## 1. Concepts of Vectors and scalars ,magnitude and direction of a vector

There are two types of physical quantities vector quantity and scalar quantity

Scalar quantity

The quantities which are the only magnitude are called scalar quantities for example time

Vector quantity

The physical quantities which have both direction and magnitude are called vector quantities for example velocity

How can we find the magnitude of a vector

Let A = 2i+3j+4k

Magnitude of vector = ✓2²+3²+4²

= ✓ 4+9+16 = ✓ 29

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Course : CBSE Class 12

Start Date : 26.06.2022

End Date : 31.12.2022

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 Subject M T W T F S S Mathmatics(10 hours) 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM Physics(10 hours) 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM

## 1. Concepts of Vectors and scalars ,magnitude and direction of a vector

Chapter 10

Concepts of Vectors and scalars ,magnitude and direction of a vector :

The Physical quantity is either a vector or a scalar. These two categories can be distinguished from one another by their distinct definitions:

Scalars are Physical quantities that has magnitude but no particular direction is described as scalar.

Vectors are Physical quantities that are fully described by both a magnitude and a direction and also satisfy the triangle law of vector addition.

Examples of Scalar Quantities

Some examples of scalar include:

• Mass
• Speed
• Distance
• Time
• Area
• Volume
• Density
• Temperature

### Examples of Vector Quantities

Examples of vector quantity include:

• Linear momentum
• Acceleration
• Displacement
• Momentum
• Angular velocity
• Force
• Electric field
• Polarization

Initial Points – The point A where from the vector starts is known as initial point.

Terminal Point – The point B, where it ends is said to be the terminal point.

Magnitude The distance between initial point and terminal point of a vector is the

Position Vector – Consider a point p (x, y, z) in space. The vector with initial point, origin O and terminal point P, is called the position vector of P.

## 1. Direction cosines and direction ratios of a line joining two points.

Chapter 11

Three-Dimensional Geometry

Direction Cosines

In three-dimensional geometry, we have three axes: namely, the x, y, and z-axis. Let us assume a line OP passes through the origin in the three-dimensional space. Then, the line  will make an angle each with the x-axis, y-axis, and z-axis respectively.

The cosines of each of these angles  that the line makes with the x-axis, y-axis, and z-axis respectively are called direction cosines of the line in three-dimensional geometry. Normally, it is tradition to denote these direction cosines using the letters l, m, n respectively.

Note that these cosines can be found only once we have found the angles that the line makes with each of the axes. Also, it is interesting to note that if we reverse the direction of this line, the angles will obviously change.

Consequently, the direction cosines i.e. the cosines of these angles will also be different once the direction of the line is reversed. We will now look at a slightly different situation where our line does not pass through the origin (0,0,0).

1.Direction cosines and direction ratios of a line joining two points:

Let a line AB in 3D space make angles α, β, γ respectively with the +ve direction of coordinate axes X, Y, Z. Therefore, we express cosα, cosβ, cosγ as direction cosines of the line AB in the 3D space. Clearly; direction cosines fix the direction cosines of a line in space. Also, parallel lines have the same direction cosines. Direction cosines are denoted by l, m, n respectively. i.e. l = cosα, m = cosβ and n = cosγ. The direction of a line cannot be fixed in space by knowing anyone or any two angles.

If a line in space makes angles α, β, γ respectively with the +ve direction of X, Y, Z axes then we can assume that the line will make angles π – α, π – β, π – γ with the -ve direction of axes. Therefore, the direction cosines can also be written as cos π -α, cos π -β, cos π –γ. Thus, direction cosines of the same line may also be taken as – cosα, –cosβ, –cosγ.

Three numbers a, b, c proportional to direction cosine l, m, n of a line in space. These numbers are called direction ratios OR direction numbers of the line. Therefore, the direction cosines of a line will be fixed but not the direction ratios:

Let direction cosine of any line in space be l, m, n and direction ratios(d.r.’s) are a:b:c.

Let P(x,y,z) be any point on the line and PM is perpendicular from P on X-axis. Then, in right triangle PMO,

We can see that cos⁡α=OMOP=x/r . In this case x = OM and r = OP or l=x/r

Similarly, we take the perpendicular point P on y and axis to obtain: m=y/r ; n=z/r

Hence, we conclude that

r2=x2+y2+z2=(lr)2+(mr)2+(nr)2

• r2=[l2+m2+n2] r2
• l2+m2+n2=1

Let AB be a line that is inclined at angles α, β, γ with positive x, y, z-axis at points A (x1, y1, z1) and B at (x2, y2, z2) to form direction ratios. Line AB makes angle γ with the z-axis which is a part of a right angle triangle ∆ARB where,

## Solved Examples

Example : Find the direction ratios and direction cosines of a line joining the points (3, -4, 6) and (5, 2, 5).

Solution:

Given points are A(3, -4, 6) and B(5, 2, 5)

The direction ratios of the line joining AB is

a = x– x1 = 5 – 3 = 2

b = y– y1 = 2 + 4 = 6

c = z– z1 = 5 – 6 = -1

AB =

So direction cosines of the line = 2/√41, 6/√41, -1/√41.

Example : Find the direction cosines of the line joining the points (2,1,2) and (4,2,0).

Solution:

Let the points are A(2,1,2) and B(4,2,0).

x2-x1 = 4-2 = 2

y2-y1 = 2-1 = 1

z2-z1 = 0-2 = -2

AB = √(22+12+(-2)2)= 3

Hence the direction cosines are ⅔, ⅓, -⅔.

Example : Find the direction cosines of the line joining the points (2,3,-1) and (3,-2,1).

Solution:

Let the points are A(2,3,-1) and B(3,-2,1).

x2-x1 = 3-2 = 1

y2-y1 = -2-3 = -5

z2-z1 = 1- (-1) = 2

AB = √(12+(-5)2+22)= √30

Hence the direction cosines are 1/√30, -5/√30, 2/√30.

## 2. Direction cosines and direction ratios of a vector

Three dimensional Geometry

i ,j and k 3 unit vectors in direction of x axis y axis and z axis

Direction cosine give the angle with which align makes an angle with the axis formula for the direction cosine is

l = a cos ¢

m = b cos ¢

n = c cos ¢

Where a b and c are the direction ratios

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## Batch List

#### On line class 12

Course : CBSE Class 12

Start Date : 26.06.2022

End Date : 31.12.2022

Types of Batch : Classroom

 Subject M T W T F S S Mathmatics(10 hours) 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM Physics(10 hours) 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM 4:45 PM

## 2. Direction cosines and direction ratios of a vector

Direction cosines and direction ratios of a vector

Consider the position vector of a point P(x, y, z) The angles α, β, γ made by the vector  with the positive directions of x, y and z-axes respectively,are called its direction angles. The cosine values of these angles, i.e., cos α, cos β and cos γ are called direction cosines of the vector , and usually denoted by l, m and n, respectively.

i.e. l = cosα, m = cosβ and n = cosγ. The direction of a line cannot be fixed in space by knowing anyone or any two angles.

one may note that the triangle OAP is right angled, and in it, we have l = cosα=x/r, Similarly, from the right angled triangles OBP and OCP, we may write  m = cosβ=y/r and

• n = cosγ=z/r
• x=lr,y=mr,z=nr.

Thus, the coordinates of the point P may also be expressed as (lr, mr,nr). The numbers lr, mr and nr, proportional to the direction cosines are called as direction ratios of vector , and denoted as a, b and c, respectively. Where r  denotes the magnitude of the vector and it is given by,

Example : Find the direction ratios and direction cosines of a line joining the points (3, -4, 6) and (5, 2, 5).

Solution:

Given points are A(3, -4, 6) and B(5, 2, 5)

The direction ratios of the line joining AB is

a = x– x1 = 5 – 3 = 2

b = y– y1 = 2 + 4 = 6

c = z– z1 = 5 – 6 = -1

AB =

So direction cosines of the line = 2/√41, 6/√41, -1/√41.

Example : Find the direction cosines of the line joining the points (2,1,2) and (4,2,0).

Solution:

Let the points are A(2,1,2) and B(4,2,0).

x2-x1 = 4-2 = 2

y2-y1 = 2-1 = 1

z2-z1 = 0-2 = -2

AB = √(22+12+(-2)2)= 3

Hence the direction cosines are ⅔, ⅓, -⅔.

Example : Find the direction cosines of the line joining the points (2,3,-1) and (3,-2,1).

Solution:

Let the points are A(2,3,-1) and B(3,-2,1).

x2-x1 = 3-2 = 1

y2-y1 = -2-3 = -5

z2-z1 = 1- (-1) = 2

AB = √(12+(-5)2+22)= √30

Hence the direction cosines are 1/√30, -5/√30, 2/√30.

## 4. Algebra of Vectors( Addition and Multiplication of Vectors)

Associative property:

[1] Multiplication of Vector by a Scalar

[2] Components of Vector

[3] Vector joining two points

Let P1(x1, y1, z1) and P2(x2, y2 z2) be the two points. Then vector joining the

points P1 and P2 is P1P2 . Join P1, P2 with O. Now OP2 = OP1 + P1P2   (by triangle law)

## 2. Concepts of Lines and Cartesian equation and vector equation of a line

Concepts of Lines and Cartesian equation and vector equation of a line:

It is known that we can uniquely determine a line if:

1. It passes through a particular point in a specific direction, or
2. It passes through two unique points

## Equation of a Line passing through a point and parallel to a vector:

Let us consider that the position vector of the given point be     with respect to the origin.

The line passing through point A is given by l and it is parallel to the vector

## as shown below. Let us choose any random point R on the line l and its position vector with respect to origin of the rectangular co-ordinate system is given by

It is vector equation of line passing through a point and parallel to a vector.

Cartesian form:

If the three-dimensional co-ordinates of the point ‘A’ are given as (x1, y1, z1) and the direction cosines of this point is given as a, b, c then considering the rectangular co-ordinates of point P as (x, y, z):

Let

It is cartesian equation of line passing through a point and parallel to a vector.

Equation of a Line passing through two unique given point:

Let us consider that the position vector of the given two point be      and  with respect to the origin.

Let us choose any random point P on the line and its position vector with respect to origin of the rectangular co-ordinate system is given by  .

Point P lies on the line AB if and only if the vectors

It is vector equation of line passing through two point .

Cartesian Form:

It is the Cartesian equation of a line. passing through two given points.

Solution:

# Write vector and the cartesian equations of the lines that passes through the origin and (5,−2,3)

Solution:

The line passing through (0,0,0) and (5,−2,3)

Example:

Find the cartesian equation of the line which passes through the points (7,4,6) and (9,1,8).

Solution:

Here , (x1,y1,z1) = (7,4,6) and (x2,y2,z2) = (9,1,8)

Direction ratios of the line are x2x1 = 2, y2y1 = -3 , z2z1​​=2

Equation of line :

## 3. Angle between two lines,coplanar and skew lines

Angle Between Two lines

Cartesian Form:

be two lines then angle between two lines is

Note:- If a₁,b₁,c₁, and a₂,b₂,c₂, are the direction ratios of two lines then

i) the lines are parallel if

ii) the lines are perpendicular if

Skew lines: Skew lines are lines in space which are neither parallel nor

intersecting. They lie in different planes.

Angle between two skew lines: Angle between skew lines is the angle between

two intersecting lines drawn from any point (preferably through the origin)

parallel to each of the skew lines.

Example: Find the angle between the pair of lines given by:

(x + 3)/3 = (y – 1)/5 = (z + 3)/4 and (x + 1)/1 = (y – 4)/1 = (z – 5)/2

Answer: We can solve this problem by finding the cosine of the angle between the two lines and then taking an inverse of the cosine. Let q be the angle between the two lines. We know from the formula that:

# Coplanarity of Two Lines:

Coplanar lines are the lines that lie on the same plane. Prove that two lines are coplanar using the condition in vector form and Cartesian form.

### Condition for Coplanarity in Vector Form

In vector form, let us consider the equations of two straight lines to be as under:

What do these equations mean? It means that the first line passes through a point, say L, whose position vector is given by  and is parallel to . Similarly, the second line is said to pass through another point whose position vector is given by  and is parallel to

The condition for coplanarity is that the line joining the two points must be perpendicular to the product of the two vectors, m1 and m2. To illustrate this, we know that the line joining the two said points can be written in vector form as . So, we have:

### Condition for Coplanarity in Cartesian Form

The derivation of the condition for coplanarity in Cartesian form stems from the vector form. Let us consider two points L (x1, y1, z1) & M (x2, y2, z2) in the Cartesian plane. Let there be two vectors  and . Their direction ratios are given by a1, b1, c1 and a2, b2, crespectively.

The vector equation of the line joining L and M can be given as:

Now use the above condition in vector form to derive our condition in Cartesian form.

Question : Are the lines (x + 3)/3 = (y – 1)/1 = (z – 5)/5 and (x + 1)/ -1 = (y – 2)/2 = (z – 5)/5 coplanar?

Answer: Comparing the equations with the general form, we have:

(x1, y1, z1) = (-3, 1, 5) and (x2, y2, z2) = (-1, 2, 5).

Note that a1, b1, c= -3, 1, 5 and a2, b2, c2 = -1, 2, 5.

So, by Cartesian form, we must solve the matrix:

= 2(5 – 10) – 1(-15 + 5) + 0(-6 + 1) = -10 + 10 = 0

Since the solution of the matrix gives us 0, we can say that the given lines are coplanar

This can be used as is for calculation purposes. By the condition above, the two lines would be coplanar if  LM. (mx m2) = 0. Therefore, in Cartesian form, the matrix representing this equation is given as 0.

## 4. shortest distance between two lines

Shortest Distance between Two Lines:

1. If  lines are intersecting then the shortest distance is zero.
2. If  lines are skew lines , the line of the shortest distance will be perpendicular to both the lines.
3. If  lines are parallel  lines , the line of the shortest distance will be perpendicular distance.

Distance between two skew lines:

In  a space, there are lines which are neither intersecting nor parallel. In fact,such pair of lines are non coplanar and are called skew lines.

## Vector Form

We shall consider two skew lines, say l1 and l­2 and we are to calculate the distance between them. The equations of the lines are:

where d measures the distance or the length of the perpendicular.

Cartesian form:

The shortest distance between the lines

Distance between parallel lines:

If two lines l1 and l2 are parallel, then they are coplanar.

Let the lines be given by

Shortest distance between LINES

## 5. Concepts of Planes and Cartesian and vector equation of a plane

Concepts of Planes and Cartesian and vector equation of a plane

Plane:

A plane is determined uniquely if any one of the following is known:

(i) the normal to the plane and its distance from the origin is given, i.e., equation of

a plane in normal form.

(ii) it passes through a point and is perpendicular to a given direction.

(iii) it passes through three given non collinear points.

Equation of a plane in normal form.:

Consider a plane whose perpendicular distance from the origin is d (d ¹ 0).

Cartesian form

Example :Find the coordinates of the foot of the perpendicular drawn from the

origin to the plane 2x – 3y + 4z – 6 = 0.

Solution Let the coordinates of the foot of the perpendicular P from the origin to the

plane is (x1, y1,z1)

Then, the direction ratios of the line OP are

x1, y1, z1.

Since the direction ratios of the normal to the plane are 2, –3, 4;

Writing the equation of the plane in the normal

Note: If d is the distance from the origin and l, m, n are the direction cosines of the normal to the plane through the origin, then the foot of the perpendicular is (ld, md, nd).

Equation of a plane perpendicular to a given vector and passing through a given point:

Cartesian Form:

This is the cartesian equation of the plane.

Example : Find the vector and cartesian equations of the plane which passes through the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1.

which is the Cartesian equation of the plane.

Equation of a plane passing through three non collinear points:

Let R, S and T be three non collinear points on the plane with position vectors

Respectively.

The vectors RS and RT are in the given plane. Therefore, the vector

is perpendicular to the plane containing points R, S and T. Let   be the position vector

of any point P in the plane. Therefore, the equation of the plane passing through R and perpendicular to the vector

This is the equation of the plane in vector form passing through three non-collinear points.

Cartesian form

Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of the points R, S and T respectively. Let (x, y, z) be the coordinates of any point P on the plane with position

Intercept form of the equation of a plane:

Let the equation of the plane be

Ax + By + Cz + D = 0 (D ¹ 0) .................... (1)

Let the plane make intercepts a, b, c on x, y and z axes, respectively .

Hence, the plane meets x, y and z-axes at (a, 0, 0),(0, b, 0), (0, 0, c), respectively.

Plane passing through the intersection of two given planes:

Two planes can intersect in the three-dimensional space. Imagine two adjacent pages of a book. These two pages are nothing but an intersection of planes, intersecting each other and the line between them is called the line of intersection. A new plane i.e. a third plane can be given to be passing through this line of intersection of planes. We are to find out the equation of this plane.

Let us assume that the equation of the first plane is π1 and that of the second is π2. The equation of our required plane is π and we are to find out this equation itself. This equation is given by –

The position vector of any point on the line of intersection must

satisfy both the equations.

### Vector form

If the equation of the two planes is given in Vector form –

So the equation of the required plane by using (1) can be written as –

is equation of plane in vector form

Cartesian form:

is equation of plane in cartesian form.

If a point on the required plane is given, you must substitute the coordinates for the values of x, y and z to obtain the value of λ in order to then replace it in the above equation.

Question:

# The equation of the plane passing through the intersection of the planes x+y+z=6 and 2x+3y+4z+5=0, and the point (1,1,1) is

Solution:

Given equation of planes are

x+y+z−6+λ(2x+3y+4z+5)=0
Given, it also passes through (1,1,1)

1+1+1+−6+λ(2+3+4+5)=0λ=143​
Hence, required plane is, 14(x+y+z−6)+3(2x+3y+4z+5)=0

20x+23y+26z−69=0

Question:

# The d.r's of the line of intersection of the planes x+y+z−1 =0 and 2x+3y+4z−7 =0 are

## Section Formula

To begin with, take a look at the figure given below:

As shown above, P and Q are two points represented by position vectors OP and OQ, respectively, with respect to origin O. We can divide the line segment joining the points P and Q by a third point R in two ways:

• Internally
• Externally

If we want to find the position vector OR for the point R with respect to the origin O, then we should take both the cases one by one.

### Case 1 – R Divides Segment PQ Internally

Take a look at Fig. 1 again. In this figure, if the point R divides such that,

where ‘m’ and ‘n’ are positive scalars, then we can say that R divides PQ internally in the ratio m:n. Now, from the triangles ORQ and OPR, we have

Therefore, replacing the values of and  in equation (1) above, we get

Hence, the position vector formula of the point R which divides PQ internally in the ratio m:n is,

### Case 2 – R Divides Segment PQ Externally

Look at the figure given below:

In Fig. 2, point R divides the segment PQ externally in the ratio m:n. Hence, we can say that point Q divides PR internally in the ratio: (m – n) : n. Therefore,

PQQR = (mn)/n

Now, by using equation (2), we have

### Note: If R is the Mid-Point of PQ

If R is the mid-point of PQ, then m = n. Therefore, from equation (2) above, we have

Therefore, r  = (b +a)/ 2. Hence, the position vector formula of mid-point R of PQ is,

OR= (b +a )/2 … (4)

Let’s look a solved example now:

Example 1

Consider two points P and Q with position vectors

Find the position vector formula of a point R which divides the line joining P and Q in the ratio 2:1, (i) internally, and (ii) externally.

Solution: Since point R divides PQ in the ratio 2:1. we have, m = 2 and n = 1

(i) R divides PQ internally

From equation (2), we have

Example:

Consider two points A and B with position vectors and

And

Find the position vector of a point C which divides the line joining A and B in the ratio 3 : 2,

(i) internally

(ii) externally.

Solution:

(i) The position vector of the point C dividing the join of A and B internally in the ratio 3 : 2 is:

Expanding the terms in the numerator,

(ii) The position vector of the point C dividing the join of A and B externally in the ratio 3 : 2 is:

Expanding the terms in the numerator,

Projection of one vector on a line:

If a vector  makes an angle θ with a given directed line l, in the anticlockwise direction, then the projection of  on l is a vector p  with magnitude |cosθ.

Also, the direction of p  is the same (or opposite) to that of the line l, depending upon whether cosθ is positive or negative. The vector p  is the projection vector and has magnitude |p |. It is also called the projection of vector   on the directed line l.In each of the figures shown above, the projection vector of  along the line l is the vector: .

The vector projection of one vector over another vector is the length of the shadow of the given vector over another vector. It is obtained by multiplying the magnitude of the given vectors with the cosecant of the angle between the two vectors. The resultant of a vector projection formula is a scalar value.

Problem:-

Find the projection of the vector î – ĵ  on the vector î + ĵ.

Let a  = ( î – ĵ) and b  = ( î + ĵ)

Now, the projection of vector a  on b  is given by,

(1)/ (I b I) (a . b ) = (1)/ (√1+1) ({1.1 + (-1) (1)})

= (1/√2) (1-1) =0

Hence, the projection of vector a   on b  is  0.

Question: Represent graphically a displacement of 40 km, 30° east of north.

Solution:

The vector represents the displacement of 40 km, 30o east of north.

Question:  Find the unit vector in the direction of vector , where P and Q are the points

(1, 2, 3) and (4, 5, 6), respectively

Solution:

We know that,

QUESTION: Find a vector in the direction of vector which has magnitude 8 units.

Solution:

Firstly,

## 6. Product of Two Vectors and Scalar (or dot) product of two vectors

Product of Two Vectors and Scalar (or dot) product of two vectors:

1. scalar product of vectors or dot product
2. vector product of vectors or cross product

## Dot Product of Vectors:

The scalar or dot product of two non-zero vectors a  and b , denoted by a .b  is

a .b  = |a | |b cosθ

where θ is the angle between a  and b  and 0 ≤ θ ≤ π as shown in the figure below.

It is important to note that if either a  = 0  or b  = 0 , then θ is not defined, and in this case

a .b  = 0

We can express the scalar product as:

where || and || represent the magnitude of the vectors  and  while cos θ denotes the cosine of the angle between both the vectors and .  indicate the dot product of the two vectors.

### Projection of Vectors:

BP is known to be the projection of a vector a on vector b in the direction of vector b given by |a| cos θ.

Similarly, the projection of vector b on a vector a in the direction of the vector a is given by |b| cos θ.

The projection of vector    on a vector

Question: Find the angle between two vectors a  and b  with magnitudes 3 and 2 ,
respectively having
a .b  = 6.

Solution: By definition of the scalar or dot product of vectors, we know that

a .b  = |a | |b cosθ

where θ is the angle between a  and b . In this question, we have,

|a | = 3, |b | = 2, and a .b  = 6

Replacing these values in the formula, we get

a .b  = |a | |b cosθ
6 = 3 x 2 x cosθ
3 x 232 x 2cosθ

Cancelling the common terms on both sides, we get

1 = 2 x cosθ
Or, cosθ = 12
Therefore, θ = π/4
Hence, the angle between the vectors a  and b  is π/4.

Dot Product Properties of Vector:

• Property 1: Dot product of two vectors is commutative i.e. a.b = b.a = ab cos θ.
• Property 2: If a.b = 0 then it can be clearly seen that either b or a is zero or cos θ = 0
•   q = π/2
• Property 3: Also we know that using scalar product of vectors (pa).(qb)=(pb).(qa)=pq (a.b)
• Property 4: The dot product of a vector to itself is the magnitude squared of the vector i.e. a.a = a.a  cos 0 = a2
• Property 5: The dot product follows the distributive law also i.e. a.(b + c) = a.b + a.c
• Property 6: In terms of orthogonal coordinates for mutually perpendicular vectors it is seen that

Example : Let there be two vectors [6, 2, -1] and [5, -8, 2]. Find the dot product of the vectors.

Solution:

Given vectors: [6, 2, -1] and [5, -8, 2] be a and b respectively.

a.b = (6)(5) + (2)(-8) + (-1)(2)

a.b = 30 – 16 – 2

a.b = 12

Example 2: Let there be two vectors |a|=4 and |b|=2 and θ = 60°. Find their dot product.

Solution:

a.b = |a||b|cos θ

a.b = 4.2 cos 60°

a.b = 4.2 × (1/2)

a.b = 4

Example :

Find the angle between two vectors

And

Solution:

Given:

And

The formula to find the angle between two vectors is given by:

Hence,

= 1-1-1

= -1

Therefore,

Now, substituting the value in the formula, we get

Cos θ = -⅓

Hence, the angle between two vectors is θ = cos-1(-⅓).

Question:  Ifare such thatis perpendicular to, then find the value of λ.

Solution:

We know that the

# Vector (or Cross) Product of Two Vectors

Vectors can be multiplied in two ways, a scalar product where the result is a scalar and cross or vector product where is the result is a vector. In this article, we will look at the cross or vector product of two vectors.

## Explanation

We have already studied the three-dimensional right-handed rectangular coordinate system. As shown in the figure below, when the positive x-axis is rotated counter-clockwise into the positive y-axis, then a right-handed standard screw moves in the direction of the positive z-axis.

As can be seen above, in a three-dimensional right-handed rectangular coordinate system, the thumb of the right-hand points in the direction of the positive z-axis when the fingers are curled from the positive x-axis towards the positive y-axis.

## Definition

The cross or vector product of two non-zero vectors a  and b , is

a  x b  = |a | |b sinθn^

Where θ is the angle between a  and b , 0 ≤ θ ≤ π. Also, n^ is a unit vector perpendicular to both a  and b  such that a b , and n^ form a right-handed system as shown below.

As can be seen above, when the system is rotated from a  to b , it moves in the direction of n^. Also, if either a  = 0 or b  = 0, then θ is not defined and we can say,

a  x b  = 0

## Important Observations

• a  x b  is a vector.
• If a  and b  are two non-zero vectors, then a  x b  = 0, if and only if a  and b  are parallel (or collinear) to each other, i.e.

a  x b  = 0  a   b

Hence, a  x a  = 0 and a  x (−a)→ = 0. This is because in the first case θ = 0. Also, in the second case θ = π, giving the value of sinθ = 0.

• If θ = π2, then a  x b  = |a | |b |
• Considering observations 2 and 3 above, for mutually perpendicular vectors i j , and k , we have

• The angle between the two vectors a  and b  is,

sinθ = |a ×b ||a ||b |

• A cross or vector product is not commutative. We know this because a  x b  = b x a . Now, we know that,

a  x b  = |a | |b sinθn^.

Where a b , and n^ form a right-handed system. Or, θ is traversed from a  to b . On the other hand,

b  x a  = |b | |a sinθn1^.

Where b a , and n1^ form a right-handed system. Or, θ is traversed from b  to a . So, if a  and b  lie on a plane of paper, then n^ and n1^ are both perpendicular to the plane of the paper. However, n^ is directed above the paper and n1^ is directed below it. Or, n^ = – n1^. Hence,

a  x b  = |a | |b sinθn^ = – |a ||b |sinθ n1^

= – b  x a

• From the observations 4 and 6 above, we have

j x i = – k
k x j^ = – i
i x k = – j^

• If a  and b  represent the two sides of a triangle, then its area is |a  x b |. To understand this, look at the figure given below.

By the definition of the area of a triangle, we have area of ΔABC =  (AB).(CD). We know that, AB = |b | and CD = |a |sinθ. Therefore,

• If a  and b  represent the two adjacent sides of a parallelogram, then its area is |a  x b |. To understand this, look at the figure given below.

By the definition of the area of a parallelogram, we have area of parallelogram ABCD = (AB).(DE). We know that, AB = |b | and DE = |a |sinθ. Therefore,

Area of parallelogram ABCD = |a ||b |sinθ = |a  x b |

## Property: Distributivity of a cross or vector product over addition

If a b , and c  are any three vectors and λ is a scalar, then

• a  x (b  + c ) = a  x b  + a  x c
• λ(a  x b ) = (λa ) x b  = a  x (λb )

Question 1: Find the area of the parallelogram whose adjacent sides are determined by the following vectors,

• a  = i^ – j^ + 3k  and
• b  = 2i^ – 7j^ + k .

Answer : We know that if a  and b  represent the two adjacent sides of a parallelogram, then its area is |a  x b |. Also,

Substituting the values of a1,a2,a3,b1,b2,and b3, we get

Solving the determinant, we get

• a  x b  = {[(-1) x 1)] – [(-7) x 3]} – {[1 x 1)] – [2 x 3]} + {[1 x (-7))] – [2 x (-1)]}
= 20
i^ + 5j^ – 5k^.

Also, the magnitude of a  x b  is,

• |a  x b | = [20+5+(−5)]450 25×9×2 = 152.

Therefore, the area of the parallelogram is 152.

Question : Explain the characteristics of vector product?

Answer: The characteristics of vector product are as follows:

• Vector product two vectors always happen to be a vector.
• Vector product of two vectors happens to be noncommutative.
• Vector product is in accordance with the distributive law of multiplication.
• If a • b = 0 and a ≠ o, b ≠ o, then the two vectors shall be parallel to each other.

## 8. Scalar triple product and properties of scalar triple product

Scalar triple product and properties of scalar triple product

The scalar triple product of three vectors a, b, c is the scalar product of ector a with the

cross product of the vectors b and c, i.e., a · (b × c). Symbolically, it is also written as [a b c] = [a, b, c] = a · (b × c). The scalar triple product [a b c] gives the volume of a parallelepiped with adjacent sides a, b, and c. If we are given three vectors a, b, c, then their scalar triple products [a b c] are:

a · (b × c)

• a · (c × b)
• b · (a × c)
• b · (c × a)
• c · (b × a)
• c · (a × b)

Now, before moving to the formula of the scalar triple product, we need to note that:

• [a, b, c] = a · (b × c) = b · (c × a) = c · (a × b)
• a · (b × c) = - a · (c × b)
• b · (c × a) = - b · (a × c)
• c · (a × b) = - c · (b × a)
• a · (b × c) = (a × b) · c

the scalar triple product of vectors means the product of three vectors. It means taking the dot product of one of the vectors with the cross product of the remaining two. It is denoted as

[a b c ] = ( a × b) . c

The following conclusions can be drawn, by looking into the above formula:

i) The resultant is always a scalar quantity.

ii) Cross product of the vectors is calculated first, followed by the dot product which gives the scalar triple product.

iii) The physical significance of the scalar triple product formula represents the volume of the parallelepiped whose three coterminous edges represent the three

vectors a, b and c. The figure will make this point more clear.

According to this figure, the three vectors are represented by the coterminous edges as shown. The cross product of vectors a and b  gives the area of the base, and also, the direction of the cross product of vectors is perpendicular to both the vectors. As volume is the product of area and height, the height, in this case, is given by the component of vector c along the direction of the cross product of a and b. The component is given by c cos α.

Thus, we can conclude that for a Parallelepiped, if the coterminous edges are denoted by three vectors and a, b and c then,

Volume of parallelepiped = ( a × b) c cos α =  ( a × b) . c

Where α is the angle between  ( a × b)  and c.

We are familiar with the expansion of cross products of vectors. Keeping that in mind,

## Properties of Scalar Triple Product:

i) If the vectors are cyclically calculated, then

( a × b) . c = a.( b × c)

ii)  The product is cyclic in nature, i.e.,

a.(b × c) = b.(c × a) = c.(a × b)

Thus,

[ a b c ] = [ b c a ] = [ c a b ] = – [ b a c ] = – [ c b a ] = – [ a c b ]

Example : Evaluate the volume of a parallelepiped whose coterminous edges are i - j + k, 2i + 3j - k, and -i - j + 5k.

Solution: To determine the volume of the parallelepiped with edges i - j + k, 2i + 3j - k, and -i - j + 5k, we will determine its scalar triple product.

[ i - j + k, 2i + 3j - k, -i - j + 5k] =

= 1(15 - 1) + 1(10 - 1) + 1(-2 + 3)

= 14 + 9 + 1

= 24

Answer: The volume of the parallelepiped is 24 cubic units.

## 6. Angle between two planes a line and a plane

Angle between Two Planes:

The angle between normal to two planes is the angle between the two planes.

Cartesian form:

Let A1 x + By + C1z + D1 = 0 and A2x + B2y + C2 z + D2 = 0  be the equation of two planes aligned to each other at an angle θ where A1, B1, Cand A2, B2, C2 are the direction ratios of the normal to the planes, then the cosine of the angle between the two planes is given by:

Example: Determine the angle between the two planes whose vector equations are given as r.(2i + 2j - 3k) = 4 and r.(3i - 3j + 5k) = 3.

Solution: The equations of the planes are given in vector formNow to find the angle between the planes r.(2i + 2j - 3k) = 4 and r.(3i - 3j + 5k) = 3, we will use the formula cos θ = |(n1 . n2)|/(|n1|.|n2|). We have,

n1 = 2i + 2j - 3k, n2 = 3i - 3j + 5k

|n1| = √(22 + 22 + (-3)2) = √(4 + 4 + 9) = √17

|n2| = √(32 + (-3)2 + 52) = √(9 + 9 + 15) = √43

Scalar product of the normal vectors is given by, n1 . n2 = (2i + 2j - 3k) . (3i - 3j + 5k) = 2 × 3 + 2 × (-3) + (-3) × 5 = 6 - 6 - 15 = -15

Substituting these values into the formula, we have

cos θ = |(-15)|/(√17 . √43)

= 15/√731

θ = cos-1(15/√731)

Example: Find the angle between two planes with equations 2x + y - 2z = 5 and 3x - 6y - 2z = 7.

Solution: Since the equations of the two planes are given in the cartesian form, we will determine the angle between two planes in cartesian form using the formula cos θ = |(A1A2 + B1B2 + C1C2)|/[√(A12 + B12 + C12)√(A22 + B22 + C22)]. The equations of the planes are 2x + y - 2z = 5 and 3x - 6y - 2z = 7. Here, A1 = 2, B1 = 1, C1 = -2, A2 = 3, B2 = -6, C2 = -2. Substituting these values into the formula, we have

cos θ = (2×3 + 1×(-6) + (-2)×(-2))/[√(22 + 12 + (-2)2)√(32 + (-6)2 + (-2)2)]

= (6 + (-6) + 4)/[√(4 + 1 + 4)√(9 + 36 + 4)]

= 4/(√9 √49)

= 4/(3×7)

= 4/21

θ = cos-1(4/21)

Important Notes on Angle Between Two Planes

• The angle between two planes is equal to the angle between the normal vectors to the two planes and is called the dihedral angle.
• For planes, r.n1 = d1 and r.n2 = d2, the angle between them is given by, cos θ = |(n1 . n2)/(|n1|.|n2|)
• For planes, A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0, the angle between two planes in cartesian form is given by, cos θ = |(A1A2 + B1B2 + C1C2)|/[√(A12 + B12 + C12)√(A22 + B22 + C22)]

## 7. Distance of a point from a plane and Distance between two planes

Distance of a Point from a Plane

The distance between point and plane is the length of the perpendicular to the plane passing through the given point.

If we want to determine the distance between the point P with coordinates (xo, yo, zo) and the given plane with equation Ax + By + Cz = D, then the distance between point P and the given plane is given by d= |Axo + Byo+ Czo + D|/√(A2 + B2 + C2).

Example: Determine the distance between the point P = (1, 2, 5) and the plane π: 3x + 4y + z + 7 = 0

Solution: We know that the formula for distance between point and plane is: d = |Axo + Byo + Czo + D |/√(A2 + B2 + C2)

Here, A = 3, B = 4, C = 1, D = 7, xo = 1, yo = 2, zo = 5

Substituting the values in the formula, we have

d = |Axo + Byo + Czo + D |/√(A2 + B2 + C2)

= |3 × 1 + 4 × 2 + 1 × 5 + 7|/√(32 + 42 + 12)

= |3 + 8 + 5|/√(9 + 16 + 1)

= |16|/√26

= 8√26/13 units

Important Notes on Distance Between Point and Plane

• Distance Between Point and Plane Formula: |Axo + Byo + Czo + D |/√(A2 + B2 + C2)
• Distance Between Point and Plane is zero if the given point lies on the given plane.

Angle between a Line and a Plane:

Let θ be the angle between the line and the normal to the plane. Its value can be given by the following equation:

Φ is the angle between the line and the plane which is the complement of θ or 90 – θ. We know that cos θ is equal to sin (90 – θ). So Φ can be given by:

sin (90 – θ) = cos θ

or

or

Problem: A line has an equation

The equation of a plane is 3+ 4y – 12z = 7. Find the angle between them.

Solution: Let θ be the angle between the line and the normal to the plane. In the vector form, the equations can be written as:

The equation of the plane in the vector form can be given by:

So we have

Finding the value of the Φ between the line and the plane:

The value of Φ can be found by

Definition : The angle between a line and a plane is the complement of the angle between the line and normal to the plane