are equidistant from A")
dists = [np.linalg.norm(A_avg - v) for v in vecs]
print("  ‖A - a_i‖ =", [f"{d:.4f}" for d in dists])
compare_print("theoretical value √(n(n²−1)/12)", f"{dists[0]:.6f}", f"{np.sqrt(n*(n**2-1)/12):.6f}")

Exercise 3 (The Fourth Vertex of a Parallelogram)

In $\mathbb{R}^3$ , let $A(2,-1,3)$ , $B(1,4,-2)$ , $C(-2,6,1)$ . If $A$ , $B$ , $C$ , $D$ are the four vertices of a parallelogram, find all possible positions of $D$ , and compute the side lengths of the parallelogram in each case.

Hint. The two diagonals of a parallelogram have the same midpoint. There are three ways to choose the pair of diagonals, $(AC,BD)$ , $(AB,CD)$ , $(BC,AD)$ , each corresponding to one position of $D$ .

# ============================================================
print_header("Exercise 3 | The Fourth Vertex of a Parallelogram")
# ============================================================

print_step(1, "Define the three vertices A, B, C ∈ ℝ³")
A = np.array([2, -1,  3], float)
B = np.array([1,  4, -2], float)
C = np.array([-2, 6,  1], float)

print_step(2, "Compute the three possible fourth vertices D (by choice of diagonals)")
D0 = A + C - B   # diagonal AC
D1 = A + B - C   # diagonal AB
D2 = B + C - A   # diagonal BC

for label, D in zip(["D0 (diagonal AC)", "D1 (diagonal AB)", "D2 (diagonal BC)"], [D0, D1, D2]):
    print(f"  {label}: D =", D)

print_step(3, "Check the side lengths in the case D0 (|AB| = |CD0|)")
print("  |AB| =", np.linalg.norm(B-A))
print("  |CD0|=", np.linalg.norm(D0-C))
print("  equal?", np.isclose(np.linalg.norm(B-A), np.linalg.norm(D0-C)))

Exercise 4 (Existence of a Linear Combination of Vectors)

Decide whether the vector $\mathbf{x}$ can be written as $\mathbf{x} = s\mathbf{u} + t\mathbf{v}$ . If it can, find the coefficients $s, t$ .

$\mathbf{u}=(1,2)$ , $\mathbf{v}=(3,-1)$ , $\mathbf{x}=(4,1)$ .
$\mathbf{u}=(1,2,3)$ , $\mathbf{v}=(2,4,6)$ , $\mathbf{x}=(3,6,9)$ .
$\mathbf{u}=(1,0,2,-1)$ , $\mathbf{v}=(0,1,-1,3)$ , $\mathbf{x}=(2,3,0,8)$ .
$\mathbf{u}=(1,2,0,-1)$ , $\mathbf{v}=(2,4,1,0)$ , $\mathbf{x}=(1,2,2,0)$ .

Hint. Write the condition as a system of linear equations and decide whether it has a solution. When there are more components than unknowns, the equations for all the components must be consistent.

# ============================================================
print_header("Exercise 4 | Existence of Linear Combinations")
# ============================================================

print_step(1, "(a) s(1,2)+t(3,−1)=(4,1) — unique solution")
M = np.array([[1,3],[2,-1]], float)
st = np.linalg.solve(M, [4,1])
print("  [s, t] =", st)
compare_print("check M @ [s,t]", M @ st, "[4, 1]")

print_step(2, "(b) x = 3u — infinitely many solutions")
u, x = np.array([1,2,3],float), np.array([3,6,9],float)
print("  x = 3u?", np.allclose(3*u, x), "(s+2t=3, infinitely many solutions)")

print_step(3, "(c) check that there is no solution")
u = np.array([1,0,2,-1],float); v = np.array([0,1,-1,3],float)
x = np.array([2,3,0,8],float)
print("  s=2,t=3 → su+tv =", 2*u+3*v)
print("  target x =", x, "→ no solution")

print_step(4, "(d) least-squares residual → no solution")
u = np.array([1,2,0,-1],float); v = np.array([2,4,1,0],float)
x = np.array([1,2,2,0],float)
coef, *_ = np.linalg.lstsq(np.column_stack([u,v]), x, rcond=None)
residual = np.linalg.norm(np.column_stack([u,v]) @ coef - x)
compare_print("residual ‖Ac−x‖", f"{residual:.6f}", "> 0 → no solution")

Exercise 5 (Rewriting a System of Equations as a Linear Combination of Vectors)

Rewrite each of the following systems of linear equations as a linear combination of vectors, $x_0\mathbf{u}_0 + x_1\mathbf{u}_1 + \cdots = \mathbf{b}$ (indices start at 0).

$\begin{cases} x_0 + 2x_1 = 3 \\ 2x_0 - x_1 = 1 \end{cases}$
$\begin{cases} x_0 + x_1 + x_2 = 4 \\ 2x_0 + 3x_2 = 5 \\ -x_0 + 2x_1 + x_2 = 1 \end{cases}$
$\begin{cases} 2x_1 - x_2 = 0 \\ 2x_0 + 3x_2 = 0 \\ -x_0 + 4x_1 = 0 \end{cases}$
$\begin{cases} x_0 + 2x_1 + x_3 = 3 \\ 2x_0 - x_1 = 1 \\ -x_0 + 3x_1 = 7 \\ 3x_0 = 4 \end{cases}$

Hint. Regard each column of the coefficient matrix as a vector; $x_i$ is the coefficient of the corresponding column.

# ============================================================
print_header("Exercise 5 | Rewriting Systems of Equations as Linear Combinations of Vectors")
# ============================================================

print_step(1, "(a) x₀(1,2)+x₁(2,−1) = (3,1)")
A_mat = np.array([[1,2],[2,-1]], float)
x_sol = np.linalg.solve(A_mat, [3,1])
print("  [x₀, x₁] =", x_sol)
compare_print("check A·x", A_mat @ x_sol, "[3, 1]")

print_step(2, "(b) 3×3 system")
A_mat = np.array([[1,1,1],[2,0,3],[-1,2,1]], float)
x_sol = np.linalg.lstsq(A_mat, [4,5,1], rcond=None)[0]
print("  solution =", x_sol)
compare_print("check A·x", A_mat @ x_sol, "[4, 5, 1]")

Parametric Equations of Lines and Division Points¶

Exercise 6 (Division Points of a Segment)

Find the point $P$ that divides the segment between two points $A$ , $B$ (or its extension), and express it as a linear combination of vectors.

$A=(1,2)$ , $B=(5,6)$ , and $P$ divides $\overline{AB}$ internally with $\overline{AP}:\overline{PB}=2:1$ .
$A=(2,-1)$ , $B=(6,3)$ , and $P$ divides $\overline{AB}$ externally with $\overline{AP}:\overline{PB}=3:1$ .
$A=(1,0,2)$ , $B=(4,3,-1)$ :
- (a) $P$ divides $\overline{AB}$ internally with $\overline{AP}:\overline{PB}=3:5$ ; find $P$ .
- (b) $A$ divides $\overline{QB}$ externally with $\overline{QA}:\overline{AB}=5:2$ ; find $Q$ .

Hint.

Internal division ( $m:n$ ): $P = \dfrac{n\mathbf{A}+m\mathbf{B}}{m+n}$ .
External division ( $m:n$ , $m\neq n$ ): $P = \dfrac{-n\mathbf{A}+m\mathbf{B}}{m-n}$ ; pay attention to the direction of the signs.

# ============================================================
print_header("Exercise 6 | Division Points of a Segment")
# ============================================================

print_step(1, "(1) ℝ²: A=(1,2), B=(5,6), internal division 2:1")
A, B = np.array([1,2],float), np.array([5,6],float)
P = (1*A + 2*B) / 3
print("  P =", P)

print_step(2, "(2) ℝ²: A=(2,−1), B=(6,3), external division 3:1")
A, B = np.array([2,-1],float), np.array([6,3],float)
P = (-1*A + 3*B) / 2
print("  P =", P)

print_step(3, "(3a) ℝ³: A=(1,0,2), B=(4,3,−1), internal division 3:5")
A, B = np.array([1,0,2],float), np.array([4,3,-1],float)
P = (5*A + 3*B) / 8
print("  P =", P)

print_step(4, "(3b) external division point Q: QA:AB = 5:2")
Q = (5*B - 3*A) / 2
print("  Q =", Q)
t_A = np.dot(A-Q, B-Q) / np.dot(B-Q, B-Q)
ratio = np.linalg.norm(Q-A) / np.linalg.norm(A-B)
assert np.allclose(A, Q + t_A*(B-Q))
assert np.isclose(ratio, 5/2) and not (0 <= t_A <= 1)
print("  QA/AB =", ratio, "; parameter of A on the line QB =", t_A, "(external division)")

Exercise 7 (Division Points on a Parametrized Line)

Let the line be given parametrically by $\mathbf{x}(t) = \mathbf{A} + t(\mathbf{B}-\mathbf{A})$ , $t\in\mathbb{R}$ .

$A=(0,-10,1,-3)$ , $B=(5,0,6,7)$ . Let $P$ lie on this line so that $P$ divides $\overline{AB}$ internally with $\overline{AP}:\overline{PB}=1:4$ ; find $\mathbf{P}$ .
$A=(2,1,-1,0)$ , $B=(6,3,2,4)$ . Let $Q$ divide $\overline{AB}$ externally with $\overline{AQ}:\overline{QB}=4:3$ ; find $\mathbf{Q}$ .
$A=(1,1,\ldots,1)\in\mathbb{R}^n$ , $B=(2,2,\ldots,2)\in\mathbb{R}^n$ :
- (a) Find $P$ such that $P$ divides $\overline{AB}$ internally in the ratio $1:k$ ( $k>0$ ).
- (b) Which point does $P$ approach as $k\to\infty$ ? What about $k\to 0$ ?

Hint. Internal division in the ratio $1:k$ corresponds to $t=\dfrac{1}{1+k}$ ; for external division, $t>1$ or $t<0$ .

# ============================================================
print_header("Exercise 7 | Division Points on a Parametrized Line")
# ============================================================

print_step(1, "(1) ℝ⁴: internal division 1:4, t = 1/5")
A = np.array([0,-10,1,-3],float); B = np.array([5,0,6,7],float)
P = A + (1/5)*(B - A)
print("  P =", P)

print_step(2, "(2) ℝ⁴: external division 4:3, t = 4")
A = np.array([2,1,-1,0],float); B = np.array([6,3,2,4],float)
Q = A + 4*(B - A)
print("  Q =", Q)

print_step(3, "(3) internal division 1:k: behavior as k approaches its extremes")
for k in [0.1, 1, 10, 1000]:
    t = 1/(1+k)
    A_ = np.ones(3); B_ = 2*np.ones(3)
    P = A_ + t*(B_ - A_)
    print(f"  k={k:6}: P[0] = {P[0]:.4f}  (theory: (2+k)/(1+k) = {(2+k)/(1+k):.4f})")
print("  k→∞: approaches A=(1,…,1); k→0: approaches B=(2,…,2)")

The Centroid of a Triangle¶

Exercise 8 (Discovering the Centroid in

\mathbb{R}^4

)

In $\mathbb{R}^4$ , consider the three vertices (position vectors) of a triangle:

\mathbf{A}=(1,2,0,-1),\quad \mathbf{B}=(3,0,2,1),\quad \mathbf{C}=(-1,4,1,3)

(25)

Verify that $\mathbf{M}_C=\dfrac{\mathbf{A}+\mathbf{B}}{2}$ is the midpoint of $AB$ ( $\|\mathbf{A}-\mathbf{M}_C\|=\|\mathbf{B}-\mathbf{M}_C\|$ ), and find the midpoints $\mathbf{M}_A$ , $\mathbf{M}_B$ of $BC$ and $AC$ .
Write down the parametric equations of the medians $\mathbf{L}_A(t)=(1-t)\mathbf{A}+t\mathbf{M}_A$ and $\mathbf{L}_B(s)=(1-s)\mathbf{B}+s\mathbf{M}_B$ .
Setting $\mathbf{L}_A(t)=\mathbf{L}_B(s)$ , find the coordinates of the intersection point $\mathbf{G}$ of the two medians.
Verify that $\mathbf{C}$ , $\mathbf{G}$ , $\mathbf{M}_C$ are collinear.
Verify that $\mathbf{G}=\dfrac{\mathbf{A}+\mathbf{B}+\mathbf{C}}{3}$ .
Verify that $\|\mathbf{A}-\mathbf{G}\|:\|\mathbf{M}_A-\mathbf{G}\|=2:1$ .

Hint. In part 3, compare components to obtain $t$ and $s$ , then substitute into either expression to find $\mathbf{G}$ . In part 4, verify that $\mathbf{G}-\mathbf{C}=\lambda(\mathbf{M}_C-\mathbf{C})$ .

# ============================================================
print_header("Exercise 8 | Discovering the Centroid in ℝ⁴")
# ============================================================

print_step(1, "Define the three vertices and compute the midpoints of the sides")
A = np.array([1,2,0,-1],float)
B = np.array([3,0,2, 1],float)
C = np.array([-1,4,1, 3],float)
MC = (A+B)/2; MA = (B+C)/2; MB = (A+C)/2
print("  MA =", MA, "  MB =", MB, "  MC =", MC)

print_step(2, "Check that MC is the midpoint of AB (A and B are equidistant from MC)")
compare_print("|A−MC| vs |B−MC|",
    f"{np.linalg.norm(A-MC):.6f}",
    f"|B−MC| = {np.linalg.norm(B-MC):.6f} (should be equal)")

print_step(3, "Centroid G = (A+B+C)/3")
G = (A+B+C)/3
print("  G =", G)
print("  ratio of the distances to the centroid |A−G|/|MA−G| =",
      np.linalg.norm(A-G)/np.linalg.norm(MA-G))
compare_print("|A−G|/|MA−G|",
    f"{np.linalg.norm(A-G)/np.linalg.norm(MA-G):.6f}", "2")

print_step(4, "The three points C, G, MC are collinear: λ = 2/3")
diff = MC - C
lam = np.dot(G-C, diff) / np.dot(diff, diff)
print(f"  λ = {lam:.6f}  error = {np.linalg.norm((G-C) - lam*(MC-C)):.2e}")

Exercise 9 (A General Proof for the Centroid of a Triangle in

\mathbb{R}^n

)

Let $\mathbf A,\mathbf B,\mathbf C$ be the position vectors of three non-collinear points in $\mathbb R^n$ , and let

\mathbf{G}=\frac{\mathbf{A}+\mathbf{B}+\mathbf{C}}{3},\quad \mathbf{M}_A=\frac{\mathbf{B}+\mathbf{C}}{2},\quad \mathbf{M}_B=\frac{\mathbf{A}+\mathbf{C}}{2},\quad \mathbf{M}_C=\frac{\mathbf{A}+\mathbf{B}}{2}

(30)

Prove that $\mathbf{A}$ , $\mathbf{G}$ , $\mathbf{M}_A$ are collinear, that is, there exists $\lambda$ such that $\mathbf{G}-\mathbf{A}=\lambda(\mathbf{M}_A-\mathbf{A})$ ; prove similarly that $\mathbf{B}$ , $\mathbf{G}$ , $\mathbf{M}_B$ are collinear and that $\mathbf{C}$ , $\mathbf{G}$ , $\mathbf{M}_C$ are collinear.
Prove that $\|\mathbf A-\mathbf G\|=2\|\mathbf M_A-\mathbf G\|$ ; only when $\mathbf M_A\neq\mathbf A$ is this written as the ratio $2:1$ . The other two cases are similar.

Hint. Compute $\mathbf{G}-\mathbf{A}$ and $\mathbf{M}_A-\mathbf{A}$ directly and find the proportionality between them.

# ============================================================
print_header("Exercise 9 | General Proof for the Centroid in ℝⁿ (numerical check)")
# ============================================================

print_step(1, "Random checks of collinearity and the ratio for n = 3, 5, 10")
rng = np.random.default_rng(42)
for n in [3, 5, 10]:
    A = rng.standard_normal(n); B = rng.standard_normal(n); C = rng.standard_normal(n)
    G  = (A+B+C)/3
    MA = (B+C)/2
    err = np.linalg.norm((G-A) - (2/3)*(MA-A))
    ratio = np.linalg.norm(A-G) / np.linalg.norm(MA-G)
    print(f"  n={n:2d}: collinearity error={err:.2e}  |A−G|/|MA−G| = {ratio:.6f} (should be 2)")

Inner Products, Norms, and Inequalities¶

Exercise 10 (Computing Inner Products)

Compute the following inner products:

$(1,2,-1,3)\cdot(1,0,1,0)$
$(66,24,19)\cdot(23,-17,12)+(23,-17,12)\cdot(-66,-23,-19)$
$((3,4)+(5,12))\cdot((3,4)-(5,12))$
$\left(5\left(\tfrac{1}{3},-\tfrac{1}{3},\tfrac{2}{3}\right)\right)\cdot\left(3\left(\tfrac{1}{5},\tfrac{1}{5},\tfrac{1}{5}\right)\right)$

Hint. For part 2, use the linearity of the inner product; for part 3, use the difference-of-squares expansion $(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}-\mathbf{b})=\|\mathbf{a}\|^2-\|\mathbf{b}\|^2$ .

# ============================================================
print_header("Exercise 10 | Computing Inner Products")
# ============================================================

print_step(1, "(a) (1,2,−1,3)·(1,0,1,0)")
val_a = np.dot([1,2,-1,3],[1,0,1,0])
compare_print("(a) inner product", val_a, "1·1 + 2·0 + (−1)·1 + 3·0 = 0")

print_step(2, "(b) using linearity v·(u+w)")
u = np.array([66,24,19]); v = np.array([23,-17,12]); w = np.array([-66,-23,-19])
val_b = np.dot(v,u) + np.dot(v,w)
compare_print("(b) v·u + v·w = v·(u+w)", val_b, f"{np.dot(v, u+w)}")

print_step(3, "(c) (a+b)·(a−b) = ‖a‖²−‖b‖²")
a_ = np.array([3,4]); b_ = np.array([5,12])
val_c = np.dot(a_+b_, a_-b_)
compare_print("(c) difference-of-squares formula", val_c, f"‖a‖²−‖b‖² = {np.dot(a_,a_)-np.dot(b_,b_)}")

print_step(4, "(d) scalars can be factored out")
val_d = np.dot(5*np.array([1/3,-1/3,2/3]), 3*np.array([1/5,1/5,1/5]))
compare_print("(d) 5·3·(1/3,…)·(1/5,…)", val_d, "5·3·(1/15−1/15+2/15)")

# ============================================================
print_header("Exercise 11 | Unit Vectors")
# ============================================================

print_step(1, "Define u = (1,1,2,−1) and compute ‖u‖")
u = np.array([1,1,2,-1], float)
norm_u = np.linalg.norm(u)
compare_print("‖u‖", f"{norm_u:.6f}", f"√7 = {np.sqrt(7):.6f}")

print_step(2, "Compute the unit vector û = u / ‖u‖")
u_hat = u / norm_u
print("  û =", u_hat)
compare_print("‖û‖ (should be 1)", f"{np.linalg.norm(u_hat):.6f}", "1")

Exercise 12 (Verifying the Cauchy Inequality and the Triangle Inequality)

In $\mathbb{R}^4$ , let $\mathbf{u}=(1,2,1,-1)$ , $\mathbf{v}=(2,-1,0,3)$ , $\mathbf{w}=(0,1,-1,2)$ .

Verify the Cauchy-Schwarz inequality: $|\mathbf{u}\cdot\mathbf{v}|\leq\|\mathbf{u}\|\|\mathbf{v}\|$ .
Verify the triangle inequality: $\|\mathbf{u}+\mathbf{v}\|\leq\|\mathbf{u}\|+\|\mathbf{v}\|$ .

Hint. Compute the numerical values directly and compare the two sides.

# ============================================================
print_header("Exercise 12 | Cauchy–Schwarz and the Triangle Inequality")
# ============================================================

print_step(1, "Define the vectors u, v ∈ ℝ⁴")
u = np.array([1, 2, 1,-1], float)
v = np.array([2,-1, 0, 3], float)
print("  u =", u, "  v =", v)

print_step(2, "Check the Cauchy–Schwarz inequality |u·v| ≤ ‖u‖‖v‖")
cs_lhs = abs(np.dot(u,v))
cs_rhs = np.linalg.norm(u)*np.linalg.norm(v)
compare_print("|u·v|", f"{cs_lhs:.4f}", f"‖u‖‖v‖ = {cs_rhs:.4f}  ✓ {cs_lhs<=cs_rhs}")

print_step(3, "Check the triangle inequality ‖u+v‖ ≤ ‖u‖+‖v‖")
tri_lhs = np.linalg.norm(u+v)
tri_rhs = np.linalg.norm(u)+np.linalg.norm(v)
compare_print("‖u+v‖", f"{tri_lhs:.4f}", f"‖u‖+‖v‖ = {tri_rhs:.4f}  ✓ {tri_lhs<=tri_rhs}")

# ============================================================
print_header("Exercise 13 | Characterizing the Zero Vector by Inner Products")
# ============================================================

print_step(1, "Nonzero vector: u·u = ‖u‖² > 0")
rng = np.random.default_rng(1)
u = rng.standard_normal(5)
compare_print("u·u = ‖u‖²", f"{np.dot(u,u):.6f}", "> 0 (nonzero vector)")

print_step(2, "Zero vector: 0·0 = 0, confirming uniqueness")
compare_print("0·0", np.dot(np.zeros(5), np.zeros(5)), "0 (necessary and sufficient condition: u=0)")

Exercise 14 (The Law of Cosines and the Inner Product)

Let $\mathbf{A}=(a_0,\ldots,a_{n-1})$ , $\mathbf{B}=(b_0,\ldots,b_{n-1})\in\mathbb{R}^n$ , and $\mathbf{A}\cdot\mathbf{B}=\sum a_ib_i$ .

Prove by computation that $\|\mathbf{A}-\mathbf{B}\|^2 = \|\mathbf{A}\|^2+\|\mathbf{B}\|^2-2\mathbf{A}\cdot\mathbf{B}$ .
When $n=2$ or $n=3$ and $\mathbf A,\mathbf B$ are both nonzero, use the geometric law of cosines to prove that $\mathbf A\cdot\mathbf B=\|\mathbf A\|\|\mathbf B\|\cos\theta$ .
When $n=3$ , why does the law of cosines also hold for a triangle in space?
When $n>3$ and $\mathbf A,\mathbf B$ are both nonzero:
- (a) For nonzero vectors $\mathbf A,\mathbf B$ , define the angle $\theta$ by $\cos\theta=\dfrac{\mathbf{A}\cdot\mathbf{B}}{\|\mathbf{A}\|\|\mathbf{B}\|}$ , $0\leq\theta\leq\pi$ . Why is the right-hand side $\in[-1,1]$ ?
- (b) Using this definition, prove that the law of cosines still holds when $n>3$ .

Hint. Part 4(a) relies on the Cauchy-Schwarz inequality; 4(b) follows simply by substituting the definition.

# ============================================================
print_header("Exercise 14 | The Law of Cosines and the Inner Product")
# ============================================================

print_step(1, "Define the vectors A, B ∈ ℝ⁴")
A = np.array([1,2,3,4],float); B = np.array([1,-1,0,2],float)

print_step(2, "Check ‖A−B‖² = ‖A‖² + ‖B‖² − 2A·B")
lhs = np.linalg.norm(A-B)**2
rhs = np.dot(A,A) + np.dot(B,B) - 2*np.dot(A,B)
compare_print("‖A−B‖²", f"{lhs:.4f}", f"‖A‖²+‖B‖²−2A·B = {rhs:.4f}  ✓ {np.isclose(lhs,rhs)}")

print_step(3, "Compute the angle cosθ (Cauchy–Schwarz guarantees ∈[−1,1])")
cos_t = np.dot(A,B)/(np.linalg.norm(A)*np.linalg.norm(B))
compare_print("cosθ", f"{cos_t:.6f}", "∈ [−1, 1]")

Orthogonal Projection¶

Exercise 15 (Computing Vector Projections)

Find the projection vector $\text{proj}_{\mathbf{u}}\mathbf{v}$ of $\mathbf{v}$ along the direction of $\mathbf{u}$ , and the scalar projection (a signed scalar).

$\mathbf{v}=(3,4)$ , $\mathbf{u}=(1,1)$ .
$\mathbf{v}=(-2,1)$ , unit vector $\mathbf{u}=\left(\tfrac{3}{5},\tfrac{4}{5}\right)$ .
$\mathbf{u}=(1,1,1)$ , $\mathbf{v}=(3,0,-3)$ .
$\mathbf{u}=\left(\tfrac{1}{2},-\tfrac{1}{2},\tfrac{1}{2},-\tfrac{1}{2}\right)$ , $\mathbf{v}=(0,6,8,0)$ .

Hint. $\text{proj}_{\mathbf{u}}\mathbf{v}=\dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|^2}\mathbf{u}$ , and the scalar projection (a scalar) is $=\dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|}$ .

# ============================================================
print_header("Exercise 15 | Computing Projection Vectors")
# ============================================================

print_step(1, "Compute the projection vector and the scalar projection for each part")
cases = [
    ("(a) v=(3,4), u=(1,1)",
     np.array([3.,4.]), np.array([1.,1.])),
    ("(b) v=(−2,1), u=(3/5,4/5)",
     np.array([-2.,1.]), np.array([3/5,4/5])),
    ("(c) u=(1,1,1), v=(3,0,−3)",
     np.array([3.,0.,-3.]), np.array([1.,1.,1.])),
    ("(d) u=(½,−½,½,−½), v=(0,6,8,0)",
     np.array([0.,6.,8.,0.]), np.array([.5,-.5,.5,-.5])),
]
for label, v, u in cases:
    p = proj(u, v)
    scalar = np.dot(u,v)/np.linalg.norm(u)
    print(f"  {label}")
    print(f"    projection vector = {p}  scalar projection = {scalar:.4f}")

Exercise 16 (Projection onto a Line through the Origin)

In each part below, compute the projection vector of $\mathbf{v}$ onto the line through the origin with direction $\mathbf{u}$ , and the distance.

$\mathbf{v}=(3,4)$ , $\mathbf{u}=(1,1)$ . Compute $\text{proj}_{\mathbf{u}}\mathbf{v}$ and the distance from the point $(3,4)$ to the line.
$\mathbf{v}=(1,2,-3)$ , $\mathbf{u}=(1,0,1)$ . Compute $\text{proj}_{\mathbf{u}}\mathbf{v}$ , and verify that $\mathbf{v}-\text{proj}_{\mathbf{u}}\mathbf{v}\perp\mathbf{u}$ .
$\mathbf{v}=(1,2,1,0)$ , $\mathbf{u}=(1,0,1,0)$ . Verify that $\mathbf{v}-\text{proj}_{\mathbf{u}}\mathbf{v}\perp\mathbf{u}$ .

Hint. The distance is $=\|\mathbf{v}-\text{proj}_{\mathbf{u}}\mathbf{v}\|$ ; to verify perpendicularity, it suffices to check that the inner product is zero.

# ============================================================
print_header("Exercise 16 | Projection onto a Line through the Origin and Distance")
# ============================================================

print_step(1, "(1) v=(3,4), line direction u=(1,1)")
v, u = np.array([3.,4.]), np.array([1.,1.])
r = v - proj(u,v)
compare_print("distance ‖v−proj‖", f"{np.linalg.norm(r):.6f}", f"1/√2 = {1/np.sqrt(2):.6f}")

print_step(2, "(2) v=(1,2,−3), line direction u=(1,0,1)")
v, u = np.array([1.,2.,-3.]), np.array([1.,0.,1.])
r = v - proj(u,v)
compare_print("residual·u (should be 0)", f"{np.dot(r,u):.2e}", "0")

print_step(3, "(3) v=(1,2,1,0) ∈ ℝ⁴, direction u=(1,0,1,0)")
v, u = np.array([1.,2.,1.,0.]), np.array([1.,0.,1.,0.])
r = v - proj(u,v)
compare_print("residual·u (should be 0)", f"{np.dot(r,u):.2e}", "0")

Exercise 17 (General Properties of Projection)

Let $\mathbf{u}\neq\mathbf{0}$ .

If $\|\mathbf{u}\|=1$ , prove that $\text{proj}_{\mathbf{u}}\mathbf{v}=(\mathbf{v}\cdot\mathbf{u})\mathbf{u}$ .
Prove that the vector decomposition $\mathbf{v}=\text{proj}_{\mathbf{u}}\mathbf{v}+(\mathbf{v}-\text{proj}_{\mathbf{u}}\mathbf{v})$ satisfies $\text{proj}_{\mathbf{u}}\mathbf{v}\perp(\mathbf{v}-\text{proj}_{\mathbf{u}}\mathbf{v})$ .
Using this decomposition, prove that the distance from $\mathbf{v}$ to the line through the origin with direction $\mathbf{u}$ equals

\frac{\left(\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2\right)^{1/2}}{\|\mathbf{u}\|}

(46)

Hint. For part 3: the distance equals $\|\mathbf{v}-\text{proj}_\mathbf{u}\mathbf{v}\|$ ; then expand and compute.

# ============================================================
print_header("Exercise 17 | General Properties of Projection")
# ============================================================

print_step(1, "Randomly generate u, v ∈ ℝ⁴ and compute the projection and the residual")
rng = np.random.default_rng(7)
u = rng.standard_normal(4); v = rng.standard_normal(4)
p = proj(u, v); r = v - p

print_step(2, "Check p ⊥ r")
compare_print("p·r (should be ≈0)", f"{np.dot(p,r):.2e}", "0")

print_step(3, "Check the distance formula ‖v−proj_u(v)‖")
dist1 = np.sqrt(np.dot(u,u)*np.dot(v,v) - np.dot(u,v)**2) / np.linalg.norm(u)
dist2 = np.linalg.norm(r)
compare_print("value of the distance formula", f"{dist1:.6f}", f"direct computation ‖r‖ = {dist2:.6f}  ✓ {np.isclose(dist1,dist2)}")

Exercise 18 (Distance from a Point to a Line)

Compute the distance from each point to the given line.

The distance from the point $(2,3)$ to the line $x+y=0$ .
The distance from the point $(3,1)$ to the line through $(-1,2)$ with direction $\mathbf{u}=(1,1)$ .
The distance from the point $(1,2,3)$ to the line through the origin with direction $\mathbf{u}=(1,0,1)$ .
The distance from the point $(2,1,0)$ to the line $\begin{cases}x+y+z=0\\y=1\end{cases}$ .
The distance from the point $P=(1,1,1,1)$ to the line through the origin with direction $\mathbf{u}=(1,0,0,1)$ (in $\mathbb{R}^4$ ).

Hint. For a line that does not pass through the origin, first translate the point so that the line passes through the origin, then use the projection formula.

# ============================================================
print_header("Exercise 18 | Distance from a Point to a Line")
# ============================================================

print_step(1, "(1) (2,3) to the line through the origin with direction (1,−1)")
p = np.array([2.,3.]); u = np.array([1.,-1.])
compare_print("distance", f"{np.linalg.norm(p - proj(u,p)):.6f}", f"5√2/2 = {5*np.sqrt(2)/2:.6f}")

print_step(2, "(2) (3,1) to the line through (−1,2) with direction (1,1)")
p = np.array([3.,1.]) - np.array([-1.,2.]); u = np.array([1.,1.])
compare_print("distance", f"{np.linalg.norm(p - proj(u,p)):.6f}", f"5√2/2 = {5*np.sqrt(2)/2:.6f}")

print_step(3, "(3) (1,2,3) to the line through the origin with direction (1,0,1)")
p = np.array([1.,2.,3.]); u = np.array([1.,0.,1.])
compare_print("distance", f"{np.linalg.norm(p - proj(u,p)):.6f}", f"√6 = {np.sqrt(6):.6f}")

print_step(4, "(4) (2,1,0) to the line {x+y+z=0, y=1}")
base = np.array([-1.,1.,0.]); u = np.array([1.,0.,-1.])
p = np.array([2.,1.,0.]) - base
compare_print("distance", f"{np.linalg.norm(p - proj(u,p)):.6f}", f"3√2/2 = {3*np.sqrt(2)/2:.6f}")

print_step(5, "(5) P=(1,1,1,1) ∈ ℝ⁴, direction (1,0,0,1)")
p = np.array([1.,1.,1.,1.]); u = np.array([1.,0.,0.,1.])
compare_print("distance", f"{np.linalg.norm(p - proj(u,p)):.6f}", f"√2 = {np.sqrt(2):.6f}")

Exercise 19 (Projection onto a Plane (

\mathbb{R}^3

))

In $\mathbb{R}^3$ , the plane $\Sigma$ through the origin has the parametric equation $\mathbf{p}(s,t)=s(1,-1,0)+t(0,1,-1)$ . Given $\mathbf{v}=(0,0,4)$ , find $\text{proj}_\Sigma\mathbf{v}$ by the following steps.

Find a normal vector (or unit normal vector) $\mathbf{n}$ of the plane.
Find the projection of $\mathbf{v}$ in the direction of $\mathbf{n}$ , then set $\text{proj}_\Sigma\mathbf{v}=\mathbf{v}-\text{proj}_\mathbf{n}\mathbf{v}$ .
Verify that $\text{proj}_\Sigma\mathbf{v}\in\Sigma$ (that is, it is perpendicular to $\mathbf{n}$ ).

Hint. The normal vector is $\mathbf{n}=\mathbf{u}_0\times\mathbf{u}_1$ (cross product), where $\mathbf{u}_0=(1,-1,0)$ and $\mathbf{u}_1=(0,1,-1)$ are the two direction vectors of the plane; alternatively, $\mathbf{n}$ can be determined from the two direction vectors by requiring $\mathbf{n}\perp\mathbf{u}_0$ , $\mathbf{n}\perp\mathbf{u}_1$ .

# ============================================================
print_header("Exercise 19 | Projection onto a Plane")
# ============================================================

print_step(1, "ℝ³: Σ = span{(1,−1,0),(0,1,−1)}, v=(0,0,4)")
u0 = np.array([1.,-1.,0.]); u1 = np.array([0.,1.,-1.])
n  = np.cross(u0, u1)
v  = np.array([0.,0.,4.])
proj_sigma = v - proj(n, v)
print("  normal vector n =", n)
print("  proj_Σ(v) =", proj_sigma)
compare_print("residual·n (should be 0)", f"{np.dot(proj_sigma, n):.2e}", "0")

print_step(2, "ℝ³: distance from v=(3,−1,2) to the plane x+2y−z=0")
n = np.array([1.,2.,-1.]); v = np.array([3.,-1.,2.])
ps = v - proj(n,v)
compare_print("distance ‖v−proj_Σ‖", f"{np.linalg.norm(v-ps):.6f}", "|n·v|/‖n‖")

print_step(3, "ℝ⁴: Σ = span{(1,0,1,0),(0,1,0,1)}, v=(1,1,1,1)")
u0 = np.array([1.,0.,1.,0.]); u1 = np.array([0.,1.,0.,1.])
v  = np.array([1.,1.,1.,1.])
G_mat = np.array([[np.dot(u0,u0), np.dot(u0,u1)],
                   [np.dot(u1,u0), np.dot(u1,u1)]])
b_ = np.array([np.dot(v,u0), np.dot(v,u1)])
ab = np.linalg.solve(G_mat, b_)
ps4 = ab[0]*u0 + ab[1]*u1
print("  proj_Σ(v) =", ps4)
compare_print("residual ⊥ u0,u1",
    f"‖r·u0‖={abs(np.dot(v-ps4,u0)):.2e}, ‖r·u1‖={abs(np.dot(v-ps4,u1)):.2e}", "0, 0")

Exercise 20 (Distance from a Vector to a Plane (Higher Dimensions))

Compute the distance from each point to the given plane.

In $\mathbb{R}^3$ , the distance from the point $P=(2,1,3)$ to the plane $x-2y+2z=0$ .
In $\mathbb{R}^3$ , the plane has the parametric equation $\mathbf{p}(s,t)=s(0,1,1)+t(0,1,-1)$ ; find the distance from the point $P=(1,2,1)$ to this plane.
In $\mathbb{R}^3$ , the plane has the parametric equation $\mathbf{p}(s,t)=(2,1,-1)+s(1,1,1)+t(1,-1,0)$ ; find the distance from the origin to this plane.
In $\mathbb{R}^4$ , for the plane $\mathbf{p}(s,t)=s(1,1,0,0)+t(0,0,1,1)$ , find the distance from the point $P=(2,1,1,1)$ to this plane.

Hint. In $\mathbb{R}^3$ , use a normal vector: the distance is $=\dfrac{|\mathbf{n}\cdot(\mathbf{P}-\mathbf{a})|}{\|\mathbf{n}\|}$ . In $\mathbb{R}^4$ the normal vector is not unique, so you need the projection method of solving a system of equations.

# ============================================================
print_header("Exercise 20 | Distance from a Point to a Plane")
# ============================================================

print_step(1, "(1) P=(2,1,3) to the plane x−2y+2z=0 given by its normal vector")
P = np.array([2.,1.,3.]); n = np.array([1.,-2.,2.])
compare_print("distance |n·P|/‖n‖", f"{abs(np.dot(P,n))/np.linalg.norm(n):.6f}", "2")

print_step(2, "(2) P=(1,2,1) to span{(0,1,1),(0,1,−1)}")
u0=np.array([0.,1.,1.]); u1=np.array([0.,1.,-1.]); P=np.array([1.,2.,1.])
n = np.cross(u0,u1)
compare_print("distance", f"{abs(np.dot(P,n))/np.linalg.norm(n):.6f}", "1")

print_step(3, "(3) the origin to the affine plane (2,1,−1)+s(1,1,1)+t(1,−1,0)")
u0=np.array([1.,1.,1.]); u1=np.array([1.,-1.,0.]); a=np.array([2.,1.,-1.])
n = np.cross(u0,u1)
dist = abs(np.dot(np.zeros(3)-a, n))/np.linalg.norm(n)
compare_print("distance", f"{dist:.6f}", f"5/√6 = {5/np.sqrt(6):.6f}")

print_step(4, "(4) P=(2,1,1,1) ∈ ℝ⁴ to span{(1,1,0,0),(0,0,1,1)}")
u0=np.array([1.,1.,0.,0.]); u1=np.array([0.,0.,1.,1.]); P=np.array([2.,1.,1.,1.])
G_mat=np.array([[np.dot(u0,u0),np.dot(u0,u1)],[np.dot(u1,u0),np.dot(u1,u1)]])
ab=np.linalg.solve(G_mat,[np.dot(P,u0),np.dot(P,u1)])
ps=ab[0]*u0+ab[1]*u1
compare_print("distance", f"{np.linalg.norm(P-ps):.6f}", f"√2/2 = {np.sqrt(2)/2:.6f}")

Orthogonal Bases¶

Exercise 21 (An Orthogonal Basis in

\mathbb{R}^2

)

In $\mathbb{R}^2$ , consider the vectors $\mathbf{u}_0=\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)$ , $\mathbf{u}_1=\left(\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}\right)$ .

Verify that $\|\mathbf{u}_0\|=\|\mathbf{u}_1\|=1$ and $\mathbf{u}_0\cdot\mathbf{u}_1=0$ .
If $\mathbf{v}=c_0\mathbf{u}_0+c_1\mathbf{u}_1$ , use inner products to explain why $c_i=\mathbf{v}\cdot\mathbf{u}_i$ .
For $\mathbf{v}=(3,1)$ , compute $c_0,c_1$ , and verify that $\mathbf{v}=c_0\mathbf{u}_0+c_1\mathbf{u}_1$ .

Hint. For part 2: take the inner product of both sides of $\mathbf{v}=c_0\mathbf{u}_0+c_1\mathbf{u}_1$ with $\mathbf{u}_i$ , and simplify using orthogonality.

# ============================================================
print_header("Exercise 21–23 | Orthogonal Bases")
# ============================================================

print_step(1, "ℝ²: expansion of v=(3,1) in the Hadamard orthogonal basis")
u0=np.array([1,1])/np.sqrt(2); u1_=np.array([1,-1])/np.sqrt(2)
v=np.array([3.,1.])
c0,c1=np.dot(v,u0),np.dot(v,u1_)
print(f"  c₀ = {c0:.4f},  c₁ = {c1:.4f}")
compare_print("reconstruction error", np.allclose(c0*u0+c1*u1_,v), "True")
compare_print("Parseval's identity", np.isclose(c0**2+c1**2, np.dot(v,v)), "True")

print_step(2, "ℝ³: check of the orthonormal matrix U")
U=np.array([[2/3,2/3,1/3],[1/3,-2/3,2/3],[-2/3,1/3,2/3]])
print("  UU^T ="); print(np.round(U@U.T,8))
v=np.array([2.,0.,0.])
c=U@v
compare_print("Parseval (ℝ³)", np.isclose(np.dot(c,c),np.dot(v,v)), "True")

print_step(3, "ℝ⁴: Hadamard matrix (scaled by ½)")
H=np.array([[1,1,1,1],[1,1,-1,-1],[1,-1,1,-1],[1,-1,-1,1]])/2.0
print("  HH^T ="); print(np.round(H@H.T,8))
v=np.array([3.,0.,4.,0.])
c=H@v
compare_print("Parseval (ℝ⁴)", np.isclose(np.dot(c,c),np.dot(v,v)), "True")

Exercise 22 (An Orthogonal Basis in

\mathbb{R}^3

)

In $\mathbb{R}^3$ , consider

\mathbf{u}_0=\left(\tfrac{2}{3},\tfrac{2}{3},\tfrac{1}{3}\right),\quad \mathbf{u}_1=\left(\tfrac{1}{3},-\tfrac{2}{3},\tfrac{2}{3}\right),\quad \mathbf{u}_2=\left(-\tfrac{2}{3},\tfrac{1}{3},\tfrac{2}{3}\right)

(54)

Verify that this is an orthogonal basis ( $\|\mathbf{u}_i\|=1$ , $\mathbf{u}_i\cdot\mathbf{u}_j=0$ for $i\neq j$ ).
Express $\mathbf{v}=(2,0,0)$ as $c_0\mathbf{u}_0+c_1\mathbf{u}_1+c_2\mathbf{u}_2$ .
Verify that $c_0^2+c_1^2+c_2^2=\|\mathbf{v}\|^2$ .

Exercise 23 (An Orthogonal Basis in

\mathbb{R}^4

(the Hadamard Matrix))

In $\mathbb{R}^4$ , consider

\mathbf{u}_0=\tfrac{1}{2}(1,1,1,1),\quad \mathbf{u}_1=\tfrac{1}{2}(1,1,-1,-1),\quad \mathbf{u}_2=\tfrac{1}{2}(1,-1,1,-1),\quad \mathbf{u}_3=\tfrac{1}{2}(1,-1,-1,1)

(56)

Verify that this is an orthogonal basis.
Express $\mathbf{v}=(3,0,4,0)$ as a linear combination of this basis.
Verify that $\|\mathbf{v}\|^2=c_0^2+c_1^2+c_2^2+c_3^2$ .

Hint. These are the column vectors (after scaling) of the $4\times4$ Hadamard matrix, which is extremely important in signal processing.

The Cross Product¶

Exercise 24 (Basic Properties of the Cross Product)

Let $\mathbf{u}=(u_0,u_1,u_2)$ , $\mathbf{v}=(v_0,v_1,v_2)$ , and define the cross product by

\mathbf{u}\times\mathbf{v}=\left( \begin{vmatrix}u_1&u_2\\v_1&v_2\end{vmatrix},\; -\begin{vmatrix}u_0&u_2\\v_0&v_2\end{vmatrix},\; \begin{vmatrix}u_0&u_1\\v_0&v_1\end{vmatrix} \right)

(59)

Compute and prove that $\mathbf{u}\times\mathbf{v}$ is perpendicular to $\text{span}\{\mathbf{u},\mathbf{v}\}$ .
Prove the area formula: $\|\mathbf{u}\times\mathbf{v}\|^2=\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2$ .
Assume $\mathbf u,\mathbf v$ are both nonzero. Using the angle definition from the inner product, prove that $\|\mathbf{u}\times\mathbf{v}\|=\|\mathbf{u}\|\|\mathbf{v}\||\sin\theta|$ .
If $\mathbf{u}$ and $\mathbf{v}$ are linearly independent, explain why the area of the parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$ equals $\|\mathbf{u}\times\mathbf{v}\|$ .

Hint. For part 1, compute $(\mathbf{u}\times\mathbf{v})\cdot\mathbf{u}$ directly and expand; for part 2, expand both sides and compare.

# ============================================================
print_header("Exercise 24 | Basic Properties of the Cross Product")
# ============================================================

print_step(1, "Compute u×v and check perpendicularity")
u = np.array([1.,2.,3.]); v = np.array([1.,0.,1.])
w = np.cross(u, v)
print("  u×v =", w)
compare_print("(u×v)·u (should be 0)", f"{np.dot(w,u):.2e}", "0")
compare_print("(u×v)·v (should be 0)", f"{np.dot(w,v):.2e}", "0")

print_step(2, "Check the area formula ‖u×v‖² = ‖u‖²‖v‖²−(u·v)²")
lhs = np.dot(w,w)
rhs = np.dot(u,u)*np.dot(v,v) - np.dot(u,v)**2
compare_print("‖u×v‖²", f"{lhs:.4f}", f"‖u‖²‖v‖²−(u·v)² = {rhs:.4f}  ✓ {np.isclose(lhs,rhs)}")

print_step(3, "The area agrees with para_area")
compare_print("‖u×v‖", f"{np.linalg.norm(w):.6f}", f"para_area(u,v) = {para_area(u,v):.6f}")

Exercise 25 (Deciding Cross Product Identities)

For each of the following cross product identities, prove it if it is true, and give a counterexample if it is false.

$\mathbf{u}\times\mathbf{u}=\mathbf{0}$
$\mathbf{u}\times\mathbf{v}=\mathbf{v}\times\mathbf{u}$
$\mathbf{u}\times(\mathbf{v}+\mathbf{w})=\mathbf{u}\times\mathbf{v}+\mathbf{u}\times\mathbf{w}$
$\mathbf{u}\times(\mathbf{v}\times\mathbf{w})=(\mathbf{u}\times\mathbf{v})\times\mathbf{w}$
$\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})=-\mathbf{w}\cdot(\mathbf{v}\times\mathbf{u})$

# ============================================================
print_header("Exercise 25 | Deciding Cross Product Identities")
# ============================================================

u=np.array([1.,0.,0.]); v=np.array([0.,1.,0.]); w=np.array([0.,0.,1.])

print_step(1, "(1) u×u = 0 (alternating property)")
compare_print("u×u", np.cross(u,u), "[0, 0, 0]")

print_step(2, "(2) u×v ≠ v×u (counterexample showing antisymmetry)")
print("  u×v =", np.cross(u,v), "  v×u =", np.cross(v,u))
compare_print("u×v = v×u? (should be False)", np.allclose(np.cross(u,v), np.cross(v,u)), "False")

print_step(3, "(3) distributivity u×(v+w) = u×v + u×w")
compare_print("distributivity holds?", np.allclose(np.cross(u,v+w), np.cross(u,v)+np.cross(u,w)), "True")

print_step(4, "(4) the cross product is not associative (counterexample)")
u4 = np.array([1.,0.,0.]); v4 = u4.copy(); w4 = np.array([0.,1.,0.])   # counterexample: u=v=e₀, w=e₁
print("  u×(v×w) =", np.cross(u4,np.cross(v4,w4)))
print("  (u×v)×w =", np.cross(np.cross(u4,v4),w4))
compare_print("associativity holds? (should be False)",
    np.allclose(np.cross(u4,np.cross(v4,w4)), np.cross(np.cross(u4,v4),w4)), "False")

print_step(5, "(5) cyclic symmetry: D(u,v,w) = −D(w,v,u)")
compare_print("D(u,v,w) vs −D(w,v,u)",
    f"{triple(u,v,w):.4f}", f"{-triple(w,v,u):.4f}")

Exercise 26 (Cross Product Computation Practice)

Compute:

$(1,2,3)\times(3,2,1)$
$(1,2,3)\times((1,2,3)+(3,2,1))$
$(1,2,3)\times((1,2,3)\times(3,2,1))$
$(3,2,1)\cdot((1,2,3)\times(3,2,1))$

Hint. For part 2, use distributivity; for part 4, use $\mathbf{v}\cdot(\mathbf{v}\times\mathbf{w})=0$ .

# ============================================================
print_header("Exercise 26 | Cross Product Computations")
# ============================================================

u=np.array([1.,2.,3.]); v=np.array([3.,2.,1.])

print_step(1, "(1) u×v")
print("  u×v =", np.cross(u,v))

print_step(2, "(2) u×(u+v) (should equal u×v)")
compare_print("u×(u+v)", np.cross(u,u+v), f"u×v = {np.cross(u,v)}")

print_step(3, "(3) u×(u×v) (double cross product, BAC–CAB)")
print("  u×(u×v) =", np.cross(u,np.cross(u,v)))

print_step(4, "(4) v·(u×v) (should be 0)")
compare_print("v·(u×v)", f"{np.dot(v,np.cross(u,v)):.2e}", "0")

Exercise 27 (Proof of the BAC-CAB Formula)

Prove the vector triple product formula:

\mathbf{u}\times(\mathbf{v}\times\mathbf{w})=(\mathbf{u}\cdot\mathbf{w})\mathbf{v}-(\mathbf{u}\cdot\mathbf{v})\mathbf{w}

(65)

Hint. Expand component 0 directly, then cycle the indices to obtain components 1 and 2; at no point do you need to divide by any inner product.

# ============================================================
print_header("Exercise 27 | Checking the BAC–CAB Formula")
# ============================================================

print_step(1, "Randomly generate u, v, w ∈ ℝ³")
rng = np.random.default_rng(0)
u=rng.standard_normal(3); v=rng.standard_normal(3); w=rng.standard_normal(3)
print("  u =", u)
print("  v =", v)
print("  w =", w)

print_step(2, "Check u×(v×w) = (u·w)v − (u·v)w")
lhs = np.cross(u, np.cross(v,w))
rhs = np.dot(u,w)*v - np.dot(u,v)*w
print("  left-hand side u×(v×w) =", lhs)
print("  right-hand side (u·w)v−(u·v)w =", rhs)
compare_print("left-hand side ≈ right-hand side?", np.allclose(lhs,rhs), "True")

Exercise 28 (A Vector Proof of the Law of Sines)

In $\mathbb{R}^3$ , consider the triangle spanned by linearly independent vectors $\mathbf{a}$ and $\mathbf{b}$ , and let $\mathbf{c}=\mathbf{a}-\mathbf{b}$ . Starting from $\mathbf{c}\times\mathbf{c}=\mathbf{0}$ , prove that the three sides of this triangle satisfy the law of sines.

Hint. Expanding $(\mathbf{a}-\mathbf{b})\times(\mathbf{a}-\mathbf{b})=\mathbf{0}$ gives $\mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}$ ... Then derive the result from the equation $\|\mathbf{a}\times\mathbf{b}\|=\|\mathbf{a}\|\|\mathbf{b}\|\sin C$ .

Exercise 29 (The Three Parallelograms Determined by Three Points Have Equal Areas)

Three non-collinear points $P$ , $Q$ , $R$ in space determine three different parallelograms (there are three choices for the fourth vertex). Use the cross product formula to prove that these three parallelograms all have the same area.

Hint. The three possible fourth vertices are $S_0=P+Q-R$ , $S_1=P+R-Q$ , $S_2=Q+R-P$ , and the corresponding parallelograms are formed from different combinations of side vectors.

Exercise 30 (Volume of a Parallelepiped)

In $\mathbb{R}^3$ , consider $\mathbf{u}=(1,1,1)$ , $\mathbf{v}=(-1,0,1)$ , $\mathbf{w}=(0,0,2)$ .

Compute $\text{proj}_{\mathbf{v}\times\mathbf{w}}\mathbf{u}$ .
Use the formula base area $\times$ height to compute the volume of the parallelepiped spanned by the three vectors.

Hint. The base is the parallelogram spanned by $\mathbf{v}$ and $\mathbf{w}$ , with area $\|\mathbf{v}\times\mathbf{w}\|$ ; the height is the scalar projection of $\mathbf{u}$ in the direction of the normal vector, $=|\mathbf{u}\cdot\hat{\mathbf{n}}|$ .

# ============================================================
print_header("Exercise 30 | Volume of a Parallelepiped")
# ============================================================

print_step(1, "Define the three vectors u, v, w ∈ ℝ³")
u=np.array([1.,1.,1.]); v=np.array([-1.,0.,1.]); w=np.array([0.,0.,2.])

print_step(2, "Compute v×w (normal vector of the base)")
vxw = np.cross(v, w)
print("  v×w =", vxw)

print_step(3, "Compute the volume (base area × height vs scalar triple product)")
p = proj(vxw, u)
vol_geo = np.linalg.norm(vxw)*np.linalg.norm(p)
vol_trip = abs(triple(u,v,w))
compare_print("base area×height", f"{vol_geo:.6f}", f"|u·(v×w)| = {vol_trip:.6f}")

Exercise 31 (Properties of the Signed Volume)

This exercise is in fact a three-dimensional preview of the determinant axioms of Chapter 7: the following five properties (normalization, multilinearity, antisymmetry, cyclic symmetry, and the degeneracy criterion) are exactly the axiomatic framework that Chapter 7 uses to define the determinant. Once you understand them thoroughly in familiar three-dimensional geometry, the abstract definition in Chapter 7 will come with a ready-made geometric anchor.

Let $D(\mathbf{u},\mathbf{v},\mathbf{w})=\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})$ for $\mathbf{u},\mathbf{v},\mathbf{w}\in\mathbb{R}^3$ . Prove:

$D(\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2)=1$
$D(\alpha\mathbf{u}_0+\beta\mathbf{u}_1,\mathbf{v},\mathbf{w})=\alpha D(\mathbf{u}_0,\mathbf{v},\mathbf{w})+\beta D(\mathbf{u}_1,\mathbf{v},\mathbf{w})$ (linearity in the first argument)
$D(\mathbf{u},\mathbf{v},\mathbf{w})=-D(\mathbf{v},\mathbf{u},\mathbf{w})$
$D(\mathbf{u},\mathbf{v},\mathbf{w})=D(\mathbf{v},\mathbf{w},\mathbf{u})=D(\mathbf{w},\mathbf{u},\mathbf{v})$ (cyclic symmetry)
$\mathbf{u},\mathbf{v},\mathbf{w}$ are linearly dependent if and only if $D(\mathbf{u},\mathbf{v},\mathbf{w})=0$

Hint. For part 5: linear dependence is equivalent to one vector being expressible as a linear combination of the other two; substitute this and use properties 1–3.

# ============================================================
print_header("Exercise 31 | Properties of the Signed Volume D(u,v,w)")
# ============================================================

print_step(1, "Baseline check: D(e₀,e₁,e₂) = 1")
e = np.eye(3)
compare_print("D(e₀,e₁,e₂)", triple(e[0],e[1],e[2]), "1")

print_step(2, "Linearity: D(2u+3v, a, b) = 2D(u,a,b) + 3D(v,a,b)")
rng=np.random.default_rng(5)
u=rng.standard_normal(3); v=rng.standard_normal(3)
a=rng.standard_normal(3); b=rng.standard_normal(3)
lhs = triple(2*u+3*v,a,b)
rhs = 2*triple(u,a,b)+3*triple(v,a,b)
compare_print("linearity", f"{lhs:.6f}", f"2D+3D = {rhs:.6f}")

print_step(3, "Antisymmetry: D(u,v,a) = −D(v,u,a)")
compare_print("antisymmetry",
    f"D(u,v,a)={triple(u,v,a):.6f}",
    f"−D(v,u,a)={-triple(v,u,a):.6f}")

print_step(4, "Cyclic symmetry: D(u,v,a)=D(v,a,u)=D(a,u,v)")
print(f"  D(u,v,a)={triple(u,v,a):.6f}  D(v,a,u)={triple(v,a,u):.6f}  D(a,u,v)={triple(a,u,v):.6f}")

Exercise 32 (Deciding Whether Four Points Are Coplanar)

Decide whether the four points $(1,2,1)$ , $(3,1,2)$ , $(2,3,0)$ , $(4,2,3)$ are coplanar.

Hint. Taking the first point as the base point, form three difference vectors and compute the signed volume $D$ . $D=0$ means the points are coplanar.

# ============================================================
print_header("Exercise 32 | Deciding Whether Four Points Are Coplanar")
# ============================================================

print_step(1, "Define the four points P₀,P₁,P₂,P₃ ∈ ℝ³")
P0=np.array([1.,2.,1.]); P1=np.array([3.,1.,2.])
P2=np.array([2.,3.,0.]); P3=np.array([4.,2.,3.])

print_step(2, "Compute the signed volume D(P₁−P₀, P₂−P₀, P₃−P₀)")
d = triple(P1-P0, P2-P0, P3-P0)
compare_print("value of D (D=0 ⟺ coplanar)", f"{d:.6f}", "≠ 0 → the four points are not coplanar")
print("  verdict:", "coplanar" if abs(d)<1e-10 else "not coplanar")

Exercise 33 (The Perpendicular-Diagonals Condition for a Rhombus)

In $\mathbb{R}^n$ , consider the parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$ (with vertices $\mathbf{0}$ , $\mathbf{u}$ , $\mathbf{v}$ , $\mathbf{u}+\mathbf{v}$ ). Prove that the two diagonals are perpendicular if and only if this parallelogram is a rhombus ( $\|\mathbf{u}\|=\|\mathbf{v}\|$ ).

Hint. The two diagonal vectors are $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}-\mathbf{v}$ ; compute their inner product.

# ============================================================
print_header("Exercise 33 | The Perpendicular-Diagonals Condition for a Rhombus")
# ============================================================

print_step(1, "Check (u+v)·(u−v) = ‖u‖²−‖v‖² for three shapes")
cases = [
    ("square    u=(1,0), v=(0,1)",      np.array([1.,0.]), np.array([0.,1.])),
    ("rectangle u=(2,0), v=(0,1)",      np.array([2.,0.]), np.array([0.,1.])),
    ("rhombus   u=(1,0), v=(cos60°,sin60°)", np.array([1.,0.]), np.array([.5,np.sqrt(3)/2])),
]
for label, u, v in cases:
    dot = np.dot(u+v, u-v)
    is_rhombus = np.isclose(np.linalg.norm(u), np.linalg.norm(v))
    is_perp    = np.isclose(dot, 0)
    print(f"  {label}")
    print(f"    (u+v)·(u−v)={dot:.4f}  rhombus={is_rhombus}  diagonals perpendicular={is_perp}")

Exercise 34 (The Area of a Parallelogram in

\mathbb{R}^4

)

There is no cross product in $\mathbb{R}^4$ ; derive the area formula for a parallelogram by following the guided steps below.

The vectors $\overrightarrow{OP}=(1,0,1,0)$ , $\overrightarrow{OQ}=(0,1,0,1)$ :
- (a) Compute $\overline{OH}=\|\text{proj}_{\overrightarrow{OQ}}\overrightarrow{OP}\|$ .
- (b) Compute the height $\overline{PH}=\|\overrightarrow{OP}-\text{proj}_{\overrightarrow{OQ}}\overrightarrow{OP}\|$ .
- (c) Compute the area of the parallelogram.
For linearly independent vectors $\mathbf{u},\mathbf{v}\in\mathbb{R}^4$ , prove in the same way that

\text{area}=\sqrt{\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2}

(75)

Starting from area $=\|\mathbf{u}\|\|\mathbf{v}\|\sin\theta$ , derive the formula above again using the angle definition from the inner product.
If $\mathbf{v}=\lambda\mathbf{u}$ , check that the area formula gives 0.
If $\mathbf{u}\perp\mathbf{v}$ , find the area and explain its geometric meaning.
If $\|\mathbf{u}\|=\|\mathbf{v}\|$ (a rhombus), check that the area of the rhombus is $=\dfrac{pq}{2}$ ( $p,q$ are the lengths of the diagonals).

Hint. In part 2, the base is $=\|\mathbf{v}\|$ and the height is $=\|\mathbf{u}-\text{proj}_\mathbf{v}\mathbf{u}\|$ ; then expand.

Solution to Exercise 34

Part 1. $\overrightarrow{OP}\cdot\overrightarrow{OQ}=0$ , so $\text{proj}_{\overrightarrow{OQ}}\overrightarrow{OP}=\mathbf{0}$ . Then $\overline{OH}=0$ , the height is $\overline{PH}=\|\overrightarrow{OP}\|=\sqrt{2}$ , and the area is $=\|\overrightarrow{OQ}\|\cdot\overline{PH}=\sqrt{2}\cdot\sqrt{2}=2$ .

Part 2.

\text{area}=\|\mathbf{v}\|\cdot\left\|\mathbf{u}-\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{v}\|^2}\mathbf{v}\right\|

(76)

= \|\mathbf{v}\|\cdot\sqrt{\|\mathbf{u}\|^2-\frac{(\mathbf{u}\cdot\mathbf{v})^2}{\|\mathbf{v}\|^2}} = \sqrt{\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2}

(77)

Part 3. $\text{area}=\|\mathbf{u}\|\|\mathbf{v}\|\sin\theta=\|\mathbf{u}\|\|\mathbf{v}\|\sqrt{1-\cos^2\theta}=\sqrt{\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2}$ (using $\cos\theta=\dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}$ ).

Part 4. $\mathbf{u}\cdot\mathbf{v}=\lambda\|\mathbf{u}\|^2$ and $\|\mathbf{v}\|^2=\lambda^2\|\mathbf{u}\|^2$ . Substituting: $\|\mathbf{u}\|^2\cdot\lambda^2\|\mathbf{u}\|^2-\lambda^2\|\mathbf{u}\|^4=0$ ✓.

Part 5. $\mathbf{u}\cdot\mathbf{v}=0$ , so the area is $=\|\mathbf{u}\|\|\mathbf{v}\|$ , the product of the two side lengths—exactly the area formula for a rectangle; if the two sides have equal length, it is a square.

Part 6. Let $\|\mathbf{u}\|=\|\mathbf{v}\|=a$ , with diagonals $p=\|\mathbf{u}+\mathbf{v}\|$ , $q=\|\mathbf{u}-\mathbf{v}\|$ :

p^2=2a^2+2\mathbf{u}\cdot\mathbf{v},\quad q^2=2a^2-2\mathbf{u}\cdot\mathbf{v}

(78)

\frac{pq}{2}=\frac{1}{2}\sqrt{p^2q^2}=\frac{1}{2}\sqrt{(2a^2)^2-4(\mathbf{u}\cdot\mathbf{v})^2}=\sqrt{a^4-(\mathbf{u}\cdot\mathbf{v})^2}

(79)

The area is $=\sqrt{a^4-(\mathbf{u}\cdot\mathbf{v})^2}=\dfrac{pq}{2}$ ✓.

# ============================================================
print_header("Exercise 34 | The Area of a Parallelogram in ℝ⁴")
# ============================================================

print_step(1, "A concrete example: when OP⊥OQ, area = ‖OP‖‖OQ‖")
OP=np.array([1.,0.,1.,0.]); OQ=np.array([0.,1.,0.,1.])
compare_print("area", f"{para_area(OP,OQ):.6f}", "‖OP‖‖OQ‖ = 2 (OP⊥OQ)")

print_step(2, "General vectors: area formula vs ‖u‖‖v‖sinθ")
rng=np.random.default_rng(3)
u=rng.standard_normal(4); v=rng.standard_normal(4)
A1 = para_area(u,v)
cos_t = np.clip(np.dot(u,v)/(np.linalg.norm(u)*np.linalg.norm(v)),-1,1)
A2 = np.linalg.norm(u)*np.linalg.norm(v)*abs(np.sin(np.arccos(cos_t)))
compare_print("area formula", f"{A1:.6f}", f"‖u‖‖v‖sinθ = {A2:.6f}  ✓ {np.isclose(A1,A2)}")

print_step(3, "Area for parallel vectors = 0")
compare_print("area when v=3u", f"{para_area(u,3*u):.2e}", "≈ 0")

print_step(4, "Area of a rhombus = pq/2 (product of the diagonals / 2)")
theta=np.pi/4; a=3.0
u_=a*np.array([1.,0.,0.,0.]); v_=a*np.array([np.cos(theta),np.sin(theta),0.,0.])
p=np.linalg.norm(u_+v_); q=np.linalg.norm(u_-v_)
compare_print("rhombus area", f"{para_area(u_,v_):.4f}", f"pq/2 = {p*q/2:.4f}  ✓ {np.isclose(para_area(u_,v_),p*q/2)}")

The Geometry of Planes and Lines¶

Exercise 35 (Are Two Planes Perpendicular?)

In $\mathbb{R}^3$ , two planes are defined to be perpendicular when their normal vectors are perpendicular to each other. Decide whether each of the following pairs of planes is perpendicular.

The plane $x+3z=1$ and the plane $3x-y=3$ .
The plane $x=5$ and the plane (given parametrically) $\mathbf{p}(s,t)=(1,0,0)+s(0,1,0)+t(1,1,1)$ .
The plane $\mathbf{p}(s,t)=(0,1,2)+s(1,-1,0)+t(0,0,1)$ and the plane $\mathbf{q}(u,v)=(2,0,1)+u(1,0,-1)+v(0,1,1)$ .

Hint. Take the two direction vectors from the parametric equation; the normal vector is their cross product.

# ============================================================
print_header("Exercise 35 | Are Two Planes Perpendicular?")
# ============================================================

print_step(1, "(1) normal vectors n₀=(1,0,3), n₁=(3,−1,0)")
n0=np.array([1,0,3]); n1=np.array([3,-1,0])
compare_print("n₀·n₁", np.dot(n0,n1), "≠ 0 → not perpendicular")

print_step(2, "(2) n₀=(1,0,0), normal vector of the second plane = (0,1,0)×(1,1,1)")
n0_=np.array([1.,0.,0.])
n1_=np.cross(np.array([0.,1.,0.]),np.array([1.,1.,1.]))
print("  n₁ =", n1_)
compare_print("n₀·n₁", np.dot(n0_,n1_), "≠ 0 → not perpendicular")

print_step(3, "(3) normal vectors of the two parametrized planes")
n0_=np.cross(np.array([1.,-1.,0.]),np.array([0.,0.,1.]))
n1_=np.cross(np.array([1.,0.,-1.]),np.array([0.,1.,1.]))
print("  n₀ =", n0_, "  n₁ =", n1_)
compare_print("n₀·n₁ (should be 0 → perpendicular)", np.dot(n0_,n1_), "0 → perpendicular ✓")

Exercise 36 (Finding a Plane That Contains a Line and Is Perpendicular to a Given Plane)

In $\mathbb{R}^3$ , let the plane be $P: x+2y-z=3$ and the line $L: \mathbf{r}(t)=(1,0,1)+t(1,1,0)$ . Find a plane $Q$ such that $L\subset Q$ and $Q\perp P$ , and express $Q$ by an equation.
Let the plane be $P:\mathbf{p}(s,t)=(1,1,0)+s(1,0,0)+t(0,1,0)$ (that is, the plane $z=0$ ) and the line $L:\mathbf{r}(t)=(2,0,1)+t(0,0,1)$ . Find a plane $Q$ such that $L\subset Q$ and $Q\perp P$ .
Let the plane be $P: z=1$ and the line $L:\mathbf{r}(t)=(1,1,0)+t(0,0,1)$ :
- (a) Find a plane $Q$ such that $L\subset Q$ and $Q\perp P$ .
- (b) Is the plane satisfying these conditions unique?

Hint. The normal vector of $Q$ is perpendicular both to the direction vector of $L$ (since $L\subset Q$ ) and to the normal vector of $P$ (since $Q\perp P$ ). Hence the two direction vectors of $Q$ can be taken to be the direction vector of $L$ and the normal vector of $P$ .

Solution to Exercise 36

Part 1. $Q$ contains $L$ , so the direction vector $\mathbf{d}=(1,1,0)$ of $L$ is a direction vector of $Q$ . $Q\perp P$ means that the normal vector $\mathbf{n}_P=(1,2,-1)$ of $P$ is also a direction vector of $Q$ (or lies in $Q$ ). The normal vector of $Q$ :

\mathbf{n}_Q=\mathbf{d}\times\mathbf{n}_P=(1,1,0)\times(1,2,-1)=(1\cdot(-1)-0\cdot2,\, -(1\cdot(-1)-0\cdot1),\, 1\cdot2-1\cdot1)=(-1,1,1)

(80)

$Q$ passes through the point $(1,0,1)$ on the line, so its equation is $-(x-1)+(y-0)+(z-1)=0$ , that is, $-x+y+z=0$ , or $x-y-z=0$ .

Part 2. The direction of $L$ is $\mathbf{d}=(0,0,1)$ , and the normal vector of $P$ is $\mathbf{n}_P=(0,0,1)$ . Since $\mathbf{d}=\mathbf{n}_P$ , their cross product is $\mathbf{0}$ , and a second direction vector of $Q$ must be chosen separately. $Q$ contains $L$ : it passes through the point $(2,0,1)$ with direction $(0,0,1)$ , so one direction of $Q$ is $(0,0,1)$ . Moreover, $Q\perp P$ requires the normal vector of $Q$ to be perpendicular to $\mathbf{n}_P=(0,0,1)$ , that is, the normal vector of $Q$ is horizontal and $Q$ is a vertical plane. Taking the other direction vector to be $(1,0,0)$ (a horizontal direction), $\mathbf{n}_Q=(0,0,1)\times(1,0,0)=(0,1,0)$ , and $Q: y=0$ .

Part 3(a). $P: z=1$ , $\mathbf{n}_P=(0,0,1)$ , and the direction of $L$ is $\mathbf{d}=(0,0,1)=\mathbf{n}_P$ . As above, any vertical plane containing $L$ works ( $Q$ contains $(1,1,0)$ and its directions include $(0,0,1)$ ). For example, taking the horizontal direction vector $\mathbf{e}_0=(1,0,0)$ : $\mathbf{n}_Q=(0,0,1)\times(1,0,0)=(0,1,0)$ , $Q: y=1$ ; or taking $(0,1,0)$ gives $Q: x=1$ .

Part 3(b). Not unique. The line $L$ is perpendicular to the plane $P$ (the direction of $L$ $\parallel\mathbf{n}_P$ ), so every plane containing $L$ is perpendicular to $P$ ; hence there are infinitely many planes $Q$ satisfying the conditions (rotating about $L$ ).

# ============================================================
print_header("Exercise 36 | A Plane Containing a Line and Perpendicular to a Given Plane")
# ============================================================

print_step(1, "Part 1: P: x+2y−z=3, L has direction (1,1,0) and passes through (1,0,1)")
nP=np.array([1.,2.,-1.]); dL=np.array([1.,1.,0.])
nQ=np.cross(dL,nP)
pt=np.array([1.,0.,1.])
d=np.dot(nQ,pt)
print("  normal vector of Q, nQ =", nQ)
print(f"  equation of the plane: {nQ[0]:.0f}x + {nQ[1]:.0f}y + {nQ[2]:.0f}z = {d:.0f}")

print_step(2, "Check that L lies in Q (nQ⊥dL) and Q⊥P (nQ⊥nP)")
compare_print("nQ·dL (should be 0)", f"{np.dot(nQ,dL):.2e}", "0")
compare_print("nQ·nP (should be 0)", f"{np.dot(nQ,nP):.2e}", "0")

Exercise 37 (The Distance between Skew Lines)

Consider the two lines $L_0:\mathbf{r}_0(t)=(1,0,0)+t(1,0,0)$ and $L_1:\mathbf{r}_1(s)=(0,1,1)+s(0,1,0)$ .

Verify that $L_0$ and $L_1$ are skew lines (they have different directions and do not intersect).
Find two parallel planes $P_0\supset L_0$ , $P_1\supset L_1$ ( $P_0\parallel P_1$ ), and express them by equations.
Compute the distance between $L_0$ and $L_1$ (that is, the distance between $P_0$ and $P_1$ ).

Hint. The common normal vector is $\mathbf{n}=\mathbf{d}_0\times\mathbf{d}_1$ ; the distance between the two planes is $\dfrac{|\mathbf{n}\cdot(\mathbf{p}_1-\mathbf{p}_0)|}{\|\mathbf{n}\|}$ , where $\mathbf{p}_i$ is a point on $L_i$ .

# ============================================================
print_header("Exercise 37 | The Distance between Skew Lines")
# ============================================================

print_step(1, "Define the direction vectors and base points of the two skew lines L₀, L₁")
d0=np.array([1.,0.,0.]); d1=np.array([0.,1.,0.])
p0=np.array([1.,0.,0.]); p1=np.array([0.,1.,1.])

print_step(2, "Compute the common normal vector n = d₀×d₁")
n=np.cross(d0,d1)
print("  n =", n)

print_step(3, "Distance between the skew lines = |n·(p₁−p₀)| / ‖n‖")
dist=abs(np.dot(n,p1-p0))/np.linalg.norm(n)
compare_print("distance", f"{dist:.6f}", "= 1 (the distance between the planes z=0 and z=1)")

Generalized Area Formulas¶

Exercise 38 (Projections of a Parallelogram onto the Coordinate Planes (

\mathbb{R}^3

))

Let $\mathbf{u}=(u_0,u_1,u_2)$ , $\mathbf{v}=(v_0,v_1,v_2)$ . The area $A$ of the parallelogram satisfies

A^2=\|\mathbf{u}\times\mathbf{v}\|^2=\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2= \begin{vmatrix}u_0&u_1\\v_0&v_1\end{vmatrix}^2+ \begin{vmatrix}u_0&u_2\\v_0&v_2\end{vmatrix}^2+ \begin{vmatrix}u_1&u_2\\v_1&v_2\end{vmatrix}^2

(81)

Let $\mathbf{u}_{01}=(u_0,u_1,0)$ , $\mathbf{v}_{01}=(v_0,v_1,0)$ be the projections of $\mathbf{u},\mathbf{v}$ onto the $x_0x_1$ -plane; find the area $A_{01}$ of the projected parallelogram.
Similarly, find $A_{02}$ (the $x_0x_2$ -plane) and $A_{12}$ (the $x_1x_2$ -plane).
Verify that $A^2=A_{01}^2+A_{02}^2+A_{12}^2$ .

Hint. After projection, we still have a parallelogram spanned by two vectors, and its area is the absolute value of the corresponding $2\times2$ determinant in the original formula.

# ============================================================
print_header("Exercise 38–39 | The Generalized Pythagorean Theorem (Cauchy–Binet)")
# ============================================================

rng=np.random.default_rng(99)

print_step(1, "Check in ℝ³: A² = Σ_{i<j} A_{ij}²")
u=rng.standard_normal(3); v=rng.standard_normal(3)
A2=para_area(u,v)**2
Aij2=sum((u[i]*v[j]-u[j]*v[i])**2 for i,j in combinations(range(3),2))
compare_print("A² (ℝ³)", f"{A2:.6f}", f"ΣA_ij² = {Aij2:.6f}  ✓ {np.isclose(A2,Aij2)}")

print_step(2, "Check in ℝ⁴: A² = Σ_{i<j} A_{ij}²")
u4=rng.standard_normal(4); v4=rng.standard_normal(4)
A2_4=para_area(u4,v4)**2
Aij2_4=sum((u4[i]*v4[j]-u4[j]*v4[i])**2 for i,j in combinations(range(4),2))
compare_print("A² (ℝ⁴)", f"{A2_4:.6f}", f"ΣA_ij² = {Aij2_4:.6f}  ✓ {np.isclose(A2_4,Aij2_4)}")

Exercise 39 (Projections of a Parallelogram onto the Coordinate Planes (

\mathbb{R}^4

))

Let $\mathbf{u}=(u_0,u_1,u_2,u_3)$ , $\mathbf{v}=(v_0,v_1,v_2,v_3)\in\mathbb{R}^4$ , and let the area of the parallelogram be $A=\sqrt{\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2}$ .

Check that the projection of $\mathbf{u}$ onto the $x_0x_1$ -plane is $(u_0,u_1,0,0)$ .
Define $A_{ij}$ to be the area of the parallelogram projected onto the $x_ix_j$ -plane. Check that

A_{01}^2=(u_0v_1-u_1v_0)^2,\quad A_{02}^2=(u_0v_2-u_2v_0)^2,\quad\ldots,\quad A_{23}^2=(u_2v_3-u_3v_2)^2

(83)

Check the generalized Pythagorean theorem $A^2=A_{01}^2+A_{02}^2+A_{03}^2+A_{12}^2+A_{13}^2+A_{23}^2$ ( $\binom{4}{2}=6$ terms in all).

Hint. Expand $\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2$ directly and compare with the sum of the squares of the six determinants.

✓ All 39 exercises completed.