Unit 4: Vector Algebra and its Applications

Scalar and Vector Triple Products

This unit combines the dot and cross products to define products involving three vectors.

Scalar Triple Product (STP)

The STP combines three vectors (a, b, c) to produce a **scalar** result.

Definition: [a b c] = a · (b × c)

Geometric Interpretation:
The magnitude of the STP, |a · (b × c)|, represents the volume of the parallelepiped whose adjacent edges are the vectors a, b, and c.

Calculation (Determinant Form):
If a = a₁i + a₂j + a₃k, b = b₁i + b₂j + b₃k, c = c₁i + c₂j + c₃k, then:

[a b c] =

a₁	a₂	a₃
b₁	b₂	b₃
c₁	c₂	c₃

Properties of STP:

• Dot and Cross are interchangeable:
  a · (b × c) = (a × b) · c
• Cyclic property: The value is unchanged under a cyclic permutation (abc → bca → cab).
  [a b c] = [b c a] = [c a b]
• Anti-cyclic property: Swapping any two adjacent vectors negates the value.
  [a b c] = -[a c b]
• Condition for Coplanarity: If the STP is zero, the volume is zero. This means the three vectors lie on the same plane.

  [a b c] = 0 ⟺ a, b, c are coplanar.

Vector Triple Product (VTP)

The VTP combines three vectors (a, b, c) to produce a **vector** result.

Definition: a × (b × c)

Geometric Interpretation:
The vector (b × c) is perpendicular to the plane containing b and c.
The vector a × (b × c) is perpendicular to (b × c), which means it must lie back in the plane of b and c.

The "BAC-CAB" Rule:
This identity is used to expand the VTP. It is a must-memorize formula.

a × (b × c) = b(a · c) - c(a · b)

Note: The terms in brackets, (a · c) and (a · b), are scalars. The result is a linear combination of vectors b and c.

Properties of VTP:

• Not Associative: Order matters greatly.
  a × (b × c) ≠ (a × b) × c
• (a × b) × c = -c × (a × b) = -[a(c · b) - b(c · a)] = b(a · c) - a(b · c)

Vector Equations of Lines, Planes, and Spheres

We use a **position vector** r = xi + yj + zk to represent a general point P(x, y, z) in space.

Vector Equation of a Line

A line is defined by a point and a direction.

• Form 1: Passing through point A (position vector a) and parallel to vector b.
  The vector from A to P is r - a. This vector must be parallel to b.
  r - a = t*b, where t is a scalar parameter.

  r = a + t*b

• Form 2: Passing through two points A (vector a) and B (vector b).
  The line passes through a and is parallel to the vector (b - a).
  Using Form 1:

  r = a + t(b - a) or r = (1-t)a + t*b

Vector Equation of a Plane

A plane is defined by a point and a normal (perpendicular) vector.

• Form 1: Passing through point A (vector a) with normal vector n.
  Any vector in the plane, like r - a, must be perpendicular to n.
  Their dot product must be zero.

  (r - a) · n = 0 or r · n = a · n

  If d = a · n (a scalar constant), the general form is r · n = d.

• Form 2: Normal Form.
  If n̂ is the unit normal vector, and p is the perpendicular distance from the origin to the plane.

  r · n̂ = p

• Form 3: Three-Point Form (passing through a, b, c).
  The vectors (b - a) and (c - a) lie in the plane.
  The normal vector is n = (b - a) × (c - a).
  The vector (r - a) is also in the plane, so it must be coplanar with (b - a) and (c - a).
  Using the condition for coplanarity (STP = 0):

  [r - a b - a c - a] = 0

Vector Equation of a Sphere

A sphere is defined by its center and radius.

• Form 1: Center at origin (0), radius R.
  The length of the position vector r for any point on the sphere is R.
  |r| = R

  r · r = R²

• Form 2: Center at C (vector c), radius R.
  The vector from the center C to any point P is r - c. Its length is R.
  |r - c| = R
  Squaring both sides:

  (r - c) · (r - c) = R²
  or r·r - 2r·c + c·c = R²

• Form 3: Diameter Form (ends of diameter at A(a) and B(b)).
  For any point P(r) on the sphere, the vectors PA = (a - r) and PB = (b - r) form a right angle.
  Their dot product is zero.

  (r - a) · (r - b) = 0

Related Problems

This section involves using the above definitions to solve problems.

Example: Find the intersection of a line and a plane.

Line: r = a + t*b
Plane: r · n = d

Solution: The intersection point r must satisfy both equations. Substitute the line equation into the plane equation:

1. (a + t*b) · n = d
2. Distribute: a · n + t(b · n) = d
3. Solve for the parameter t:
   t(b · n) = d - a · n
t = (d - a · n) / (b · n)
4. Substitute this value of t back into the line equation r = a + t*b to find the position vector of the intersection point.

Note: If b · n = 0, the vector b (line) and n (normal) are perpendicular. This means the line is parallel to the plane and will not intersect (unless d - a · n = 0 as well, in which case the line lies *in* the plane).

Unit 4: Exam Quick Tips