# Applications of Vectors in Geometry

Vectors are such a powerful tool in mathematics and physics, that many results can be proved very easily and intuitively.

#### Statement : If the diagonals of a parallelogram are congruent, then it is a rectangle.

Proof :

Let $ABCD$ be the parallelogram. Let $A$ be the origin. So, the position vectors will be from $A$. Diagonals are $AC$ and $BD$. We have $|\vec {AC}| = |\vec {BD}|$.

Therefore, $|\vec c| = |\vec d - \vec b|$. By parallelogram law of vector addition, $\vec c = \vec d + \vec b$. So,

${|\vec c| = |\vec d - \vec b| = |\vec d + \vec b|}$

On squaring,

${\vec d \cdot \vec d - 2 \vec d \cdot \vec b + \vec b \cdot \vec b = \vec d \cdot \vec d + 2 \vec d \cdot \vec b + \vec b \cdot \vec b}$

So,

${- 2 \vec d \cdot \vec b = 2 \vec d \cdot \vec b}$

This implies $\vec d \cdot \vec b = 0$, or $\vec b$ is perpendicular to $\vec d$. Hence, $\angle A$ is a right angle. This being a parallelogram, will have all other angles equal to $90^o$ and hence it is a rectangle.

Statement : The diagonals of a kite are at right angles.

Proof :

Let $ABCD$ be the kite, with $A$ as the origin. Clearly $BC$ and $DC$ are equal sides. In terms of vectors,

${|\vec {BC}| = |\vec {DC}| \ \ i.e. \ \ |\vec c - \vec b| = |\vec c - \vec d|}$

On squaring,

${\vec c \cdot \vec c - 2 \vec c \cdot \vec b + \vec b \cdot \vec b = \vec c \cdot \vec c - 2 \vec c \cdot \vec d + \vec d \cdot \vec d}$

Canceling common terms,

${- 2 \vec c \cdot \vec b = -2 \vec c \cdot \vec d}$

Note that $AB$ and $AD$ are equal. So, $b^2 = d^2$. So,

${\vec c \cdot (\vec b - \vec d) = 0}$

Hence, $\vec {AC} \cdot \vec {DB} = 0$. Hence the diagonals are perpendicular.

Statement : If in a tetrahedron, edges in each of the two pairs of opposite edges are perpendicular then the edges in the third pair are also perpendicular.

Proof:

In a tetrahedron, each triangle shares an edge with the other. Considering any 2 triangular faces, we are left with only 1 edge. The pair of common edge and the uncommon edge is said to be a pair of opposite edges. Let $OABC$ be a tetrahedron. So, $(OA,BC)$, $(AB, OC)$ and $(OB,AC)$ are the pairs of opposite edges. Let any 2 of them be perpendicular.

$\vec {OA} \cdot \vec {BC} = 0$ and $\vec {OC} \cdot \vec {AB} =0$. Therefore,

${\vec a \cdot (\vec c - \vec b) = 0 \ \ and \ \ \vec c \cdot {\vec b - \vec a} = 0}$

Expanding the brackets and then adding the equations,

${\vec a \cdot \vec c - \vec a \cdot \vec b + \vec c \cdot \vec b - \vec c \cdot \vec a =0}$

Or

${- \vec a \cdot \vec b + \vec c \cdot \vec b = 0}$

This gives $\vec b \cdot (\vec c - \vec a ) = 0$ i.e. $\vec {OB} \cdot \vec {AC} = 0$. Hence $OB$ is perpendicular to $BC$ and these 2 form the third pair.