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 = *a*1*i*Ë† + *a*2Ë†*j *+ *a*3*k*Ë† and * *= *b*1*i*Ë† + *b*2Ë†*j *+ *b*3*k*Ë† the scalar product (or dot product) is denoted by Ã— * *and is calculated by

* *Ã— * *= *a*1*b*1 + *a*2*b*2 + *a*3*b*3 ,

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

Note

* *Ã— is a vector.

Â´ is a scalar, and

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

If and are arbitrary non-zero vectors, then | Ã— | = and so | Ã— | 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 Ã— =| | | | cosÎ¸ , where Î¸ 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 Â´ 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 Â´ b can be calculated by the formula

| Â´ | = | a | | b | | sinÎ¸ |, where Î¸ 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 *Ï€ *.

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

Property

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

â‹… * *= 0 if and only if * *and * *are perpendicular to each other.

Ã— * *= if and only if * *and * *are parallel to each other.

(2) If , * *, and *c *are any three vectors and *Î± *is a scalar, then

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) *a*2 = *b*2 + *c*2 - 2*bc *cos *A*

(ii) *b*2 = *c*2 + *a*2 - 2*ca *cos *B*

(iii) *c*2 = *a*2 + *b*2 - 2*ab *cos *C*

Solution

With usual notations in triangle *ABC*, we have

*a*2 = b2 + c2 + 2bc cos(Ï€ - A)

*a*2 = 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) *a *= *b *cos *C *+ *c *cos *B*

(i) *b *= *c *cos *A *+ *a *cos *C*

(iii) *c *= *a *cos *B *+ *b *cos *A*

Solution

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

â‡’ *a*2 = *ab *cos *C *+ *ac *cos *B*

Therefore *a *= *b *cos *C *+ *c *cos *B *. The results in (ii) and (iii) are proved in a similar way.

By vector method, prove that cos(*Î± *+ *Î² *) = cos*Î± *cos *Î² *âˆ’ sin *Î± *sin *Î² *.

Let *a*Ë† = * *and *b*Ë† = * *be the unit vectors and which make angles *Î± *and *Î² *, respectively, with positive *x *-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)

*a*Ë† â‹… *b*Ë† = (cos*Î±i*Ë† âˆ’ sin *Î± *Ë†*j*) â‹… (cos *Î² i*Ë† + sin *Î²*Ë†*j*) = cos*Î± *cos *Î² *âˆ’ sin *Î± *sin *Î²** ........*(4)

From (3) and (4),

we get cos(*Î± *+ *Î² *) = cos*Î± *cos *Î² *âˆ’ sin *Î± *sin *Î²*.

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(*Î± *âˆ’ *Î² *) = sin *Î± *cos *Î² *âˆ’ cos*Î± *sin *Î² *.

Solution

Let *a*Ë† = and = be the unit vectors making angles *Î± *and *Î² *respectively, with positive *x *-axis, where *A *and *B *are as shown in the Fig. 6.10. Then, we get *a*Ë† = cos*Î±i*Ë† + sin *Î± *Ë†*j *and *b*Ë† = cos *Î² i*Ë† + sin *Î² *Ë†*j *,

The angle between *a*Ë† and *b*Ë† is *Î± *âˆ’ *Î² *and, the vectors *b*Ë†, *a*Ë†, *k*Ë† form a right-handed system.

Hence, we get

aË† Ã— bË† | bË† | | aË† | sin(Î± - Î² )kË† = sin(Î± - Î² )kË† â€¦â€¦â€¦â€¦â€¦â€¦(1)

On the other hand,

= (sin Î± cos Î² - cosÎ± sin Î² )kË† ... (2)

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

sin(Î± - Î² ) = sin Î± cos Î² - cosÎ± sin Î² .

Example 6.6 (Apollonius's theorem)

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

Solution

Let *A *be the origin, * *be the position vector of *B *and * *be the position vector of *C*. Now *D *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 *O *. 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 * *â‹… * *âˆ’ * *â‹… * *= 0 . That is, â‹… (* *âˆ’ * *) = 0 .

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

In triangle *ABC*, the points *D*, *E*, *F *are the midpoints of the sides *BC*, *CA *, and *AB *respectively. Using vector method, show that the area of Î”DEF is equal to 1/4 (area of Î”ABC).

In triangle *ABC*, coonsider *A *as the origin. Then the position vectors of *D*, *E*, *F *are given by

respectively. Since | Ã— *AC *| is the area of the parallelogram formed by the two vectors , * *as adjacent sides, the area of Î”*ABC *is 1/ 2 | | Similarly, considering âˆ†DEF , we have

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 *w *= * *â‹… .

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 2*i*Ë† + 5 Ë†*j *+ 6*k*Ë† and âˆ’*i*Ë† âˆ’ 2 Ë†*j *âˆ’ *k*Ë† is displaced from the point (4, âˆ’3, âˆ’2) to the point (6,1, âˆ’3) . Find the total work done by the forces.

Solution

Resultant of the given forces is * *= (2*i*Ë† + 5 Ë†*j *+ 6*k*Ë†) + (âˆ’*i*Ë† âˆ’ 2 Ë†*j *âˆ’ *k*Ë†) = *i*Ë† + 3 Ë†*j *+ 5*k*Ë†.

Let *A *and *B *be the points (4, âˆ’3, âˆ’2) and (6,1, âˆ’3) respectively. Then the displacement vector of the particle is * *= * *= * *âˆ’ * *= (6*i*Ë† + Ë†*j *âˆ’ 3*k*Ë†) âˆ’ (4*i*Ë† âˆ’ 3 Ë†*j *âˆ’ 2*k*Ë†) = 2*i*Ë† + 4 Ë†*j *âˆ’ *k*Ë† .

Therefore the work done *w *= * *â‹… * *= (*i*Ë† + 3 Ë†*j *+ 5*k*Ë†) â‹…(2*i*Ë† + 4 Ë†*j *âˆ’ *k*Ë†) = 9 units.

A particle is acted upon by the forces 3*i*Ë† âˆ’ 2 Ë†*j *+ 2*k*Ë† and 2*i*Ë† + Ë†*j *âˆ’ *k*Ë† is displaced from the point (1, 3, âˆ’1) to the point (4, âˆ’1, *Î»*) . If the work done by the forces is 16 units, find the value of *Î» *.

Resultant of the given forces is = (3*i*Ë† âˆ’ 2 Ë†*j *+ 2*k*Ë†) + (2*i*Ë† + Ë†*j *âˆ’ *k*Ë†) = 5*i*Ë† âˆ’ Ë†*j *+ *k*Ë† .

The displacement of the particle is given by

= (4*i*Ë† âˆ’ Ë†*j *+ *Î»k*Ë†) âˆ’ (*i*Ë† + 3 Ë†*j *âˆ’ *k*Ë†) = (3*i*Ë† âˆ’ 4 Ë†*j *+ (*Î» *+1)*k*Ë†) .

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

* *â‹… = 16 .

That is, (5*i*Ë† âˆ’ Ë†*j *+ *k*Ë†) â‹… (3*i*Ë† âˆ’ 4 Ë†*j *+ (*Î» *+1)*k*Ë† = 16 â‡’ *Î» *+ 20 = 16 .

So, *Î» *= âˆ’4

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 = * *Ã— * *. 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, âˆ’1) of a force 2*i*Ë† + Ë†*j *âˆ’ *k*Ë†, whose line of action passes through the origin.

Let *A *be the point (2, 0, âˆ’1) . Then the position vector of *A *is * *= 2*i*Ë†* *âˆ’ Ë†*k *and therefore = * *= âˆ’2Ë†*i *+ Ë†*k *.

Then the given force is = 2*i*Ë† + Ë†*j *âˆ’ *k*Ë† . So, the torque is

The magnitude of the torque = | -*i*Ë† - 2*k*Ë† |= âˆš5 and the direction cosines of the torque are -1/âˆš5 , 0, -2/âˆš5.

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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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