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Definition, Theorem, Proof, Solved Example Problems, Solution - Scalar Product and Vector Product | 12th Mathematics : UNIT 6 : Applications of Vector Algebra

Chapter: 12th Mathematics : UNIT 6 : Applications of Vector Algebra

Scalar Product and Vector Product

Next we recall the scalar product and vector product of two vectors as follows.

Scalar Product and Vector Product

Next we recall the scalar product and vector product of two vectors as follows.

 

Definition 6.1

Given two vectors  = a1iˆ + a2ˆa3kˆ and  b1iˆ + b2ˆb3kˆ the scalar product (or dot product) is denoted by  ×  and is calculated by

 ×  a1b1 + a2b2 + a3b3 ,

and the vector product (or cross product) is denoted by  ´  , and is calculated by


Note

 ×  is a vector.

 ´  is a scalar, and

 

1. Geometrical interpretation

Geometrically, if  is an arbitrary vector and nˆ is a unit vector, then  nˆ  is the projection of the vector  on the straight line on which nˆ  lies. The quantity   nˆ   is positive if the angle between  and nˆ is acute, see Fig. 6.4 and negative if the angle between  and nˆ is obtuse see Fig. 6.5.


If  and  are arbitrary non-zero vectors, then |  ×  | =  and so  ×  | means either the length of the straight line segment obtained by projecting the vector |  |  along  the direction of   or the length of the line segment obtained by projecting  the vector |  |  along the direction of . We recall that  ×  =|  | |  | cosθ , where θ is the angle between the two vectors  and  . We  recall that the angle between    and   is defined as the measure from  to   in the counter clockwise direction.

The vector  ´  is either  or a vector perpendicular to the plane parallel to both  and  having magnitude as the area of the parallelogram formed by coterminus vectors parallel to  and . If  and  are non-zero vectors, then the magnitude of  ´ b can be calculated by the formula

 ´  | = | a | | b | | sinθ |, where θ is the angle between  and .

Two vectors are said to be coterminus if they have same initial point.

Remark

(1) An angle between two non-zero vectors  and  is found by the following formula 


(2)  and  are said to be parallel if the angle between them is 0 or π .

(3)  and  are said to be perpendicular if the angle between them is 

Property

(1) Let  and  be any two nonzero vectors. Then

 ⋅  = 0 if and only if  and  are perpendicular to each other.

×   if and only if  and  are parallel to each other.

(2) If  , and are any three vectors and α is a scalar, then 


 

2. Application of dot and cross products in plane Trigonometry

We apply the concepts of dot and cross products of two vectors to derive a few formulae in plane trigonometry.

 

Example 6.1 (Cosine formulae)

With usual notations, in any triangle ABC, prove the following by vector method.

(i) a2 = b2 + c2 - 2bc cos A 

(ii) b2 = c2 + a2 - 2ca cos B

(iii) c2 = a2 + b2 - 2ab cos C

Solution

With usual notations in triangle ABC, we have 


a2 = b2 + c2 + 2bc cos(π - A)

a2 = b2 + c2 - 2bc cos A .

The results in (ii) and (iii) are proved in a similar way.

 

Example 6.2

With usual notations, in any triangle ABC, prove the following by vector method.

(i) cos cos B 

(i) cos cos C

(iii) cos cos A

Solution

With usual notations in triangle ABC, we have = a,  =  , and


 a2 = ab cos ac cos B

Therefore cos cos . The results in (ii) and (iii) are proved in a similar way.

 

Example 6.3

By vector method, prove that cos(α β ) = cosα cos β − sin α sin β .

Solution

Let  aˆ   =  and bˆ =  be the unit vectors and which make angles  α  and  β , respectively, with positive  -axis, where  A  and  B  are as in the Fig. 6.8. Draw  AL  and  BM  perpendicular to  the x-axis.

On the other hand, from (1) and (2)

aˆ   bˆ   = (cosαiˆ − sin α ˆj (cos β iˆ + sin βˆj) = cosα cos β − sin α sin β            ........(4)

From (3) and (4),

we get cos(α β ) = cosα cos β − sin α sin β.

 

Example 6.4

With usual notations, in any triangle ABC, prove by vector method that 

Solution

With usual notations in triangle ABC, we have 



Example 6.5

Prove by vector method that sin(α − β ) = sin α cos β − cosα sin β .

Solution 

Let  aˆ =  and  =   be the unit vectors making  angles α  and  β  respectively, with positive  -axis,  where A  and  B  are  as  shown  in  the  Fig.  6.10. Then,   we get aˆ = cosαiˆ + sin α ˆand bˆ = cos β iˆ + sin β ˆ,


The angle between aˆ and bˆ is α − β and, the vectors bˆ, aˆ, kˆ  form a right-handed system.

Hence, we get

aˆ  ×  bˆ | bˆ | | aˆ | sin(α - β )kˆ = sin(α - β )kˆ                                     ………………(1)

On the other hand,

 = (sin α cos β - cosα sin β )kˆ              ... (2)

Hence, equations (1) and (2), leads to

sin(α - β ) = sin α cos β - cosα sin β .

 

3. Application of dot and cross products in Geometry

Example 6.6 (Apollonius's theorem)

If is the midpoint of the side BC of a triangle ABC, show by vector method that


Solution

Let be the origin,  be the position vector of and  be the position vector of C. Now is the midpoint of BC , and so the, position vector of D is 

Therefore , we have  


 

Example 6.7

Prove by vector method that the perpendiculars (attitudes) from the vertices to the opposite sides of a triangle are concurrent.

Solution

Consider a triangle ABC in which the two altitudes AD and BE intersect at . Let CO be produced to meet AB at F. We take O  as the origin and let  


Since  is perpendicular to , we have  is perpendicular to , and hence we get  = 0 . That is, . ( -  ) = 0 , which means


Similarly, since  is perpendicular to , we have  is perpendicular to , and hence we get


Adding equations (1) and (2), gives    −    = 0 . That is,  ⋅ ( −  ) = 0 .

That is,    = 0 . Therefore,  is perpendicular to  which implies that  is perpendicular to . Hence, the perpendicular drawn from to the side AB passes through . Thus, the altitudes are concurrent.

 

Example 6.8

In triangle ABC, the points DEare the midpoints of the sides BCCA , and AB respectively.  Using vector method, show that the area of ΔDEF is equal to 1/4 (area of ΔABC).

Solution

In triangle ABC, coonsider as the origin. Then the position vectors of DEare given by 


respectively. Since |  × AC | is the area of the parallelogram formed by the two vectors  as adjacent sides, the area of ΔABC is 1/ 2 |  |  Similarly, considering ∆DEF , we have


 

4. Application of dot and cross product in Physics 

 

Definition 6.2

If  is the displacement vector of a particle moved from a point to another point after applying a constant force  on the particle, then the work done by the force on the particle is  ⋅ .


If the force has an acute angle, perpendicular angle, and an obtuse angle, the work done by the force is positive, zero, and negative respectively.

 

Example 6.9

A particle acted upon by constant forces 2iˆ + 5 ˆ+ 6kˆ and −iˆ − 2 ˆ− kˆ is displaced from the point (4, −3, −2) to the point (6,1, −3) . Find the total work done by the forces.

Solution

Resultant of the given forces is  = (2iˆ + 5 ˆ+ 6kˆ) + (−iˆ − 2 ˆ− kˆ) = iˆ + 3 ˆ+ 5kˆ. 

Let and be the points (4, −3, −2) and (6,1, −3) respectively. Then the displacement vector of the particle is    −  = (6iˆ + ˆ− 3kˆ) − (4iˆ − 3 ˆ− 2kˆ) = 2iˆ + 4 ˆ− kˆ .

Therefore the work done  ⋅  = (iˆ + 3 ˆ+ 5kˆ) (2iˆ + 4 ˆ− kˆ) = 9 units.

 

Example 6.10

A particle is acted upon by the forces 3iˆ − 2 ˆ+ 2kˆ and 2iˆ + ˆ− kˆ is displaced from the point (1, 3, −1) to the point (4, −1, λ) . If the work done by the forces is 16 units, find the value of λ .

Solution

Resultant of the given forces is  = (3iˆ − 2 ˆ+ 2kˆ) + (2iˆ + ˆ− kˆ) = 5iˆ − ˆkˆ .

The displacement of the particle is given by

 = (4iˆ − ˆλkˆ) − (iˆ + 3 ˆ− kˆ) = (3iˆ − 4 ˆ+ (λ +1)kˆ) .

As the work done by the forces is 16 units, we have

 ⋅  = 16 . 

That is, (5iˆ − ˆkˆ)  (3iˆ − 4 ˆ+ (λ +1)kˆ = 16  λ + 20 = 16 .

So, λ = −4

 

Definition 6.3

If a force  is applied on a particle at a point with position vector , then the torque or moment on the particle is given by  ×  . The torque is also called the rotational force.


 

Example 6.11

Find the magnitude and the direction cosines of the torque about the point (2, 0, −1) of a force  2iˆ + ˆ− kˆ, whose line of action passes through the origin.

Solution

Let be the point (2, 0, −1) . Then the position vector of is  = 2iˆ − ˆand therefore  =  = −2ˆ+ ˆ.

Then the given force is  = 2iˆ + ˆ− kˆ . So, the torque is


The magnitude of the torque = | -iˆ - 2kˆ |= √5 and the direction cosines of the torque are -1/√5 , 0, -2/√5. 


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12th Mathematics : UNIT 6 : Applications of Vector Algebra : Scalar Product and Vector Product | Definition, Theorem, Proof, Solved Example Problems, Solution

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12th Mathematics : UNIT 6 : Applications of Vector Algebra


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