Vector

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Vector and Scalar

In physics (and mathematics), quantities can be classified into 2 main categories; Scalars and Vectors.

  1. Certain quantities such as mass, time, temperature, volume, density etc. can be adequately represented just by a numerical value, an amount or a magnitude. Such quantities are called Scalars.
  2. Certain other quantities such as force, velocity, displacement, acceleration etc. cannot be adequately represented by a numerical value alone. The only way to meaningfully represent such quantities is by considering them as having a direction in addition to the magnitude. Such quantities that have a magnitude as well as a direction are called Vectors.

Vectors are represented by a directed line segment, basically arrows. The length of the line segment denotes the magnitude and the arrowhead indicates the direction of the vector.

For example,
\( \overrightarrow{OA} \) or \( \textbf{OA}\) or \( \vec{a}\) or \( \textbf{a}\)

Magnitude of a Vector

Let \( \vec{a}=(x,y)\) be a vector, then its magnitude is denoted by \( | \vec{a} |\) and defined by
\( | \vec{a} |=\sqrt{x^2 +y^2 }\)
The magnitude of a is also called the norm of a and denoted ||a||.
Note that \(a·a = a^2 =||a||^2\) so that
\( ||a|| = \sqrt{a.a}\)

Direction of a Vector

Let \( \vec{a}=(x,y)\) be a vector, then its direction is denoted by \( \tan \theta \) and defined by
\( \tan \theta =\frac{y}{x}\)

Example 1

If \( \vec{a}=(3,4)\) then find its magnitude and direction.
Solution
Here,
\( \vec{a}=(3,4)\)
This means,x=3,y=4
Now, magnitude of vector is
\( | \vec{a} |=\sqrt{x^2 +y^2 }=\sqrt{3^2 +4^2 }=5\)
Next, the direction of vector is
\( \tan \theta =\frac{y}{x}=\frac{4}{3}\)
or \( \theta =\tan ^{-1}( \frac{4}{3} )\)

Types of Vector

  1. Unit Vector
    A vector with magnitude 1 is called unit vector.
    The example of unit vectors are \( (0,1), (-1,0),(1,0), \left (\frac{3}{5},\frac{4}{5} \right)\)
  2. Zero Vector
    If magnitude of a vector is zero, then it is called zero vector. The starting point and terminal point of such vector coincides.
    The magnitude of the zero vector is zero.
    The direction of zero vector is indeterminate.
    The example of zero vectors is (0,0)
  3. Position Vector
    A vector whose initial point is at origin is called position vector.
    If a point P has coordinate P(x,y), then its position vector is written as \( \vec{p}=(x,y)\)
  4. Equal Vector
    Two vectors whose magnitudes and direction both are equal (same), are called equal vector.
    The examples of equal vector are \( \vec{p}=(3,4) \) and \(\vec{q}=(3,4)\)
  5. Negative Vector
    Two vectors whose magnitudes are equal and directions are opposite are called negative vector.
    Example: In the Figure the vector \( \vec{a}=(3,4)\) and \(\vec{b}=(-3,-4)\) are negative vector.
  6. Like and Unlike Vector
    The vectors having same direction are called like vectors and those having opposite directions are called as unlike vectors.
    In the Figure two vectors \( \vec{a}\) and \(\vec{b}\) are like vector.
    In the Figure ,two vectors \( \vec{a}\) and \(\vec{c}\) are unlike vector.
  7. Coinitial Vector
    Two vectors whose initial points are same are called co-initial vectors. In the Figure,two vectors \( \vec{a}\) and \(\vec{b}\) are co-initial vectors.
  8. Resolution Vector
    One big advantage of using vectors is that they can be resolved into any no. of components. In the simplest case, a vector can be resolved into 2 component vectors lying in the same plane.
    Shown below is a vector OA, if we draw its projections onto any 2 perpendicular axes in its plane (for convenience we have picked the x and y axes), we get a set of 2 new vectors OU and OV. These 2 new vectors are called the components of vector OA.
    We can take this one step further and resolve a vector into 3 components in 3 dimension. What is the advantage of doing this? For one, resolving vectors into perpendicular components allows us to express them in terms of the Cartesian coordinates, which we are familiar with.

Triangle law of Vector Addition

Let \( \vec{a}\) and \( \vec{b}\) are two vector, then the method of adding \( \vec{a}\) and \( \vec{b}\) by head-to-tail construction is called triangle law of vector addition.

In this method, the sum of \( \vec{a}\) and \( \vec{b}\) is the third side of the triangle formed by the segment of \( \vec{a}\) and \( \vec{b}\) as adjacent sides

Parallelogram law of Vector Addition

Let \( \vec{a}\) and \( \vec{b}\) are two vector, then the method of adding \( \vec{a}\) and \( \vec{b}\) by head-to-head (or tail to tail) construction is called parallelogram law of vector addition.

In this method, the sum of \( \vec{a}\) and \( \vec{b}\) is the is the diagonal of parallelogram formed by the segment of \( \vec{a}\) and \( \vec{b}\) as adjacent sides

Vector product

Products of vectors can be defined in many ways, however, two definitions are physically of significant. These are called dot and the vector products. A third kind of product, called geometric product, unifies dot and vector products (but we dont include it in this text)
Thus, in this text, two types of product are entertained. They are

  1. Dot product
  2. Vector product

Dot Product (Scalar Product or Inner Product)

Given two vectors \(\vec{a}\) and \(\vec{b}\) their dot product is denoted by \(\vec{a}.\vec{b}\), whose value is a scalar.
For Geometrical intrepretation, the dot product is fundamentally a projection.

Thus, the dot product of \(\vec{a}\) and \(\vec{b}\) is the product of the projection of \(\vec{a}\) on the direction defined by \(\vec{b}\) with the magnitude of \(\vec{b}\), given by
\(\vec{a}. \vec{b}= (|\vec{a}| \cos \theta) . |\vec{b}|\)
which is the same as the product of the projection of \(\vec{b}\) on the direction defined by \(\vec{a}\) with the magnitude of \(\vec{a}\), given by
\(\vec{a}. \vec{b}= (|\vec{b}|\cos \theta) . |\vec{a}|\)
This demonstrates the obvious symmetry of the result
\(\vec{a}. \vec{b}= \vec{b}.\vec{a}\)
This shows that the dot product is commutative .

In Summary, as shown in figure below, the dot product of \(\vec{a}\) and \(\vec{b}\) is the projection of \(\vec{b}\) in the direction given by \(\vec{a}\) times the magnitude of \(\vec{a}\). This leads to the geometric formula
\( \vec{a} · \vec{b} = |\vec{a} ||\vec{b} | \cos \theta \) (1)

An immediate consequence of (1) is that the dot product of a vector with itself gives the square of the length, that is
\( \vec{a} · \vec{a} = |\vec{a}|^2 \)
In particular, taking the “square” of any unit vector yields 1, for example
\( \vec{i} · \vec{i} = 1 \)
where \( \vec{i}\) as usual denotes the unit vector in the x direction.

The dot product of two vectors can be positive, can be zero, or can be negative.

  1. If \( \vec{a}.\vec{b}=0\) then angle between them is 90 (right angle)
  2. If \( \vec{a}.\vec{b} < 0\) then angle between them greater than 90 (obtuse angle)
  3. If \( \vec{a}.\vec{b}>0\) then angle between them less than 90 (acute angle)

The Geometry of Dot Produst_M1

Let \( \vec{a}\) and \( \vec{b}\) are two vectors and \( \theta \) is angle between them, then
a geometric interpretation of the dot product explains that
\( \vec{a}.\vec{b}= OP. |\vec{a}| \)
or \( OP= \frac{\vec{a}.\vec{b}}{|\vec{a}|} \) (2)

Using the notion of right angled triangle, it explains that
\( OP=|\vec{b}| \cos \theta \) (3)
It is the trigonometric ratio based on the reference angle \( \theta\)
Using (2) and (3), we get
\( \frac{\vec{a}.\vec{b}}{|\vec{a}|} = |\vec{b}| \cos \theta\)
or \( \cos \theta = \frac{\vec{a}.\vec{b}}{|\vec{a}| |\vec{b}|} \)

The Geometry of Dot Produst_M2

Let \( \vec{OA}=\vec{a} ,\vec{OB}=\vec{b},\vec{AB}=\vec{c}\)
Then
According to the cosine law, we can write
\( \vec{c}^2=\vec{a}^2+\vec{b}^2-2|\vec{a}||\vec{b}| \cos \theta\) (1)
Also
\( \vec{c}^2= (\vec{b}-\vec{a})^2\)
or\( \vec{c}^2= \vec{a}^2+\vec{b}^2-2\vec{a}.\vec{b} \) (2)
Using (1) and (2), we get
\(\vec{a}.\vec{b}=|\vec{a}||\vec{b}| \cos \theta\)

The Geometry of Dot Produst_M3

Let \( \vec{OA}=\vec{a}=(a_1,a_2) ,\vec{OB}=\vec{b}=(b_1,b_2)\)
Then
According to basic trigonometry, we can assume that
\(\vec{OA}\) has magnitude \(r_1\) and direction \(\alpha\), then
\( a_1=r_1 \cos \alpha, a_2=r_1 \sin \alpha \) (1)
Also, we can assume that
\(\vec{OB}\) has magnitude \(r_2\) and direction \(\beta\), then
\( b_1=r_2 \cos \beta, b_2=r_2 \sin \beta \) (2)
Here, we can assume that \(\theta\) is the angle between \( \vec{a}\) and \( \vec{b}\), then
\(\vec{a}.\vec{b}=|\vec{a}||\vec{b}| \cos \theta\)
or \(\vec{a}.\vec{b}=r_1 r_2 \cos (\beta -\alpha ) \)
or \(\vec{a}.\vec{b}=a_1 b_1+a_2 b_2 \)

Cross Product

Given two vectors \( \vec{a}\) and \( \vec{b}\) are given then their vector or cross product is a vector whose direction is perpendicular to the plane spaned by \( \vec{a}\) and \( \vec{b}\) and magnitude is the area of parallelogram spaned by \( \vec{a}\) and \( \vec{b}\). Thus, the magnitude of the vector \( \vec{a} \times \vec{b}\) is the the area of the parallelegram with adjacent sides \( \vec{a}\) and \( \vec{b}\).
Note that
Vector product is NOT commutative

Cross Product: Geometry_1

Let \( \vec{a}\) and \( \vec{b}\) are two vectors and \( \theta \) is angle between them, then
\( |\vec{a}\times \vec{b}|=\) Area of the Parallelogram made by \( \vec{a}\) and \( \vec{b}\)
It is the area of parallelogram made by \( \vec{a}\) and \( \vec{b}\)
Thus, we get
\( |\vec{a}\times \vec{b}|=\) base . height
or \(| \vec{a}\times \vec{b}|= | \vec{a} |.( | \vec{b} |\sin \theta )\)
or \( |\vec{a}\times \vec{b}|= | \vec{a} |.| \vec{b} |\sin \theta \)
or \( \sin \theta =\frac{|\vec{a}\times \vec{b}|}{| \vec{a} |.| \vec{b} |}\)

An immediate consequence of which is that
\( \vec{a} || \vec{b} \Leftrightarrow \vec{a} \times \vec{b} = 0\)
The direction of the cross product is given by the right-hand rule, so that
\( \vec{a} \times \vec{b} =- \vec{b} \times \vec{a} \)
Another important property of the cross product is that
\( \vec{a} \times \vec{a} = 0 \)
In terms of the standard orthonormal basis, the geometric formula is
\( \vec{i} \times \vec{j} = \vec{k}\)
\( \vec{j} \times \vec{k} = \vec{i}\)
\( \vec{k} \times \vec{i} = \vec{j}\)

Cross Product: Geometry_2

Let \( \vec{OA}=\vec{a}=(a_1,a_2) ,\vec{OB}=\vec{b}=(b_1,b_2)\)
Then
According to basic trigonometry, we can assume that
\(\vec{OA}\) has magnitude \(r_1\) and direction \(\alpha\), then
\( a_1=r_1 \cos \alpha, a_2=r_1 \sin \alpha \) (1)
Also, we can assume that
\(\vec{OB}\) has magnitude \(r_2\) and direction \(\beta\), then
\( b_1=r_2 \cos \beta, b_2=r_2 \sin \beta \) (2)
Here, we can assume that \(\theta\) is the angle between \( \vec{a}\) and \( \vec{b}\), then
\(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta\)
or \(|\vec{a}\times \vec{b}|=r_1 r_2 \sin (\beta -\alpha ) \)
or \(|\vec{a} \times \vec{b}|=a_1 b_2-a_2 b_1\)

Vector Geometry

Vector is useful topic to use in Geometry. Some of the basic applications of Vector geometry are as follows

  1. Mid-point theorem
    If A and B are two points with position vector \( \vec{a}\) and \( \vec{b}\) then the position vector of midpoint of the segment AB is \( \frac{\vec{a}+\vec{b}}{2}\)

    Let \( \vec{OA}=\vec{a}, \vec{OB}=\vec{b}\) and \( \vec{OP}=\vec{p}\)
    Then
    \( \vec{OP}=\vec{OA}+\vec{AP}\) (1)
    \( \vec{OP}=\vec{OB}+\vec{BP}\) (2)
    Adding (1) and (2), we get
    \( 2 \vec{OP}=\vec{OA}+\vec{AP}+\vec{OB}+\vec{BP}\)
    or\( 2 \vec{OP}=\vec{OA}+\vec{OB}\) \(\vec{AP}=-\vec{BP}\)
    or\( \vec{OP}=\frac{\vec{OA}+\vec{OB}}{2}\)
    or\( \vec{p}=\frac{\vec{a}+\vec{b}}{2}\)
    This completes the proof

  2. Triangle midpoint theorem
    In a triangle ABC , prove that the line joining midpoint of any two sides is parallel and half of third side.

    The proof is given as
    \( \vec{PQ}=\vec{PA}+\vec{AQ}\) (1)
    \( \vec{PQ}=\vec{PB}+\vec{BC}+\vec{CQ}\) (2)
    Adding (1) and (2), we get
    \( 2 \vec{PQ}=\vec{PA}+\vec{AQ}+\vec{PB}+\vec{BC}+\vec{CQ}\)
    or\( 2 \vec{PQ}=\vec{AQ}+\vec{BC}+\vec{CQ}\) br> \(\vec{PA}=-\vec{PB}\)
    or\( 2 \vec{PQ}=\vec{BC}\) \(\vec{AQ}=-\vec{CQ}\)
    or\( \vec{PQ}=\frac{1}{2} \vec{BC}\)
    This completes the proof

  3. Quadrilateral midpoint theorem
    Show that line joining the mid points of a quadrilateral is a parallelogram.

    The proof is given as
    \( \vec{SR}=\frac{1}{2} \vec{AC}\)
    \( \vec{PQ}=\frac{1}{2} \vec{AC}\)
    Using both relations, we get
    \( \vec{SR}=\vec{PQ}\)
    Thus, PQRS is a parallelogram
    This completes the proof

  4. Pythagoras theorem
    Prove the Pythagoras theorem

    Given that triangle ABC is a right angled triangle whose right angle is at C
    Thus
    \( \vec{AC}. \vec{BC}=0\) (1)
    Now
    \( \vec{AB}=\vec{AC}+\vec{CB}\)
    Squaring gives
    \( \vec{AB}^2=\vec{AC}^2+\vec{CB}^2+2\vec{AC}.\vec{CB} \) (2)
    Using both relations (1) and (2), we get
    \( \vec{AB}^2=\vec{AC}^2+\vec{CB}^2 \)
    or\( AB^2=AC^2+CB^2 \)
    Thus, square of hypoteneus is the sum of the square of two legs
    This completes the proof

  5. Semi-circle theorem
    Prove that angle at semi-circle is right angle.

    Let \( \vec{OA}=\vec{r}\) then \( \vec{BO}=-\vec{r}\)
    Thus
    \( \vec{AP}=\vec{AO}+\vec{OP}\) (1)
    Again
    \( \vec{BP}=\vec{BO}+\vec{OP}\) (2)
    NOw
    \( \vec{AP}. \vec{BP}=(\vec{AO}+\vec{OP}).(\vec{BO}+\vec{OP}) \)
    Using both relations (1) and (2), we get
    \( \vec{AP}. \vec{BP}=(\vec{r}+\vec{OP}).(-\vec{r}+\vec{OP}) \)
    or\( \vec{AP}. \vec{BP}=\vec{OP}^2-\vec{r}^2 \)
    or\( \vec{AP}. \vec{BP}=r^2-r^2 \) \( \vec{OP}\) is also radius
    or\( \vec{AP}. \vec{BP}=0 \)
    Thus, angle at P is right angle
    This completes the proof

Solved Examples
  1. Double intercept formula
    Show that equation of straight line in double intercept form by vector method.

    Here
    \( \vec{AP}=k\vec{AB}\)
    or\( \vec{OP}-\vec{OA}=k \vec{OB}-\vec{OA}\) \)
    or\( (x-a,y)=k (-a,b) \)
    or x-a=k and y=kb
    or \( \frac{x}{a}+k=1\) and \( k=\frac{y}{b}\)
    or \( \frac{x}{a}+\frac{y}{b}=1\)
    This completes the proof

  2. IF ABCD is a parallelogram, and G is point of intersection of its diagonals then show that \( \vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=4\vec{OG}\)

    Here
    \( \vec{OG}=\frac{\vec{OB}+\vec{OD}}{2}\)(1)
    \( \vec{OG}=\frac{\vec{OA}+\vec{OC}}{2}\)(1)
    Ading (1) and (2), we get
    \( 2 \vec{OG}=\frac{\vec{OA}+\vec{OC}}{2} +\frac{\vec{OA}+\vec{OC}}{2} \)
    or\( 4 \vec{OG}=\vec{OA}+\vec{OC}+\vec{OB}+\vec{OD} \)
    This completes the proof

  3. If ABCD is a quadrilateral, show that \( \vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}=4\vec{PQ}\) where P and Q are mid points of AC and BD respectively.

    Here
    \( \vec{OP}=\frac{\vec{OA}+\vec{OC}}{2}\)(1)
    \( \vec{OQ}=\frac{\vec{OB}+\vec{OD}}{2}\)(1)
    Ading (1) and (2), we get
    \( \vec{PQ}=\vec{OQ}-\vec{OP} \)
    or\( \vec{PQ}=\frac{\vec{OB}+\vec{OD}}{2}-\frac{\vec{OA}+\vec{OC}}{2}\)
    or\( 2 \vec{PQ}=\vec{OB}+\vec{OD}-\vec{OA}-\vec{OC}\)
    or\( 4 \vec{PQ}=(2\vec{OB}+\vec{OD}-\vec{OA}-\vec{OC})\)(A)
    Similarly, we get
    \( \vec{AB}=\vec{OB}-\vec{OA} \)(3)
    \( \vec{AD}=\vec{OD}-\vec{OA} \)(4)
    \( \vec{CB}=\vec{OC}-\vec{OB} \)(5)
    \( \vec{CD}=\vec{OD}-\vec{OB} \)(6)
    Ading (3) - (6), we get
    \( \vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}= (2\vec{OB}+\vec{OD}-\vec{OA}-\vec{OC})\) (B)
    Using (A) and (B), we establish the result
    This completes the proof

  4. If AC and BD are the diagonals of a parellelogram ABCD, show that \( \vec{AC}+ \vec{BD}=2 \vec{BC}\) and \( \vec{AC}- \vec{BD}=2 \vec{AB}\)
  5. If ABCDEF is a regular hexagon , show that \( \vec{AB}+\vec{AC}+\vec{AD}+\vec{AE}+\vec{AF}=6\vec{AO}\) where O is the center of the hexagon
  6. ABC is any triangle and D,E,F are the middle points of BC, CA and AB respectively. Show that \( \vec{AD}+\vec{BE}+\vec{CF}=0\)

Application of vectors in Trigonometry

Vector is useful topic to use in Trigonometry. Some of the basic applications are as follows

\( \sin (A+B)= \sin A \cos B+ \cos A \sin B\)

Proof
Let A be a point with coordinate \( A(r_1 \cos A, r_1 \sin A)\) which makes an angle A with X-axis and whose length is \( r_1\) from origin O.
Therefore,
\( \vec{OA}=( r_1 \cos A,r_1 \sin A )\) and \( | \vec{OA} |=r_1 \)
Also let B be another point with coordinate \( B(r_2 \cos B, - r_2 \sin B)\) which makes an angle (-B) with X-axis and whose length is \( r_2\) from origin O.
Therefore,
\( \vec{OB}=( r_2 \cos ( -B ),r_2 \sin ( -B ) )\) and \( | \vec{OB} |=r_2 \)
or \( \vec{OB}=( r_2 \cos B,-r_2 \sin B \) and \( | \vec{OB} |=r_2 \)

Now, angle between OA and OB is (A+B)
Therefore,
\( \sin (A+B)=\frac{\vec{OB}\times \vec{OA}}{| \vec{OB} |.| \vec{OA} |}\)
or \( sin(A+B)=\frac{( r_2 \cos B,-r_2 \sin B )\times ( r_1 \cos A,r_1 \sin A )}{| ( r_2 \cos B,-r_2 \sin B ) |.| ( r_1 \cos A,r_1 \sin A ) |}\)
or \( \sin (A+B)= \sin A CosB+ CosA \sin B\)

\( \sin (A-B)= \sin A \cos B- \cos A \sin B\)

Proof
Let A be a point with coordinate \( (r_1 \cos A, r_1 \sin A)\) which makes an angle A with X-axis and whose length is \( r_1\) from origin O.
Therefore,
\( \vec{OA}=( r_1 \cos A,r_1 \sin A )\) and \( | \vec{OA} |=r_1 \)
Also let B be another point with coordinate \( (r_2 \cos B, r_2 \sin B)\) which makes an angle B with X-axis and whose length is \( r_2\) from origin O.
Therefore,
\( \vec{OB}=( r_2 \cos B,r_2 \sin B )\) and \( | \vec{OB} |=r_2 \)

Now, angle between OA and OB is (A-B) .
Therefore,
\( \sin (A-B)=\frac{\vec{OB}\times \vec{OA}}{| \vec{OB} |.| \vec{OA} |}\)
or \( sin(A-B)=\frac{( r_2 \cos B,r_2 \sin B )\times ( r_1 \cos A,r_1 \sin A )}{| ( r_2 \cos B,r_2 \sin B ) |.| ( r_1 \cos A,r_1 \sin A ) |}\)
or \( \sin (A-B)= \sin A CosB- CosA \sin B\)

\( \cos (A-B)= \cos A \cos B+ \sin A \sin B\)

Proof
Let A be a point with coordinate \( (r_1 \cos A, r_1 \sin A)\) which makes an angle A with X-axis and whose length is \( r_1\) from origin O.
Therefore,
\( \vec{OA}=( r_1 \cos A,r_1 \sin A )\) and \( | \vec{OA} |=r_1 \)
Also let B be another point with coordinate \( (r_2 \cos B, r_2 \sin B)\) which makes an angle B with X-axis and whose length is \( r_2\) from origin O.
Therefore,
\( \vec{OB}=( r_2 \cos B,r_2 \sin B )\) and \( | \vec{OB} |=r_2 \)

Now, angle between OA and OB is (A-B).
Therefore,
\( \cos (A-B)=\frac{\vec{OA}.\vec{OB}}{| \vec{OA} |.| \vec{OB} |}\)
or\( \cos (A-B)=\frac{( r_1 \cos A,r_1 \sin A ).( r_2 \cos B,r_2 \sin B )}{| ( r_1 \cos A,r_1 \sin A ) |.| ( r_2 \cos B,r_2 \sin B ) |}\)
or \( \cos (A-B)= \cos A \cos B + \sin A \sin B\)

\( \cos (A+B)= \cos A \cos B- \sin A \sin B\)

Proof
Let A be a point with coordinate \( (r_1 \cos A, r_1 \sin A)\) which makes an angle A with X-axis and whose length is \( r_1\) from origin O.
Therefore,
\( \vec{OA}=( r_1 \cos A,r_1 \sin A )\) and \( | \vec{OA} |=r_1 \)
Also let B be another point with coordinate \( (r_2 \cos B, -r_2 \sin B)\) which makes an angle B with X-axis and whose length is \( r_2\) from origin O.
Therefore,
\( \vec{OB}=( r_2 \cos B,-r_2 \sin B )\) and \( | \vec{OB} |=r_2 \)

Now, angle between OA and OB is (A+B).
Therefore,
\( \cos (A+B)=\frac{\vec{OA}.\vec{OB}}{| \vec{OA} |.| \vec{OB} |}\)
or\( \cos (A+B)=\frac{( r_1 \cos A,r_1 \sin A ).( r_2 \cos B,-r_2 \sin B )}{| ( r_1 \cos A,r_1 \sin A ) |.| ( r_2 \cos B,-r_2 \sin B ) |}\)
or \( \cos (A+B)= \cos A \cos B - \sin A \sin B\)

Solution of triangles

A triangle has six parts : three angles and three sides. These are called the elements of Triangle.

If any three out of six elements, if at least one side are given them the remaining three elements can be determined by the use of trigonometric functions and their tables. This process of finding the elements of triangle is called the solution of the triangle.

Therefore, solution of triangles is the trigonometric problem of finding the characteristics of a triangle (angles and sides), when some of these are known.

Sine Law

In a triangle ABC, prove that \( \frac{a} {\sin A} =\frac{b}{\sin B}=\frac{c}{\sin C}\)

Proof
Let ABC be a triangle in which the length of the sides opposite to A, B, C are a ,b , and c respectively.

Here,
\( \vec{a}+\vec{b}+\vec{c}=0\)
or \( \vec{a}\times ( \vec{a}+\vec{b}+\vec{c} )=0\)
or \( \vec{a}\times \vec{b}=\vec{c}\times \vec{a}\) (1)
Similarly, we get
\( \vec{b}\times \vec{c}=\vec{c}\times \vec{a} \) (2)
Hence, from (1) and (2) we have
\( \vec{a}\times \vec{b}=\vec{b}\times \vec{c}=\vec{c}\times \vec{a} \)
Using the formula, we get
\( a b \sin C=bc \sin A=ca \sin B \)
or \( \frac{a} {\sin A} =\frac{b}{\sin B}=\frac{c}{\sin C}\)

Cosine law

In a triangle ABC , prove that \( a^2 =b^2 +c^2 -2bc\cos A\)

Let ABC be a triangle in which the length of the sides opposite to A, B, C are a ,b, and c respectively.

Here,
\( \vec{BC}+\vec{CA}+\vec{AB}=0\)
or \( \vec{a}+\vec{b}+\vec{c}=0\)
or \( \vec{a}=-\vec{b}-\vec{c}\)
Squaring both side, we get
\( \vec{a} ^ 2={{( -\vec{b}-\vec{c} )}^2}\)
or \( \vec{a} ^ 2=( \vec{b}+\vec{c} )^2\)
or \( a^2 =\vec{b} ^ 2+2\vec{b}\vec{c}+\vec{c} ^ 2\)
or \( a^2 =b^2 +c^2 +2\vec{b}\vec{c}\)
Using the formula , we get
\( a^2 =b^2 +c^2 +2( -bc\cos A )\)
or \( a^2 =b^2 +c^2 -2bc\cos A\)
Similarly, we can prove that

  1. In a triangle ABC, prove that \( c^2 =a^2 +b^2 -2ab\cos C\)
  2. In a triangle ABC, prove that \( b^2 =c^2 +a^2 -2ca\cos B\)

Projection Law

In a triangle ABC, prove that \( a=b\cos C+c\cos B\)

Let ABC be a triangle in which the length of the sides opposite to A, B, C are a ,b , and c respectively.

Here,
\( \vec{BC}+\vec{CA}+\vec{AB}=0\)
or \( \vec{a}+\vec{b}+\vec{c}=0\)
or \( \vec{a}=-\vec{b}-\vec{c}\)
Taking dot product on both side by \( \vec{a}\) , we get
\( \vec{a}.\vec{a}=-\vec{b}.\vec{a}-\vec{c}.\vec{a}\)
Using the formula , we get
\( a^2 =-( -ba\cos C )-( -ca\cos B )\)
or \( a^2 =ba\cos C+ca\cos B\)
or \( a=b\cos C+c\cos B\)
Similarly, we can prove that

  1. In a triangle ABC, prove that \( b=c\cos A+a\cos C\)
  2. In a triangle ABC, prove that \( c=a\cos B+b\cos A\)

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