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Chapter 6

Applications of Vector Algebra

Scalar & vector products, triple products, planes and lines in 3D.

1. The scalar (dot) product

The dot product takes two vectors and returns a single number — a scalar — that measures how much the two vectors point in the same direction. It is the quiet workhorse behind angles, projections, and the test for perpendicularity.

DefinitionScalar (dot) product

For vectors a⃗\vec a and b⃗\vec b with angle θ\theta between them (0≤θ≤π)(0 \le \theta \le \pi), the dot product is a⃗⋅b⃗=∣a⃗∣ ∣b⃗∣cos⁡θ\vec a \cdot \vec b = |\vec a|\,|\vec b|\cos\theta. In components, if a⃗=a1i⃗+a2j⃗+a3k⃗\vec a = a_1\vec i + a_2\vec j + a_3\vec k and b⃗=b1i⃗+b2j⃗+b3k⃗\vec b = b_1\vec i + b_2\vec j + b_3\vec k, then a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a \cdot \vec b = a_1 b_1 + a_2 b_2 + a_3 b_3.

The result is a scalar, not a vector. Geometrically, ∣b⃗∣cos⁡θ|\vec b|\cos\theta is the length of the projection of b⃗\vec b onto a⃗\vec a, so a⃗⋅b⃗\vec a \cdot \vec b is ∣a⃗∣|\vec a| times that projection — a measure of shared direction.

Angle between two vectors
cos⁡θ=a⃗⋅b⃗∣a⃗∣ ∣b⃗∣\cos\theta = \dfrac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|}
Projection of b on a
proja⃗b⃗=a⃗⋅b⃗∣a⃗∣\text{proj}_{\vec a}\vec b = \dfrac{\vec a \cdot \vec b}{|\vec a|}
Key idea: Two non-zero vectors are perpendicular exactly when $\vec a \cdot \vec b = 0$, because $\cos 90^\circ = 0$. Open the Visualize tab and rotate the vectors until the dot product hits zero.
Properties of the dot product
  • Commutative: a⃗⋅b⃗=b⃗⋅a⃗\vec a \cdot \vec b = \vec b \cdot \vec a
  • Distributive: a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec a \cdot (\vec b + \vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c
  • a⃗⋅a⃗=∣a⃗∣2\vec a \cdot \vec a = |\vec a|^2
  • For the standard basis: i⃗⋅i⃗=j⃗⋅j⃗=k⃗⋅k⃗=1\vec i \cdot \vec i = \vec j \cdot \vec j = \vec k \cdot \vec k = 1 and i⃗⋅j⃗=j⃗⋅k⃗=k⃗⋅i⃗=0\vec i \cdot \vec j = \vec j \cdot \vec k = \vec k \cdot \vec i = 0
ExampleDot product and angle
Find a⃗⋅b⃗\vec a \cdot \vec b and the angle between a⃗=i⃗+2j⃗+2k⃗\vec a = \vec i + 2\vec j + 2\vec k and b⃗=2i⃗+j⃗−2k⃗\vec b = 2\vec i + \vec j - 2\vec k.

2. The vector (cross) product

The cross product of two vectors in 3D returns a new vector — one that is perpendicular to both inputs. Its length records the area they span, and its direction records the orientation. Unlike the dot product, order matters.

DefinitionVector (cross) product

a⃗×b⃗=∣a⃗∣ ∣b⃗∣sin⁡θ  n^\vec a \times \vec b = |\vec a|\,|\vec b|\sin\theta\;\hat n, where n^\hat n is the unit vector perpendicular to both a⃗\vec a and b⃗\vec b, pointing in the direction given by the right-hand rule. Its magnitude is ∣a⃗×b⃗∣=∣a⃗∣ ∣b⃗∣sin⁡θ|\vec a \times \vec b| = |\vec a|\,|\vec b|\sin\theta.

The magnitude ∣a⃗×b⃗∣|\vec a \times \vec b| equals the area of the parallelogram with a⃗\vec a and b⃗\vec b as adjacent sides. Half of it is the area of the triangle formed by the two vectors.

Determinant form
a⃗×b⃗=∣i⃗j⃗k⃗a1a2a3b1b2b3∣=(a2b3−a3b2)i⃗−(a1b3−a3b1)j⃗+(a1b2−a2b1)k⃗\vec a \times \vec b = \begin{vmatrix} \vec i & \vec j & \vec k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\vec i - (a_1 b_3 - a_3 b_1)\vec j + (a_1 b_2 - a_2 b_1)\vec k
Area
Area of parallelogram=∣a⃗×b⃗∣,Area of triangle=12∣a⃗×b⃗∣\text{Area of parallelogram} = |\vec a \times \vec b|,\qquad \text{Area of triangle} = \tfrac{1}{2}|\vec a \times \vec b|
Note: The cross product is anti-commutative: $\vec a \times \vec b = -(\vec b \times \vec a)$. Reversing the order flips the resulting vector to the opposite direction.
Key idea: Two non-zero vectors are parallel exactly when $\vec a \times \vec b = \vec 0$, because $\sin 0^\circ = 0$. In the 2D explorer this shows up as the $z$-component (and the parallelogram area) collapsing to zero.
Standard basis cross products
  • i⃗×j⃗=k⃗,j⃗×k⃗=i⃗,k⃗×i⃗=j⃗\vec i \times \vec j = \vec k,\quad \vec j \times \vec k = \vec i,\quad \vec k \times \vec i = \vec j
  • i⃗×i⃗=j⃗×j⃗=k⃗×k⃗=0⃗\vec i \times \vec i = \vec j \times \vec j = \vec k \times \vec k = \vec 0
  • Reversing any of the above negates the result, e.g. j⃗×i⃗=−k⃗\vec j \times \vec i = -\vec k
ExampleCross product and area
Find a⃗×b⃗\vec a \times \vec b and the area of the parallelogram for a⃗=i⃗+j⃗+k⃗\vec a = \vec i + \vec j + \vec k and b⃗=i⃗−j⃗+k⃗\vec b = \vec i - \vec j + \vec k.

3. The scalar triple product

Combine three vectors by taking the cross product of two, then the dot product with the third. The single number you get measures the volume of the box they span — and its sign tells you the orientation.

DefinitionScalar triple product

[a⃗ b⃗ c⃗]=a⃗⋅(b⃗×c⃗)[\vec a\ \vec b\ \vec c] = \vec a \cdot (\vec b \times \vec c). In components it is the 3×33\times 3 determinant of the rows of coordinates.

Determinant form
[a⃗ b⃗ c⃗]=∣a1a2a3b1b2b3c1c2c3∣[\vec a\ \vec b\ \vec c] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

The absolute value ∣[a⃗ b⃗ c⃗]∣\big|[\vec a\ \vec b\ \vec c]\big| is the volume of the parallelepiped with a⃗\vec a, b⃗\vec b, c⃗\vec c as adjacent edges.

Key idea: Three vectors are coplanar exactly when $[\vec a\ \vec b\ \vec c] = 0$ — the box they would span is flat, so it has zero volume.
Properties
  • Cyclic invariance: [a⃗ b⃗ c⃗]=[b⃗ c⃗ a⃗]=[c⃗ a⃗ b⃗][\vec a\ \vec b\ \vec c] = [\vec b\ \vec c\ \vec a] = [\vec c\ \vec a\ \vec b]
  • Swapping any two vectors changes the sign
  • If any two of the three vectors are equal, the product is 00
  • a⃗⋅(b⃗×c⃗)=(a⃗×b⃗)⋅c⃗\vec a \cdot (\vec b \times \vec c) = (\vec a \times \vec b) \cdot \vec c — dot and cross may be interchanged
ExampleCoplanarity check
Are a⃗=i⃗+j⃗\vec a = \vec i + \vec j, b⃗=j⃗+k⃗\vec b = \vec j + \vec k, c⃗=i⃗+k⃗\vec c = \vec i + \vec k coplanar?

4. The vector triple product

Cross two vectors and cross the result again with a third, and you land back inside the plane of the last two vectors. The expansion below — sometimes called the BAC–CAB rule — lets you avoid computing nested cross products.

Vector triple product identity
a⃗×(b⃗×c⃗)=(a⃗⋅c⃗) b⃗−(a⃗⋅b⃗) c⃗\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\,\vec b - (\vec a \cdot \vec b)\,\vec c
Note: The result lies in the plane of $\vec b$ and $\vec c$. Note the cross product is not associative: in general $\vec a \times (\vec b \times \vec c) \ne (\vec a \times \vec b) \times \vec c$, so the brackets matter.
ExampleApplying the identity
Simplify a⃗×(b⃗×c⃗)\vec a \times (\vec b \times \vec c) given a⃗⋅b⃗=4\vec a \cdot \vec b = 4 and a⃗⋅c⃗=−1\vec a \cdot \vec c = -1.

5. The equation of a plane

A plane in space is pinned down by a point on it and a direction perpendicular to it — the normal vector n⃗\vec n. Every different description of a plane below is really just a repackaging of the single condition: the vector from a fixed point to any point on the plane is perpendicular to n⃗\vec n.

Vector / normal form
(r⃗−a⃗)⋅n⃗=0⟺r⃗⋅n⃗=a⃗⋅n⃗(\vec r - \vec a) \cdot \vec n = 0 \quad\Longleftrightarrow\quad \vec r \cdot \vec n = \vec a \cdot \vec n
Cartesian (general) form
ax+by+cz=d,where n⃗=ai⃗+bj⃗+ck⃗ax + by + cz = d, \quad\text{where } \vec n = a\vec i + b\vec j + c\vec k

The coefficients a,b,ca, b, c are exactly the components of the normal vector. To find the plane through three points A,B,CA, B, C, take n⃗=AB→×AC→\vec n = \overrightarrow{AB} \times \overrightarrow{AC} and then use one of the points.

Plane through three points
n⃗=AB→×AC→,(r⃗−a⃗)⋅n⃗=0\vec n = \overrightarrow{AB} \times \overrightarrow{AC},\qquad (\vec r - \vec a)\cdot \vec n = 0
Key formulas for a plane
  • Angle between two planes = angle between their normals: cos⁡θ=n⃗1⋅n⃗2∣n⃗1∣ ∣n⃗2∣\cos\theta = \dfrac{\vec n_1 \cdot \vec n_2}{|\vec n_1|\,|\vec n_2|}
  • Two planes are parallel   ⟺  \iff their normals are parallel; perpendicular   ⟺  n⃗1⋅n⃗2=0\iff \vec n_1 \cdot \vec n_2 = 0
  • Distance from point (x0,y0,z0)(x_0, y_0, z_0) to plane ax+by+cz=dax + by + cz = d: ∣ax0+by0+cz0−d∣a2+b2+c2\dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}
Note: Angle between a line (direction $\vec b$) and a plane (normal $\vec n$): $\sin\theta = \dfrac{|\vec b \cdot \vec n|}{|\vec b|\,|\vec n|}$. The line is parallel to the plane when $\vec b \cdot \vec n = 0$, and meets it in a single point otherwise.

6. Distance from a point to a plane

The distance from a point to a plane is measured along the normal — it is the length of the shortest segment from the point to the plane. The formula is a direct application of projecting onto the normal vector.

D=∣ax0+by0+cz0−d∣a2+b2+c2D = \dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}
ExampleDistance from a point to a plane
Find the distance from the point (1,2,0)(1, 2, 0) to the plane 2x−y+2z=52x - y + 2z = 5.