## Medians of a Triangle are Concurrent. The Point of Concurrency is the Centroid.

A median of a triangle is the line joining the midpoint of a side and the opposite vertex.

## Altitudes of a Triangle are Concurrent. The Point of Concurrency is the Orthocenter.

An altitude is the perpendicular line drawn from a vertex on the opposite side of the triangle.

## Volume of Parallelopiped and Tetrahedron

A parallelopiped is a 3-D object, each of whose faces is a parallelogram.

A rectangular parallelopiped is the one, whose faces are rectangular.

A cube is a rectangular parallelopiped, whose edges are of equal length.

A tetrahedron is a 3-D object, whose all faces are triangular. It can be shown that a parallelopiped can be decomposed into 6 tetrahedra.

For a visual proof :  1 Parallelopipded  = 6 Tetrahedra

Let there be 2 such objects, whose coterminus edges are identical. Then, the volume of the tetrahedron is ${\dfrac 1 6}$th of that of the parallelopiped. The expression can also be obtained using the following theorem :

## Proof of Section Formula using Vectors

The section formula is one of the important results in the geometry. We will obtain the section formula for internal as well as external division, using vectors.

# Section Formula for External Division

## Vector Algebra : Theorems (I)

In this blog, we will prove fundamental theorems in vector algebra. These theorems are in accordance with the +2 curriculum.

# Theorem 4 : Non-coplanar Vectors

## Trigonometric Functions

• ### Trigonometric Functions

The trigonometric functions are functions of angles. In the blogpost ‘Angle and Its Measurement‘, we studied what an angle is and how it is measured. An important concept to be recalled is that of coterminal angles. These are the angles, whose initial and terminal rays are identical and their magnitudes differ by an integral multiple of $360^o$ or $2 \pi ^c$.

There are 6 basic trigonometric functions:

1) Sine (abbr. sin)

2) Cosine (abbr. cos)

3) Tangent (abbr. tan)

4) Cosecant (abbr. cosec)

5) Secant (abbr. sec)

6) Cotangent (abbr. cot)

• ### Definitions

Consider a circle (radius $r$) and a point $P$ on it. Fix the $X$ and $Y$ axes such that the origin is the center of the circle. Let $\theta$ be the angle made by $OP$ with positive $X$ axis.

Recall: Length of perpendicular from a point on $X$ ($Y$) axis gives the magnitude of $y$ ($x$) coordinate of the point.

Let coordinates of $P$ be $x$ and $y$. Then,

${sin (\theta) = \frac {y}{r}, \ cos (\theta) = \frac {x}{r}}$

${tan (\theta) = \frac {y}{x}, x \ne 0, \ cot (\theta) = \frac {x}{y}, y \ne 0}$

${cosec (\theta) = \frac {r}{y}, y \ne 0 , \ sec (\theta) = \frac{r}{x}, x \ne 0}$

As the location of point $P$ changes (along the circle), $\theta$ changes and so the ratios. Coterminal angles have same trigonometric ratios.

It may also be noted that the ratios are independent of radius of circle.

• ### Alternative Definitions

${tan (\theta) = \frac {sin (\theta)}{cos (\theta)}, \ cot (\theta) = \frac {cos (\theta)}{sin (\theta)}}$

${cosec (\theta) = \frac {1}{sin (\theta)}, \ sec (\theta) = \frac {1}{cos (\theta)}}$

• ### Important Identities

For any angle $\theta$,

${sin^2 (\theta)+ cos^2 (\theta)=1}$

${1 + tan^2 (\theta)= sec^2 (\theta)}$

${1 + cot^2 (\theta) = cosec^2 (\theta)}$

These can be derived from the Pythagoras’ theorem.

• ### Domains and Ranges of Trigonometric Functions

1) $Sine$ Function: Domain $\mathbb R$ and Range $[-1,1]$

2) $Cosine$ Function: Domain $\mathbb R$ and Range $[-1,1]$

3) $Tangent$ Function: Domain $\mathbb R - \{ \theta | \ \theta = (2n-1)\frac {\pi}{2}, n \in \mathbb Z \}$ and Range $\mathbb R$

4) $Cotangent$ Function: Domain $\mathbb R - \{ \theta | \ \theta = n \pi, n \in \mathbb Z \}$ and Range $\mathbb R$

5) $Cosecant$ Function: Domain $\mathbb R - \{ \theta | \ \theta = {n \pi}, n \in \mathbb Z \}$ and Range $\mathbb R - (-1,1)$

6) $Secant$ Function: Domain $\mathbb R - \{ \theta | \theta = (2n-1)\frac {\pi}{2}, n \in \mathbb Z \}$ and Range $\mathbb R - (-1,1)$

• ### Standard Angles

The angles $0, \frac {\pi}{6}, \frac {\pi}{4}, \frac {\pi}{3}, \frac {\pi}{2}, \pi, \frac {3 \pi}{2}, 2 \pi$ are termed as the standard angles. Given below are their $sine$ and $cosine$ values. The other ratios can be found using these (if they are defined).

• ### Signs of Trigonometric Ratios in the Quadrants

Note: $r$ is always positive, $x$ is positive in $1st$ and $4th$ quadrants, $y$ is positive in $1st$ and $2nd$ quadrants.

• ### Periodicity

A characteristic of trigonometric functions is their periodicity. The values get repeated after a certain fixed interval. This can also be seen from the graphs.

A function $f(x)$ is said to be periodic, if $\forall x$ in its domain, $f(x+T)= f(x)$, where $T$ is the period.

The least positive value of $T$ is known as the fundamental period.

Sine, Cosine, Cosecant and Secant are periodic with fundamental period $2 \pi$.

Tangent and Cotangent are periodic with fundamental period $\pi$.

• ### Odd and Even Trigonometric Functions

A function $f(x)$ is said to be odd, when $\forall x$ in its domain,

${f(-x) = - f(x)}$

Sine, Tangent, Cosecant, Cotangent are odd functions.

A function $f(x)$ is said to be even, when $\forall x$ in its domain,

${f(-x) = f(x)}$

Cosine and Secant are even functions.

## 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.