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For an arbitrary vector $\mathbf{x}$, expansion in the orthonormal basis gives $x_i=(a_i{a}_j+b_i{b}_j+c_i{c}_j)x_j$, with summation over $j$. Since this holds for every $\mathbf{x}$, the resolution of the identity in an orthonormal basis isLet $P_a=\mathbf{a}{\mathbf{a}}^T$, and similarly for $b,c$. The three orthogonal projection matrices sum to $I$ and have pairwise products zero. ThusAny ray meeting all three ideal perpendicular mirrors therefore returns in the opposite direction, independently of reflection order. This is the geometry of a corner reflector, a kind of retroreflector used in lunar laser ranging.

True. Applying the matrices in order gives $AB\mathbf{x}=A(b\mathbf{x})=ab\mathbf{x}$. The eigenvalue isThis conclusion only uses the shared eigenvector; it does not require $A$ and $B$ to commute on other vectors.

The cross product is $\mathbf{a}\times \mathbf{b}=(-2,0,2)$, which is perpendicular to both vectors. Any nonzero scalar multiple works, so a simple answer isIts dot product with each given vector is zero.

The matrix product of the transposed first vector with the second is their dot product:

The permutation matrix cycles the three coordinate vectors, so three applications give $C^3=I$. Consequently $C^{2}C=C{C}^2=I$, proving $C^{-1}=C^2$. Since $16=5\cdot 3+1$,

Since $\mathbf{n}\cdot \mathbf{n}=1$, the vector $K\mathbf{r}$ is perpendicular to $\mathbf{n}$:It follows that $K^2\mathbf{r}=K\mathbf{r}$. If $\mathbf{r}=t\mathbf{n}$, then $K\mathbf{r}=0$; if $\mathbf{n}\cdot \mathbf{r}=0$, then $K\mathbf{r}=\mathbf{r}$. Thus $K=I-\mathbf{n}{\mathbf{n}}^T$ is the orthogonal projection matrix onto ${\mathbf{n}}^{\perp }$: it removes exactly the component normal to the plane.

For the equally inclined plane choose $\mathbf{n}=(1,1,1)/\sqrt{3}$. Reversing its sign gives the same plane; other sign choices give equivalent reflections of the coordinate axes. One suitable orthonormal basis of this plane isTheir dot products with $\mathbf{n}$ vanish, their lengths are one, and their mutual dot product is zero. The equal first two components of ${\mathbf{e}}_1$ give its equal inclinations to $\mathbf{i},\mathbf{j}$. Because ${\mathbf{e}}_j$ is already in the projection plane, ${\mathbf{e}}_j\cdot K\mathbf{r}={\mathbf{e}}_j\cdot \mathbf{r}$. The coordinates of the projected point are thereforeFor a circle of radius $R$ centered at the origin in $z=0$, its equation $x^2+y^2=R^2$ becomes $3u^2+v^2=R^2$. This is an ellipse with semiaxes $R/\sqrt{3}$ and $R$. Translating the original center only translates its orthogonal projection, so the result applies to every circle in that plane. ConsequentlyThe general reason is the orthogonal projection of a circle between planes: the direction common to the planes retains its length, while the perpendicular direction is shortened by the cosine of their angle.

The relation rearranges to $(z_1+z_3)/2=(z_2+z_4)/2$, so the diagonals $AC$ and $BD$ have the same midpoint. Equivalently $z_2-z_1=z_3-z_4$ and $z_4-z_1=z_3-z_2$: each pair of opposite sides has the same displacement vector. Thus the given quadrilateral is a parallelogram. The identities also describe degenerate configurations if coincident or collinear vertices are allowed.

The gradient is $\nabla f=(y(1-2x+y),x(1-x+2y))$. At the three specified points it is respectivelyEach vector is perpendicular to its straight boundary edge and points out of the triangle, where $f$ increases from its negative interior values. The arrows in the preceding figure mark these directions.

The permutation matrix acts by $A(x,y,z{)}^T=(y,z,x{)}^T$. Applying this cyclic permutation three times returns every vector to itself, proving $A^3=I$. Explicitly $A^2=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$ and multiplication once more gives the identity matrix.

Since $\mathbf{a}$ is constant, componentwise differentiation gives $\nabla \cdot (\mathbf{a}\varphi )={\sum }_i{a}_i{\partial }_i\varphi =\mathbf{a}\cdot \nabla \varphi$. Applying the divergence theorem to this vector field givesThis holds for every constant vector $\mathbf{a}$, so the vector integrals agree. This is the vector gradient form of the divergence theorem. With $\varphi =z$, the gradient is the constant unit vector ${\mathbf{e}}_z$. Therefore

The dot product is $\mathbf{a}\cdot \mathbf{b}=2-1+2=3$, and both vectors have length $\sqrt{6}$. The angle between vectors therefore satisfies $\cos \theta =3/6=1/2$, giving

The line has equation $2x-y+2=0$, with normal vector $(2,-1)$. The distance from a point to a line is the absolute value of the normal component of the displacement:The perpendicular foot is $(4,-6)-(16/5)(2,-1)=(-12/5,-14/5)$, which lies on the line and realizes this shortest distance.

The direction vector is $b-a=(1,2,-2)$. The vector equation and symmetric Cartesian equation of the line areThe $(x,y)$-plane has $z=0$, which gives $t=2$. Thus its intersection with the line has position

If $y\ne 0$, the equation would imply $b=-(x/y)a$, contradicting the assumed nonparallel directions. Hence $y=0$. Then $xa=0$ and $a\ne 0$ force $x=0$. This proves that the two vectors are linearly independent.

For either sign, the dot product with $v_1$ isThe remaining pair also satisfiesHence the three eigenvectors are mutually orthogonal. This also follows from the spectral theorem for real symmetric matrices, since $M=M^T$ and its eigenvalues are distinct, but the displayed products verify the actual chosen vectors directly.

Let $d=b\cdot x$. Taking a dot product with $b$ reduces the vector equation to $(1-b\cdot c)d=b\cdot a$. If $b\cdot c\ne 1$, this determines a unique scalar and hence a unique vector:Direct substitution verifies it, so necessity has also produced sufficiency. The formula includes $c=0$, when $x=a$.

If $b\cdot c=1$, the reduced scalar equation becomes $0=b\cdot a$. When $b\cdot a\ne 0$, there is no solution. When $b\cdot a=0$, all solutions are $x=a+dc$ with arbitrary $d\in \mathbb{R}$: their dot product with $b$ is indeed $d$. Since $b\cdot c=1$ forces $c\ne 0$, this is an entire affine line, not a single point. These alternatives are the solvability of an affine rank-one vector equation.

The characteristic polynomial isFor $\lambda =2$, the eigenvector equation reduces to $x=y$; for $\lambda =8$, it reduces to $x=-y$. Choose the following unit vectors consistently for the remaining parts:Each eigenspace consists of all scalar multiples of its displayed vector, including zero; an eigenvector itself must be nonzero.

Expand the same double cross product in two ways. First expanding with $c\times d$ as the inner cross product givesAlternatively, use $(a\times b)\times w=(a\cdot w)b-(b\cdot w)a$ with $w=c\times d$ to getEquating these expressions and moving every term to one side provesThis four-vector scalar triple product identity is valid without assuming that any three of the four vectors are independent.

At an intersection, the two position vectors agree, so $a-b=tn-sm$. The scalar triple product with $m,n$ is thereforesince a determinant with repeated columns vanishes. Geometrically the displacement of the two anchor points must lie in the same plane as the two line directions.

This is a necessary condition. If $m,n$ are not parallel, it is also sufficient, because their span is exactly the plane perpendicular to $m\times n$. For parallel directions the triple product vanishes automatically, even for distinct lines, so the condition alone cannot guarantee intersection.

Hold $r$ fixed and differentiate the position vector $\mathbf{X}(\theta ,\varphi )$. Its coordinate tangent vectors areTheir dot product is zero, and their squared Euclidean norms are $r^2$ and $r^2{\sin }^2\theta$. To first order, the displacement is $\delta \mathbf{X}={\mathbf{X}}_{\theta }\delta \theta +{\mathbf{X}}_{\varphi }\delta \varphi +O((|\delta \theta |+|\delta \varphi |{)}^2)$. Squaring givesThis is the line element on a sphere. The chord distance and the short surface arc length agree to the order retained here.