Vectors. Cross Products and Planes.

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Recall: Vector Basics

Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.

Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $\vert{}\vert{}\langle 3, 2, 1 \rangle\vert{}\vert{} = \sqrt{3^2+2^2+1^2} = \sqrt{14}$, $\vert{}\vert{}\langle 3, -4, 5 \rangle\vert{}\vert{} = \sqrt{3^2+(-4)^2+5^2} = \sqrt{50} = 5\sqrt{2}$, or $\vert{}\vert{}\langle 1, 0, 0 \rangle\vert{}\vert{} = \sqrt{1^2+0^2+0^2} = \sqrt{1} = 1$.

Basic Vector Operations

• Vector Addition: The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k} = \langle a_1+b_1, a_2+b_2, a_3+b_3 \rangle$.

• Vector Subtraction: The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k} = \langle a_1-b_1, a_2-b_2, a_3-b_3 \rangle$.

• Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction (unless the scalar is negative, which reverses the direction). It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = \langle ca_1, ca_2, ca_3 \rangle$.

• The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = \langle 2, 2, -1 \rangle \cdot \langle 5, -3, 2 \rangle = (2 \cdot 5) + (2 \cdot -3) + (-1 \cdot 2) = 10 - 6 - 2 = 2.$

Geometrically, it is the product of their magnitudes multiplied by the cosine of the angle $\theta$ between them, $\boxed{\vec{A} \cdot \vec{B} = \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cos(\theta)}$

Determinant in space

Let $\vec{A}, \vec{B},$ and $\vec{C}$ be three vectors in $\mathbb{R}^3$ given by column vectors: $\vec{A} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}, \quad \vec{B} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}, \quad \vec{C} = \begin{pmatrix} c_1 \\ c_2 \\ c_3 \end{pmatrix}$

The determinant of the matrix formed by these three vectors, denoted as $det(\vec{A}, \vec{B}, \vec{C})$, is calculated using cofactor expansion: $det(\vec{A}, \vec{B}, \vec{C}) = |\begin{smallmatrix}a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\\\ c_1 & c_2 & c_3\end{smallmatrix}| = a_1det(\begin{smallmatrix}b_2 & b_3\\\ c_2 & c_3\end{smallmatrix})-a_2det(\begin{smallmatrix}b_1 & b_3\\\ c_1 & c_3\end{smallmatrix})+a_3det(\begin{smallmatrix}b_1 & b_2\\\ c_1 & c_2\end{smallmatrix})$

Where the 2 × 2 determinants (also called minors) are calculated as follows:

$det(\begin{smallmatrix}b_2 & b_3\\\ c_2 & c_3\end{smallmatrix}) = b_2c_3-b_3c_2, det(\begin{smallmatrix}b_1 & b_3\\\ c_1 & c_3\end{smallmatrix}) = b_1c_3 -b_3c_1, det(\begin{smallmatrix}b_1 & b_2\\\ c_1 & c_2\end{smallmatrix})=b_1c_2-b_2c_1$

In conclusion, $det(\vec{A}, \vec{B}, \vec{C}) = a_1(b_2c_3-b_3c_2) -a_2(b_1c_3 -b_3c_1) + a_3(b_1c_2-b_2c_1)$

Geometric Interpretation

Imagine each vector $\vec{A}, \vec{B},~\text{and}~ \vec{C}$ originating from the same point. They extend out in space, forming the edges of a three-dimensional shape called a parallelepiped (a three-dimensional figure with parallelogram-like faces).

Geometrically, the determinant $det(\vec{A}, \vec{B}, \vec{C})$ represents the volume of the parallelepiped spanned or formed by the vectors $\vec{A}, \vec{B},and~ \vec{C}$ (Figure ii).

The sign of the determinant indicates the orientation of the three vectors:

1. Positive determinant. The vectors form a right-handed system. This means that if you extend your right hand’s index and middle fingers along the direction of vector $\vec{A}$ and $\vec{B}$ respectively, your thumb points roughly in the direction of $\vec{C}$. In other words, if you align your right hand’s thumb, index, and middle fingers along the directions of $\vec{A}, \vec{B},and~ \vec{C}$, respectively, your hand will be in a natural position, e.g., the standard basic vectors $\vec{i}, \vec{j}, \vec{k}$ form a right-handed system.
2. Negative determiant. The vectors form a left-handed system. The natural orientation is reversed.

Cross Product

Definition. The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular (orthogonal) to both input vectors, has a magnitude equal to the area of the parallelogram formed by the two input vectors, and has a direction determined by the right-hand rule.

This operation is particularly useful in maths, physics, and engineering for finding a vector that is orthogonal (normal or perpendicular) to a plane defined by two other vectors, among other applications.

Direction and the Right-Hand Rule. The direction of the resulting vector $\vec{A}x\vec{B}$ is strictly dictated by the right-hand rule. If you curl the fingers of your right hand from $\vec{A}$ to $\vec{B}$ (first or index finger points in the direction of $\vec{A}$, second or middle finger points to $\vec{B}$), your thumb, when extended, points in the direction of the cross product $\vec{A} \times \vec{B}$ (Figure iii).

Calculating the Cross Product

The cross product $\vec{A}x\vec{B}$ can be computed using a determinant-like setup with unit vectors $\vec{i}, \vec{j}, \vec{k}$ in the top row: $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\\ b_1 & b_2\end{smallmatrix}|\vec{k}$

Where the 2×2 determinants (minors) are calculated as follows: $|\begin{smallmatrix}a_2 & a_3\\\ b_2 & b_3\end{smallmatrix}| = a_2b_3-a_3b_2, |\begin{smallmatrix}a_1 & a_3\\\ b_1 & b_3\end{smallmatrix}| = a_1b_3-a_3b_1, |\begin{smallmatrix}a_1 & a_2\\\ b_1 & b_2\end{smallmatrix}| = a_1b_2-a_2b_1.$

Putting it all together, we get: $\vec{A}x\vec{B} = (a_2b_3-a_3b_2)\vec{i}-(a_1b_3-a_3b_1)\vec{j}+(a_1b_2-a_2b_1)\vec{k}$

Examples of Cross Products

To better understand the concept of the cross product, let’s go through a few examples in detail.

• Cross Product of Unit Vectors. Find the cross product $\vec{i}x\vec{j}$. $\vec{i}x\vec{j} = det(\begin{smallmatrix}i & j & k\\\ 1 & 0 & 0\\\ 0 & 1 & 0\end{smallmatrix}) = (0·0-0·1)\vec{i}-(1·0-0·0)\vec{j}+(1·1-0·0)\vec{k} = \vec{k}$

• Cross product of two vectors. Find the cross product $⟨1,0,2⟩x⟨−1,−2,4⟩$. $⟨1,0,2⟩x⟨−1,−2,4⟩ = det(\begin{smallmatrix}i & j & k\\\ 1 & 0 & 2 \\\ -1 & -2 & 4\end{smallmatrix})=(0⋅4−2⋅−2)\vec{i}−(1⋅4−2⋅−1)\vec{j}+(1⋅−2−0⋅−1)\vec{k}=4\vec{i}-6\vec{j}+-2\vec{k}=⟨4, -6, -2⟩$.

• Another Example: $⟨−2,1,3⟩x⟨5,2,−1⟩ = det(\begin{smallmatrix}i & j & k\\\ -2 & 1 & 3 \\\ 5 & 2 & -1\end{smallmatrix})=-7\vec{i}+13\vec{j}+-9\vec{k}=⟨-7, 13, -9⟩$.

Alternative way of calculating the volume of a parallelepiped

To calculate the volume of a parallelepiped spanned by the vectors $\vec{A}, \vec{B},and~ \vec{C}$ (Figure v), we can use the base times heigh formula (Volume = Base Area × Height).

1. The area of the base parallelogram formed by $\vec{B}$ and $\vec{C}$ is the magnitude of their cross product: Area = $\vert{}\vert{}\vec{B} \times \vec{C}\vert{}\vert{}$.
2. The height is the projection of $\vec{A}$ onto a unit vector $\vec{n}$ perpendicular to the base. $\vec{n} = \frac{\vec{B} \times \vec{C}}{\vert{}\vert{}\vec{B} \times \vec{C}\vert{}\vert{}}$
3. The volume is the Base Area multiplied by the absolute value of the projection: $\text{Volume} = \vert{}\vert{}\vec{B} \times \vec{C}\vert{}\vert{} \cdot \left\vert{} \vec{A} \cdot \frac{\vec{B} \times \vec{C}}{\vert{}\vert{}\vec{B} \times \vec{C}\vert{}\vert{}} \right\vert{} = \vert{}\vec{A} \cdot (\vec{B} \times \vec{C})\vert{}$
4. Notice that expanding the scalar triple product $\vec{A} \cdot (\vec{B} \times \vec{C})$ gives us the exact same cofactor expansion as the determinant of the three vectors: $\vec{A} \cdot (\vec{B} \times \vec{C}) = \vec{A} \cdot \left( \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix}\vec{i} - \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix}\vec{j} + \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix}\vec{k} \right) = a_1 \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} = \det(\vec{A}, \vec{B}, \vec{C})$
5. This shows that the volume of the parallelepiped can also be found using the determinant of the matrix formed by the vectors $\vec{A}, \vec{B},and~ \vec{C}$. Volume = $\vert{}\det(\vec{A}, \vec{B}, \vec{C})\vert{}$.

Properties of the Cross Product

The cross product, a fundamental operation in vector algebra, has several important properties. Let $\vec{u}, \vec{v}$, and $\vec{w}$ be vectors in space, and let c be a scalar, here are the key properties of the cross product:

1. Anticommutative property: $\vec{u}x\vec{v}=-(\vec{v}x\vec{u})$. This property means that the order in which you cross two vectors matters. Switching the order yields a vector of the same length pointing in the opposite direction, e.g., if $\vec{u}x\vec{v}$ points upwards, then $\vec{v}x\vec{u}$ point downwards.
2. Distributive property over Addition: $\vec{u}x(\vec{v}+\vec{w})=(\vec{u}x\vec{v})+(\vec{u}x\vec{w})$. It allows us to distribute the cross product over vector addition.
3. Scalar Compatibility: $c(\vec{u}x\vec{v})=(c\vec{u})x\vec{v}=\vec{u}x(c\vec{v})$. Scaling either vector by a constant c before taking the cross product is the same as scaling the resulting cross product by c.
4. Cross product of the zero vector: $\vec{u}x\vec{0} = \vec{0}x\vec{u} = \vec{0}$. Crossing any vector with the zero vector results in the zero vector. This makes complete sense because the zero vector has no magnitude and does not point in any specific direction, so it cannot contribute to (or form) a perpendicular vector.
5. Self-Cross Product: $\vec{u}x\vec{u}=\vec{0}$. The cross product of any vector with itself is always the zero vector. This happens because there is no area spanned by a vector with itself (a vector cannot form a parallelogram with itself), hence the resulting orthogonal vector has zero magnitude.
6. Triple scalar product: $\vec{u}·(\vec{v}x\vec{w})=(\vec{u}x\vec{v})·\vec{w}$. This property shows the relationship between the dot product and the cross product. The triple scalar product gives the volume of the parallelepiped formed by the three vectors.
7. Triple cross product (Lagrange’s Formula): $\vec{u}x(\vec{v}x\vec{w})=(\vec{u}·\vec{w})\vec{v}-(\vec{u}·\vec{v})\vec{w}$. This identity helps to simplify expressions involving the cross product of a cross product by turning double cross products into dot products.

Solved exercises

Area of a Triangle in 3D Space

• Find the area of the triangle with vertices $P = (1, 0, 0)$, $Q = (0, 1, 0)$, and $R = (0, 0, 1)$ (Figure 2).

1. First, we need to find the vectors representing two sides of the triangle. Let’s use the vectors $\vec{PQ}$ and $\vec{PR}$.
   $\vec{PQ} =⟨0−1,1−0,0−0⟩=⟨−1,1,0⟩, \vec{PR} = ⟨0−1,0−0,1−0⟩=⟨−1,0,1⟩.$
2. The area of the parallelogram with adjacent sides $\vec{PQ}$ and $\vec{PR}$ is given by the magnitude of their cross product $|\vec{PQ}x\vec{PR}| = \Biggl \vert\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\\ -1 & 1 & 0\\\ -1 & 0 & 1\end{smallmatrix}\Biggr \vert$
3. We calculate this determinant by expanding across the top row: $\vec{PQ}x\vec{PR} = (1·1-0·0)\hat{\mathbf{i}}-(-1·1-0·(-1))\hat{\mathbf{j}}+(-1·0-(-1)·1) = \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
4. Next, we find the magnitude of the resulting vector: $\vec{PQ}x\vec{PR} = \sqrt{1^2+1^2+1^2} = \sqrt{3}$
5. The area of the triangle with vertices $P$, $Q$, and $R$ is exactly half the area of the parallelogram formed by its adjacent sides, which gives us $\vec{PQ}$ and $\vec{PR}$, which gives us $\frac{\sqrt{3}}{2}$

• Verify that the points in space P(1, 0, 1), Q(2, 4, 6), R(3, -1, 2) and S(6, 2, 8) are coplanar.

1. ✴️ Fundamental idea ✴️ If the points in space P(1, 0, 1), Q(2, 4, 6), R(3, -1, 2) and S(6, 2, 8) are coplanar (lying on the same flat plane), the volume of the 3D parallelepiped spanned by the vector $\vec{PQ}, \vec{PR}$ and $\vec{PS}$ will be zero (Figure 1).
2. This volume can be calculated as follows: Volume = $|det(\vec{PQ}, \vec{PR}, \vec{PS})| = |\vec{PQ}·(\vec{PR} x \vec{PS})|$
3. First, we need to find the vectors $\vec{PQ}, \vec{PR}$ and $\vec{PS}$ from our initial point P: $\vec{PQ} = \langle 2-1, 4-0, 6-1 \rangle = \langle 1, 4, 5 \rangle$, $\vec{PR} = \langle 3-1, -1-0, 2-1 \rangle = \langle 2, -1, 1 \rangle$, and $\vec{PS} = \langle 6-1, 2-0, 8-1 \rangle = \langle 5, 2, 7 \rangle$
4. Compute the Cross Product $\vec{PR} \times \vec{PS}$: $\vec{PR} \times \vec{PS} = \left\vert{} \begin{matrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -1 & 1 \\ 5 & 2 & 7 \end{matrix} \right\vert{} = (-7 - 2)\hat{\mathbf{i}} - (14 - 5)\hat{\mathbf{j}} + (4 - (-5))\hat{\mathbf{k}} = \langle -9, -9, 9 \rangle$
5. Finally, calculate the volume via the dot product: $\text{Volume} = \vert{}\vec{PQ} \cdot (\vec{PR} \times \vec{PS})\vert{} = \vert{}\langle 1, 4, 5 \rangle \cdot \langle -9, -9, 9 \rangle\vert{}$
   $\text{Volume} = \vert{}-9 - 36 + 45\vert{} = \vert{}0\vert{} = 0$
6. Because the volume is zero, the points $P$, $Q$, $R$, and $S$ are indeed coplanar.

Orthogonal Vectors and Triangle Areas

• Find a vector orthogonal to the plane containing $P = (1, 0, -1)$, $Q = (2, 4, 5)$, and $R = (3, 1, 7)$, and determine the area of the triangle with these vertices (Figure 3).

1. Only a single plane can pass through any set of three non-colinear points. The plane must contain the vectors $\vec{PQ}$ and $\vec{PR}$.
2. First, we need to find the vectors $\vec{PQ}$ and $\vec{PS}$ using the coordinates of the points P, Q, and R.
   $\vec{PQ} = Q - P = \langle 2-1, 4-0, 5-(-1) \rangle = \langle 1, 4, 6 \rangle, \vec{PR} = R - P = \langle 3-1, 1-0, 7-(-1) \rangle = \langle 2, 1, 8 \rangle$.
3. The cross product $\vec{PQ} \times \vec{PR}$ yields a vector orthogonal (perpendicular) to both surface vectors, and thus to the plane containing the points.
   $\vec{PQ} \times \vec{PR} = \left\vert{} \begin{matrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 4 & 6 \\ 2 & 1 & 8 \end{matrix} \right\vert{}$
   $\vec{PQ} \times \vec{PR} = (4 \cdot 8 - 6 \cdot 1)\hat{\mathbf{i}} - (1 \cdot 8 - 6 \cdot 2)\hat{\mathbf{j}} + (1 \cdot 1 - 4 \cdot 2)\hat{\mathbf{k}} = \langle 26, 4, -7 \rangle$
   This vector $\langle 26, 4, -7 \rangle$ is normal to the plane.
4. The area of the triangle formed by the points P, Q, and R is half the area of the parallelogram formed by $\vec{PQ}x\vec{PR}$. The magnitude of the cross product $|\vec{PQ}x\vec{PR}|$ gives the area of this parallelogram.
   $\text{Area} = \frac{1}{2}\vert{}\vec{PQ} \times \vec{PR}\vert{} = \frac{1}{2}\sqrt{26^2 + 4^2 + (-7)^2} = \frac{1}{2}\sqrt{676 + 16 + 49} = \frac{\sqrt{741}}{2} \approx 13.61$

Defining a Plane by its Equation

• What is the set of points defined by the equation $x + 2y + 3z = 0$?

1. The general form of a plane in 3D space is given by: $Ax + By + Cz = D$. In our case, the equation can be written explicitly with coefficients: $1 \cdot x + 2 \cdot y + 3 \cdot z = 0$
2. A plane is entirely defined by a point on the plane and a normal vector (a vector perpendicular to the plane’s surface). The coefficients of $x$, $y$, and $z$ in the standard Cartesian equation represent exactly this normal vector. Therefore, the normal vector is $\vec{A} = \langle 1, 2, 3 \rangle$.
3. Any point $P = \langle x, y, z \rangle$ that satisfies the equation lies on the plane. If we plug in the origin $O = \langle 0, 0, 0 \rangle$, we get $0 + 0 + 0 = 0$. Since this holds true, the plane passes through the origin.
4. Geometrically, let $\vec{OP} = \langle x, y, z \rangle$ be the position vector of a point on the plane. The dot product of $\vec{OP}$ and the normal vector $\vec{A}$ must equal zero, enforcing their perpendicular relationship:
   $\boxed{\vec{OP} \cdot \vec{A} = 0 \iff x \cdot 1 + y \cdot 2 + z \cdot 3 = 0 \iff \vec{OP} \perp \vec{A}}$ (Figure A).

Finding a Perpendicular Plane through an Intersecting Line

• The planes $x - z = 1$ (i) and $y + 2z = 3$ (ii) intersect in a line. Find a third plane that contains this line and is perpendicular to the plane $x + y - 2z = 1$.

1. Find the line of intersection of the two given planes. The line of intersection of the planes x − z = 1 and y + 2z = 3 can be found by parameterizing the solution to the system of equations. Let’s solve for x and y in terms of z.
   From the first plane (i), $x = z + 1$. From the second plane (ii), $y = 3 - 2z$.
   We can let z = t, where t is our parameter. The parametric form of the line of intersection is $\vec{r}(t) = \langle t + 1, 3 - 2t, t \rangle$

2. Calculate the Normal Vector to the given Plane.
   Recall that the general form of the equation of a plane in $\mathbb{R}^3$ is $ax + by + cz + d = 0$ where a, b, and c, are the components of the normal vector $\vec{N} = \langle a, b, c \rangle$, which is perpendicular to the plane. For our given plane $x + y - 2z = 1$, the normal vector is: $\langle 1, 1, -2 \rangle$

3. Define the Normal Vector for our unknown third plane.
   Let the new plane have a normal vector $\vec{M} = \langle a, b, c \rangle$. Because our new plane must be perpendicular to $x + y - 2z = 1$, their normal vectors must also be perpendicular.
   Therefore, their dot product must equal zero: $\vec{M} \cdot \vec{N} = 0 \implies (a)(1) + (b)(1) + (c)(-2) = 0 \implies a + b - 2c = 0$
   Solving for $a$, we get: $a = 2c - b$.

4. Form the Equation of the Third Plane. A point P with coordinates (x, y, z) is in the plane $\iff \vec{P_1P}·\vec{M} = 0$ where $P_1$ is a point in this third plane containing the line, $\vec{r}(t) = \langle t + 1, 3 - 2t, t \rangle$.
   Choose $t = 0 \implies P_1 = (1, 3, 0)$, therefore the equation of the plane is: $\vec{P_1P}·\vec{M} = \langle x-1, y-3, z\rangle \cdot \langle a, b, c \rangle = (x-1)a +(y-3)b +zc = 0.$
   Let’s consider another point on the line $\vec{r}(t) = \langle t + 1, 3 - 2t, t \rangle$. Choose $t = 1 \implies P_2 = (2, 1, 1)$.
   Substituting these coordinates in the plane equation: $(2−1)a + (1−3)b +1c = 0 \implies a -2b + c = 0 \implies [\text{Using the relation: } a = 2c−b] 2c -b -2b +c = 0 \implies 3c -3b = 0 \implies c = b \implies a = 2c -b = c \implies a = b = c.$
   If we choose the simplest non-zero scaling where $a = 1 \implies b = c = 1$, and the equation of the plane becomes $(x -1) + (y -3) + z = 0 \implies x + y + z = 4$.

5. Conclusion. Using the point $P_1 = \langle 1, 3, 0 \rangle$ and the normal vector $\vec{M}=⟨1, 1, 1⟩$, the equation of the third plane is $x + y + z = 4$.

Bibliography

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