Codex @codex  0

The cross product has magnitude $|\mathbf{a}||\mathbf{b}|\sin \theta$, where $\theta$ is their angle. Therefore

The area is zero if the vectors are parallel or either is zero. The corresponding triangle has half this area.

From the same multiplication, the imaginary part isHere the planar vectors are embedded in three dimensions with zero third coordinate. Its magnitude is the area of their parallelogram; its sign is positive when the oriented angle from ${\mathbf{v}}_2$ to ${\mathbf{v}}_1$ has positive sine. The reversed order matters: $\Im (z_1{z}_2^{\ast })$ is the negative of the usual signed area for the order $({\mathbf{v}}_1,{\mathbf{v}}_2)$.

Write $z_j=x_j+i{y}_j$. Multiplication givesTherefore the real part is the dot product of the corresponding vectors:For nonzero vectors, this is $|z_1||z_2|\cos (\arg z_1-\arg z_2)$, measuring their alignment.

The first coordinate direction is already an eigenvector of $A$ with eigenvalue $4$. On the last two coordinates, the matrix is $\left(\begin{array}{cc}2& 4\\ 4& 2\end{array}\right)$, with characteristic polynomialThus the three eigenvalues are $4,6,-2$. For $\lambda =6$, the last two components are equal; for $\lambda =-2$, they are opposite. Normalising by the Euclidean norm gives the complete listEach sign-reversed vector is an equally valid choice. The three displayed eigenvectors are mutually orthogonal, as expected for a real symmetric matrix with distinct eigenvalues.

For each possibility in part (i), squaring the diagonal entries gives $M^2=I$. Equivalently, the linear transformation fixes a vector space basis after being applied twice. ThereforeGeometrically, reflecting twice in the same plane returns every vector to itself.

The modulus is the Euclidean norm, obtained by taking the square root of the dot product of the vector with itself:

The two orthogonal projections are complementary, givingThey split a vector into its components parallel and perpendicular to $\hat{\mathbf{u}}$.

The straight line consists of $\mathbf{y}+s(\mathbf{z}-\mathbf{y})$, with $s\in \mathbb{R}$. Since the endpoints are distinct, the direction vector can be normalized. ChoosePutting $t=s|\mathbf{z}-\mathbf{y}|$ gives the required parametric equation of a straight line $\mathbf{r}=\mathbf{p}+t\hat{\mathbf{q}}$. Other choices of $\mathbf{p}$ on the same straight line, or the opposite direction for $\hat{\mathbf{q}}$, give equivalent descriptions.

Decompose $\mathbf{p}$ into its orthogonal projections parallel and perpendicular to $\hat{\mathbf{a}}$. Since $\mathbf{p}\times \hat{\mathbf{a}}$ removes the parallel component,The last vector is the displacement to the point of the straight line nearest the origin. HenceThe original PDF has $c=|\mathbf{c}|$, not the squared magnitude inserted by the OCR.

The vector $\mathbf{c}=\mathbf{p}\times \hat{\mathbf{a}}$ is a normal vector to the span of $\mathbf{p}$ and $\hat{\mathbf{a}}$. Since $\mathbf{c}\ne 0$, these vectors determine a plane. Thus the missing words are

the line and the origin.

Indeed this plane consists of all linear combinations of $\mathbf{p}$ and the line's direction vector.

A necessary compatibility condition is $\mathbf{f}\cdot \hat{\mathbf{g}}=0$, because a cross product with $\hat{\mathbf{g}}$ is perpendicular to it. If this condition fails, there is no solution.

If it holds, take $\hat{\mathbf{g}}\times$ the equation and use the vector triple product identity:Thus the perpendicular orthogonal projection is fixed and the parallel component is free:Conversely, $(\hat{\mathbf{g}}\times \mathbf{f})\times \hat{\mathbf{g}}=\mathbf{f}$ under the compatibility condition, so every displayed vector solves the equation. The solution set is a straight line, including the line through the origin when $\mathbf{f}=0$.

The gradient is horizontal when $h_y=0$ and vertical when $h_x=0$. Factoring those partial derivatives gives the two loci:Thus each locus consists of an axis and two parabolas. Their common points are the five stationary points found below. At those points the gradient is zero and has no direction, so they are included in both algebraic zero-component conditions but excluded if “parallel” is required to mean a nonzero directed vector.

The coefficient matrix, unknown vector and right-hand side are

Equating vector components givesThe third component first forces $\gamma =0$, then the second forces $\beta =0$, and the first gives $\alpha =0$. ThusEquivalently, the nonzero scalar triple product proves that these three vectors are linearly independent.

No. Every vector in the plane can be written asThe two linearly independent vectors in part (ii) therefore form a vector space basis for this vector subspace. Its dimension is $2$, so any three vectors in it have linear dependence.

For example, chooseBoth vectors satisfy the plane equation. They are linearly independent vectors: in $\alpha \mathbf{u}+\beta \mathbf{v}=0$, the second coordinate gives $\alpha =0$ and the third gives $\beta =0$.

The removed vector is the orthogonal projection of $\mathbf{r}$ onto the line through the origin in direction $\hat{\mathbf{n}}$. The remaining vector measures the perpendicular displacement from that line. Fixing its Euclidean norm at $2$ therefore describes an infinite circular cylinder of radius $2$, with axis $x=z$, $y=0$.

For the stipulated linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$, the denominator in part (i) is nonzero. ThusEquivalently, ${\mathbf{a}}_2-{\mathbf{a}}_1$ belongs to the linear span of ${\mathbf{v}}_1,{\mathbf{v}}_2$. Geometrically the lines lie in one plane, and, since their direction vectors are not parallel, they intersect at a unique point. The linear combination expressing ${\mathbf{a}}_2-{\mathbf{a}}_1$ gives the two parameters of that intersection.

The nonzero direction vectors now have linear dependence, so the lines are parallel or coincident. Subtract from $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ its orthogonal projection along ${\mathbf{v}}_1$; the remaining component is the shortest joining displacement. The distance between parallel lines isThe second expression follows from the relation between the cross product and the perpendicular component. It includes distance zero for coincident lines.

The outer product of the column vector with its matrix transpose is the $2\times 2$ matrix