Multivariable Calculus: 3-D Space, Vectors, the Dot Product, and the Cross Product

Octant

One of eight sections split off in a 3-D space.

Distance formula

√((x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²)

Equation of a sphere

(x-h)² + (y-k)² + (z-l)² = r²

Vector

Quantity with a magnitude (length) and a direction. They are represented with an arrow. They have an initial and terminal point. Equivalent vectors are equal in magnitude and direction. The zero vector has a magnitude of 0 and no direction. Scalars can change the magnitude. If a vector is negative, it points in the opposite direction. The position vector starts at the origin.

Vector Addition

Vector form from the initial point of a to the terminal point of b. The sum can be found by combining the components of each vector. It is commutative.

Magnitude of a vector

||a|| = √(a₁² + a₂² + a₃²)

Unit vector

Vector of length one, usually written in i j k form. They can be found by dividing each component of the vector by the magnitude of the vector.

Components given magnitude and direction

a = ||v||cosθ, b = ||v||sinθ

Direction angle given magnitude and components

cosθ_a = (a)/(||v||); cosθ_b = (b)/(||v||); cosθ_c = (c)/(||v||)

Dot Product

Way to multiply two vectors. Yields a scalar. a • b = a₁b₁ + a₂b₂ + a₃b₃ = ||a|| ||b|| cosθ, where θ is the angle between the vectors. If a • b = 0, the vectors are orthogonal (perpendicular). a • a = (||a||)². It is commutative, distributive, and associative with a scalar. The zero vector dotted with any vector is 0. Just make sure you never dot a vector with a scalar.

Scalar projection of b onto a

comp_ab = (a • b)/(||a||)

Vector projection of b on a

proj_ab = ((a • b)/(a • a)) * a = ((a • b)/(||a||)²) * a

Cross product

Product of two vectors that creates another vector. That vector is perpendicular to the two vectors. It is given by a x b = <a₂b₃ - a₃b₂, a₃b₁ - a₁b₃, a₁b₂ - a₂b₁>. The direction can be found by using the right-hand rule. It only works in 3 dimensions.

Magnitude of the cross product

||a x b|| = ||a||||b||sinθ. It is equal to the area of a parallelogram defined by a and b.

Properties of the cross product

• a x b = -b x a

• (ca) x b = c(a x b) = a x (cb)

• a x (b + c) = a x b + a x c

• a • (b x c) = (a x b) • c

• a x (b x c) = (a • c)b - (a • b)c

Triple Product

a • (b x c). Gives the volume of the parallelepiped defined by a, b, and c. If it equals zero, the three vectors are coplanar.

Vector equations of lines in 3-D space

An equation of a line can be found with a direction vector and a point. The vector form of the equation is given by <x, y, z> = <x_o, y_o, z_o> + t<a, b, c>, where (x_o, y_o, z_o) is the original point and <a, b, c> is the direction vector.

Parametric equations of lines 3-D space

x = x_o + at, y = y_o + bt, z = z_o + ct, where (x_o, y_o, z_o) is the original point and <a, b, c> is the direction vector.

Symmetric equations of lines in 3-D space

t = (x-x_o)/a = (y-y_o)/b = (z-z_o)/c, where (x_o, y_o, z_o) is the original point and <a, b, c> is the direction vector.

Equation of a plane

ax + by + cz = 0. The vector <a, b, c> is the normal vector to the plane. The equation can be extended to <a, b, c> • <x-x_o, y-y_o, z-z_o> = 0, where (x_o, y_o, z_o) is any point on the plane.

Drawing planes

Can be drawn by using traces. Find the equation of the line traced by the plane through the xy, yz, and xz planes, and that will give you an image of the plane.

Line where two planes intersect

The cross product of the normal vectors of each plane will give the direction vector of the line, and a point on the plane can be found by using a combination of system methods.