Applications in geometry

Vector algebra is a coordinate-free language for geometric statements. Once you can write a problem in vector form, the manipulation often becomes mechanical: a few dot products, a cross product, a triple product , and you're done.

Collinearity of three points

Three points $A, B, C$ are collinear iff the vectors $\vec{AB}$ and $\vec{AC}$ are parallel, i.e. $\vec{AB} \times \vec{AC} = \vec{0}.$ Equivalently, $\vec{AC} = \lambda \vec{AB}$ for some scalar.

Coplanarity of four points

Four points $A, B, C, D$ are coplanar iff $[\vec{AB}\;\vec{AC}\;\vec{AD}] = 0.$

Coplanarity of three vectors

Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar iff $[\vec a\;\vec b\;\vec c] = 0$ . (Each lies in the plane spanned by the others.)

Parallel and perpendicular vectors

Parallel: $\vec a \times \vec b = \vec 0$ , or $\vec b = \lambda \vec a$ .
Perpendicular: $\vec a \cdot \vec b = 0$ .

Distance from a point to a line

A line passing through $A$ with direction vector $\vec{d}$ , i.e. parametrised as $\vec{r} = \vec a + t \vec d$ . The distance from a point $P$ to this line is $D = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}.$

Why? $\vec{AP}\times\vec d$ has magnitude $|\vec{AP}||\vec d|\sin\theta$ , where $\theta$ is the angle between $\vec{AP}$ and $\vec d$ . The distance from $P$ to the line is the component of $\vec{AP}$ perpendicular to $\vec d$ , which is $|\vec{AP}|\sin\theta = |\vec{AP}\times\vec d|/|\vec d|$ .

Angle between two vectors / lines

Vectors: $\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ , $\sin\theta = \dfrac{|\vec a\times\vec b|}{|\vec a||\vec b|}$ .
Lines (defined by direction vectors $\vec d_1, \vec d_2$ ): same formula. The angle is taken in $[0, \pi/2]$ by convention (use absolute value of cosine).

Worked examples

Example 1. Show that the points $A = (1, 2, 3)$ , $B = (3, 5, 7)$ , $C = (5, 8, 11)$ are collinear.

$\vec{AB} = (2, 3, 4)$ . $\vec{AC} = (4, 6, 8) = 2\vec{AB}$ . So they are collinear.

Example 2. Find the angle between the diagonals of a cube of unit side.

A cube with one vertex at the origin has main diagonal from $(0, 0, 0)$ to $(1, 1, 1)$ , direction $(1, 1, 1)$ . Another main diagonal from $(1, 0, 0)$ to $(0, 1, 1)$ , direction $(-1, 1, 1)$ . Dot product: $-1 + 1 + 1 = 1$ . Magnitudes: $\sqrt 3$ each. So $\cos\theta = 1/3$ .

Example 3. Find the distance from $P = (1, 2, 3)$ to the line $\vec r = \hat i + 2\hat j + t(\hat i + \hat j + \hat k)$ .

$A = (1, 2, 0)$ , $\vec d = (1, 1, 1)$ . $\vec{AP} = (0, 0, 3)$ . $\vec{AP}\times\vec d = \begin{vmatrix}\hat i & \hat j & \hat k\\0 & 0 & 3\\1 & 1 & 1\end{vmatrix} = \hat i(0 - 3) - \hat j(0 - 3) + \hat k(0) = -3\hat i + 3\hat j$ . Magnitude $= 3\sqrt 2$ . $|\vec d| = \sqrt 3$ . Distance $= \dfrac{3\sqrt 2}{\sqrt 3} = \sqrt 6$ .

Example 4. Show that the four points $A = (1, 1, 1)$ , $B = (2, 1, 3)$ , $C = (3, 2, 2)$ , $D = (3, 3, 4)$ are coplanar.

$\vec{AB} = (1, 0, 2)$ , $\vec{AC} = (2, 1, 1)$ , $\vec{AD} = (2, 2, 3)$ .

$\det \begin{pmatrix} 1 & 0 & 2 \\ 2 & 1 & 1 \\ 2 & 2 & 3 \end{pmatrix} = 1(3 - 2) - 0 + 2(4 - 2) = 1 + 4 = 5 \ne 0$ .

So they are not coplanar. (Sometimes problems are designed to fool you , always compute.)

Example 5. Show using vectors that the perpendiculars from the vertices of a triangle to the opposite sides are concurrent (i.e. the orthocentre exists).

Let triangle $ABC$ have position vectors $\vec a, \vec b, \vec c$ with respect to some origin. Let $H$ be a point such that $\vec{AH}\cdot\vec{BC} = 0$ and $\vec{BH}\cdot\vec{CA} = 0$ (i.e. $H$ lies on two altitudes). Then we must show $\vec{CH}\cdot\vec{AB} = 0$ .

Let $\vec h$ be the position vector of $H$ . $\vec{AH} = \vec h - \vec a$ , $\vec{BC} = \vec c - \vec b$ , etc. The first two conditions: $(\vec h - \vec a)\cdot(\vec c - \vec b) = 0$ , $(\vec h - \vec b)\cdot(\vec a - \vec c) = 0$ . Expand:

$\vec h \cdot \vec c - \vec h\cdot\vec b - \vec a\cdot\vec c + \vec a\cdot\vec b = 0$ .
$\vec h\cdot\vec a - \vec h\cdot\vec c - \vec a\cdot\vec b + \vec b\cdot\vec c = 0$ .

Add: $\vec h\cdot\vec a - \vec h\cdot\vec b - \vec a\cdot\vec c + \vec b\cdot\vec c = 0$ , i.e. $(\vec h - \vec c)\cdot(\vec a - \vec b) = 0$ . So $\vec{CH}\cdot\vec{AB} = 0$ . Concurrence proved.

Example 6. Find the unit vector in the plane of $\hat i + 2\hat j$ and $\hat i - \hat j$ that is perpendicular to $2\hat i + \hat j + \hat k$ .

Let the desired vector be $\vec r = \alpha (\hat i + 2\hat j) + \beta(\hat i - \hat j) = (\alpha + \beta)\hat i + (2\alpha - \beta)\hat j$ . Perpendicularity to $(2, 1, 1)$ : $2(\alpha + \beta) + (2\alpha - \beta) = 0$ , so $4\alpha + \beta = 0$ , i.e. $\beta = -4\alpha$ . Substituting: $\vec r = -3\alpha\hat i + 6\alpha\hat j = 3\alpha(-\hat i + 2\hat j)$ . Magnitude: $3|\alpha|\sqrt 5$ . Unit vector: $\dfrac{-\hat i + 2\hat j}{\sqrt 5}$ .

Try it yourself

Show that $A = (2, 3, 4)$ , $B = (3, 4, 5)$ , $C = (4, 5, 6)$ are collinear.
Show that the diagonals of a rectangle have equal lengths.
Find the distance from $P = (2, 3, 4)$ to the line through $(1, 2, 3)$ with direction $(1, 1, 1)$ .
Are $(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)$ coplanar?
Find the angle between the body diagonal of a cube and one of its edges.
Show that the medians of a triangle are concurrent at the centroid (using vectors).
Find a vector orthogonal to both $2\hat i + 3\hat j$ and $\hat j - \hat k$ , with magnitude $7$ .
The midpoints of sides $AB$ and $AC$ of a triangle are $M$ and $N$ . Show $\vec{MN} = \frac{1}{2}\vec{BC}$ .
Find $\lambda$ so that $A = (1, -1, 2)$ , $B = (3, 2, 1)$ , $C = (\lambda, 5, 3)$ are collinear.
The position vectors of two points are $\vec a$ and $\vec b$ . Find the position vector of a point that divides the segment $AB$ in ratio $\sqrt 2 : 1$ internally.
Find $\hat n$ perpendicular to the plane of $\hat i + \hat j - 2\hat k$ and $2\hat i - \hat j + \hat k$ .
Show that the line joining the midpoints of two sides of a triangle is parallel to the third side.
A particle moves with position $\vec r(t) = t\hat i + t^2\hat j + t^3\hat k$ . Find its velocity and acceleration at $t = 1$ .
The vertices of a quadrilateral are $A = (1, 2, 3)$ , $B = (4, 5, 6)$ , $C = (7, 8, 10)$ , $D = (4, 5, 7)$ . Find the area.

Pitfalls and tricks

Vectors first, then coordinates. Many proofs are cleaner if you stay in pure vector form before introducing coordinates.
Use the right tool: dot product for angles and perpendicularity, cross product for areas and normals, triple product for volumes and coplanarity.
Distance from a point to a line is the magnitude of the perpendicular component of $\vec{AP}$ , captured by $|\vec{AP}\times\vec d|/|\vec d|$ .
Position vectors are relative to an origin , but most identities (involving differences of position vectors) are origin-independent.
Practice sketching even rough 3D pictures , they keep you honest about which vector points where.