VECTORS

A vector is a quantity with both magnitude and direction, there are two operations defined on vectors and these both have a very direct geometric interpretation. We draw a vector as a line with an arrow, for now I will call the end without the arrow the 'start' of the vector and the end with the arrow the 'end' of the vector.

Vector addition: to add two vectors we take the start of the second vector and move it to the end of the first vector. The addition of these two vectors is the vector from the start of the first vector to the end of the second vector.

Scalar multiplication changes the length of a vector without changing its direction. That is we 'scale' it by the multiplying factor. So scalar multiplication involves multiplying a scalar (single number) by a vector to give another number.

Vectors are represented in a natural way by directed line segments, the length of the segment measuring the magnitude of the vector and the arrowhead indicating the direction of the vector. The arrowhead is usually placed at one end (or near the end) of the line segment; this end is called the head of the vector and the other end is called the tail of the vector. Such vectors are called free vectors because they are distinguished only by their length and direction, and not by their position in space.

The vector with the same magnitude as u but opposite direction, the negative of u, is labelled -u.

The zero vector (a vector with zero length and no direction) is written 0.

Position vectors

The line segment , where A and B are points in space, is called the position vector of B relative to A. It is also known as the vector from A to B as well as the displacement of B from A.

The vector from B to A has the same magnitude as the vector from A to B but opposite direction. That is, is the negative of .

The zero vector is the position vector of a point A with respect to itself; that is, 0 = . The length of the zero vector is 0 but it has no direction.

(TYPES OF VECTORS) Zero vector , Proper vector (non-zero vector), Unit vector, Equal vector, Collinear vector(two or more vectors have same or parallel supports, irrespective of their magnitudes & directions), Coplanar vectors(three or more vectors lie in the same plane or parallel to the same plane ).Equality of vectors

Given points A and B, we may think of the position vector v = as the vector that acts on the point A to get B.

If the line segments AB and DC have the same direction and the same length, then ABCD is a parallelogram and the position vectors and are equal; we write this as

= .

Using a head-to-tail procedure, a vector v can be added to itself to give the vector v + v.

We have v = and v = . It’s natural to write + = as 2v. The vector 2v has length 2|v| and is in the same direction as v. The notion of multiplication of a vector by a positive integer is then generalised to define the vector sv for all scalars s, as follows.

For example, -3v has three times the magnitude of v but

points in the opposite direction; v (also written ) has

magnitude and has the same direction as v.

Note that -1v has the same magnitude as v but has the opposite direction, and so is the same vector as the negative of v, that is, -1v = -v.

When are vectors parallel?

Two vectors u and v are said to be parallel if they have either the same direction or opposite direction. This means that each is a scalar multiple of the other: for some non-zero

scalar s, v = su and so u = v.

Since the zero vector can be written 0 = 0v, the zero vector is considered to be parallel to every other vector v.

Unit vectors and how to get them

It is often useful to be able to find a unit vector (a vector of length 1) in the direction of a given non-zero vector v. A common notation used for this unit vector is .

For example, if v has length 3, then = v has length 1 and is in the same direction as v.

In general, the unit vector in the same direction as v is

Since = , we also have v = |v| .

Introduction

Because non-zero vectors have direction as well as magnitude, adding vectors involves more than simply adding numbers. The sum of two vectors is another vector, and so the definition of addition must give a process for determining both the magnitude and the direction of the sum vector.

There are two equivalent procedures for addition of vectors, called the parallelogram rule and the triangle rule.

The parallelogram rule for addition

Suppose u and v are two vectors (in the plane or in space). Translate them so that they are tail-to-tail at point O.

From the head of each vector, draw a copy of the other vector to complete a parallelogram OAPB. In this parallelogram, u = = and v = = .

The triangle rule for addition

Another way to define addition of two vectors is by a head-to-tail construction that creates two sides of a triangle. The third side of the triangle determines the sum of the two vectors, as shown below.

Place the tail of the vector v at the head of the vector u. That is, u = and v = .

Now construct the vector to complete the third side of the triangle OAP.

This method is equivalent to the parallelogram law of addition, as can be easily seen by drawing a copy of v tail-to-tail with u, to obtain the same parallelogram as before.

Using position vector notation, the triangle rule of addition is written as follows: for any three points X, Y , Z,

Both the triangle and the parallelogram rules of addition are procedures that are independent of the order of the vectors; that is, using either rule, it is always true that u + v = v + u for all vectors u and v. This is known as the commutative law of addition. There are other rules like this one, and they are discussed in the component Vector Algebra.

Adding a vector to its negative

To add u to -u, place the tail of -u = at the head of u = .

Then u + (-u) = = 0, the zero vector. By the commutative law,

Subtraction of two vectors is just a special case of addition. The vector u - v is defined to be

Using the parallelogram law of addition, u - v = u + (-v) = +

= .

The vector v - u can be found in the same way.

The vectors u + v and u - v are both diagonals of the parallelogram formed by u and v.

Adding the zero vector 0 to any vector v gives the vector v again. That is,

for all vectors v. For if v = , then

and

The head-minus-tail rule

Given a vector and any point O, the triangle rule for

addition shows us that = + . We can rewrite this equation as

Since points to the head of the vector and points to its tail, we call this the “head-minus-tail” rule.

The advantage of this rule is that it allows us to rewrite all the vectors occurring in a problem as combinations of vectors emanating from a given point O (often called the origin).

Internal division of a line segment

With an assigned point O as origin, the position of any point

P is given uniquely by the vector , which is called the position vector of P relative to O.

Let P1 and P2 be any points, and let R be a point on the line P1P2 such that R divides the line segment P1P2 in the ratio m :

n. That is, R is the point such that = . Our task is to find the position vector of R (relative to O) in terms of the position vectors of P1 and P2.

As = , we have n = m and therefore

(1)

which rearranges to give

When m and n are both positive, the vectors and

have the same direction, since = . This corresponds to the situation where R lies between P1 and P2, as shown in the diagram above. R is then said to divide the line segment P1P2 internally in the ratio m : n.

As a special case of the general formula (1) we can obtain a

formula for the position vector of the midpoint M between two points P1 and P2.

In this case, m : n = 1 : 1 and so

External division of a line segment

We look at a situation similar to the one on the previous page, but allow m and n to have opposite sign. Again, R is

the point such that = , and we want to find the position vector of R (relative to O) in terms of the position vectors of P1 and P2.

Since is a negative multiple of , the vectors and have opposite direction.

The point R then lies outside the line segment P1P2 (but still on the line joining P1 and P2). In these cases, R is said to divide the line segment P1P2 externally in the ratio m : n. The formula is, as in case of an internal division,

¿ =nOP1−mOP2

m−n , m-n≠ 0 .

Coordinate axes and Cartesian coordinates

By three dimensional space we mean the space we live in. To fix a point P in three dimensional space requires a system of axes and three numbers. First select any point, call it the origin and mark it as O. All measurements will from now on originate from this point O. Next place three mutually

perpendicular axes OX, OY , OZ through O. This axis system is drawn on a page like this:

Note that although it may not look it, the angles XOY , XOZ, Y OZ are all right-angles!

To fix any given point P in three dimensional space, we refer it to the axis system. Let us first show the point P and the coordinate axes.

1. Drop a line from P perpendicular to the XOY plane (think of this plane as the floor), meeting the XOY plane at a point Q (the foot of the perpendicular).

2. Now we are in the XOY plane with a point Q and an axis system in two dimensions.

3. Drop a perpendicular from Q to OX and OY .

4. Transfer the two-dimensional picture from 3 into the three dimensional diagram from 1.

(Some of the right angles are marked: you should mark all the other right angles in the picture.)

5. Measure the lengths OR, OS, QP, and denote them by x, y, z respectively. We call the three numbers (x,y,z), in the order given, the Cartesian coordinates of the point P.

Notice that the order in which the numbers are written is important: (1, 2, 1) and (2, 1, 1) are the Cartesian coordinates of different points.

To make the diagrams uncluttered, only half of each axis has been drawn. However, if the axes are extended infinitely in both directions, it can be seen that this axis system creates eight octants in space, just as the two dimensional axis system creates four quadrants in the plane. Points in the eight different octants have Cartesian coordinates corresponding to all possible combinations of positive and

negative values of x, y and z. The figures above illustrate points in the octant corresponding to positive values of x, y and z.

Cartesian form of a vector

We begin with two dimensions. We have the following picture illustrating how to construct the Cartesian form of a point Q in the XOY plane.

Vectors i and j are vectors of length 1 in the directions OX and OY respectively. The vector is

xi. The vector is yj. The vector is the sum of and ,

that is, We now extend this to three dimensions to show how to construct the Cartesian form of a point P.

Define k to be a vector of length 1 in the direction of OZ. We now have the following picture .Draw a perpendicular PT

from P to the OZ axis.

In the rectangle OQPT,PQ and OT both have length z. The

vector is zk. We know that = xi + yj. The vector ,

being the sum of the vectors and , is therefore

This formula, which expresses in terms of i, j, k, x, y and

z, is called the Cartesian representation of the vector in

three dimensions. We call x, y and z the components of along the OX, OY and OZ axes respectively.

The formula

applies in all octants, as x, y and z run through all possible real values.

Length of a vector in terms of components

The length of the vector in the XOY plane is .

From the three dimensional picture below,

“cut out” the triangle OPQ to obtain:

It can be seen that the length of the vector is

= .

Addition and subtraction of vectors in terms of components

Given two vectors in Cartesian form

the sum + = is obtained by completing the parallelogram.

It can be proved that this is the same as the following calculation:

That is, the components of a sum are the sums of the components.

Notice that the parallelogram OQRP is part of a two dimensional plane sitting within three dimensional space (in a tilted way like the slanting face of the roof of a house).

Subtraction

The rule for subtraction works in exactly the same way.

Writing for the vector - , the rule above gives

The subtraction is illustrated below. Recall that = - =

.

An example of the use of this rule is the calculation of the Cartesian form of the position vector of a point P2 relative to a point P1. Suppose that in Cartesian form, = x1i + y1j + z1k and = x2i + y2j + z2k.

Then

Multiplication by scalars in terms of components

If a vector is expressed in Cartesian form, then it’s easy to calculate any scalar multiple of that vector in Cartesian form.

The general rule is that given any vector v = xi + yj + zk and any scalar a, then

For example, if v = 2i + j - k, then -4v = -8i - 4j + 4k.

The same formula applies in two dimensions. The vector v = xi + yj can be thought of as v = xi + yj + 0k. Then if a is any scalar,

Thus if u = 2i + j then 2u = 4i + 2j, as illustrated in the following picture.

Example 1

Find the magnitude of the vector x = 4i - 2j + 3k, the unit vector in the direction of x and the vector of magnitude 5 in the direction opposite to x.

Solution

We have |x| = = = .

Therefore the unit vector in the direction of x is the vector

and the vector of magnitude 5 in the direction opposite to x is

Example 2

Given the vectors u = 3i - j + 2k and v = 2i + 3j - k, find Cartesian forms of the vectors u + v, u - v, 2u + 3v.

Solution

First,

Next,

and finally,

Example 3

Given that P1(-1, 2, 3) and P2(3, 3, 4) are two points in space, find the Cartesian form of the vector .

Solution

Example 4

Show that the vectors

and

have the same magnitude.

Solution

Both vectors are in Cartesian form and their lengths can be calculated using the formula

We have

and

Therefore two given vectors have the same length.

Vector algebra in geometric form

We discuss properties of the two operations, addition of vectors and multiplication of a vector by a scalar. We do this first for free vectors.

Equality of Vectors Two vectors u and v are equal if they have the same magnitude (length) and direction.

The Negative of a Vector The negative of the vector u is written -u, and has the same magnitude but opposite direction to u. If u = , then -u = .

Commutative Law of Addition

for all vectors v and u.

Associative Laws

for all vectors u, v and w and for all scalars s and t.

As an illustration of the first of these associative laws, we translate the three vectors u, v and w so that they are drawn head to tail, and then draw (u + v) + w in the first figure below and u + (v + w) underneath it, demonstrating that both equal .

We may then simply write u + v + w, without using brackets. This associative law extends to sums of any number of vectors taken in any order, so that the expression u1 + u2 + u3 + ...... + un is well defined.

Distributive Laws

for all vectors v and u and for all scalars s and t.

The first of these distributive laws is illustrated below in the case s = 2.

Laws Involving the Zero Vector

for all vectors u. Vector algebra in Cartesian form

We discuss properties of the two operations, addition of vectors and multiplication of a vector by a scalar, but now for vectors in Cartesian form.

Suppose vectors v1 and v2 are written in Cartesian form,

Equality of Vectors Vectors v1 and v2 are equal if and only if their components are equal, that is, if and only if x1 = y1, x2 = y2 and x3 = y3.

The Negative of a Vector The negative of the vector v1 = x1i + x2j + x3k is the vector

Multiplication by a Scalar If s is any scalar, then

Addition of Vectors

The Zero Vector The zero vector is the vector 0 = 0i + 0j + 0k.

Example 1

Draw the vectors a + b and a - b in the diagrams below.

Solution

Example 2

Express the unknown vector v in terms of a and b.

Solution

v = b - a. v = a - b.

Example 3

Express the vector v in terms of a, b and c.

Solution

v = a + b - 2c.

Example 4

Simplify the following expressions.

a + 2b + 3( b - 5a) a - 7b + 7(b - a)

2a - 3b + c - (a - 3b)

Solution

-15a + b -6a

a - 2b + c

**Proof of the formula

If one (or both) vectors are zero vectors then the above formula is obvious. Hence assume that u and v are non-zero.

The vectors u, v and u - v then form a triangle.

The edges of the above triangles have length |u|, |v| and |u - v|. By the law of cosines (see below)

By definition u · v = |u||v| cos , so

(1)

By the formula for the magnitude of a vector

Multiplying the right hand side out we get

If we substitute the above into (1 ) we get

so u · v = u1v1 + u2v2 + u3v3 as claimed. Appendix: The law of cosines

The law of cosines asserts that in an arbitrary triangle a2

= b2 + c2 - 2bc cos , where a, b, c and are as shown below.

The formula follows by applying the Theorem of Pythagoras to the triangles ADC and DBC. Note that the red line AD has length |b cos | and the green line DB has length c - b cos . Hence by the Theorem of Pythagoras

Multiplying the right hand side out we get

Adding b2 cos 2 and rearranging we get a2 = b2 + c2 - Vector addition and subtraction is simple in that we just add or subtract corresponding terms. Vector multiplication is more complicated, there is not a general mutiplication that multiplies two vectors and produces another vector, however there are two types of multiplication that combine two vectors and produce another quantity.

A dot product takes two vectors and produces a scalar. A cross product takes two vectors and produces a vector.

Vector Add

Adding two vectors is equivalent to putting the tail of one vector against the head of the other. The order is not important, i.e. A + B = B + A.

This can be easily calculated by separately adding the x,y and z components.Vector Subtract

Subtracting two vectors is equivalent to putting the head of the vectors together, the result is then the vector between the tail of the vectors. An alternative way to think of it is to reverse the direction of the vector to be subtracted, then add the vectors.

The resulting vector can be easily calculated by separately subtracting the x,y and z components.

Dot Product

The dot product operation combines two vectors and produces a scalar output.

Dot Product

It is also given by A•B = |A| |B| cos(θ)

where:

|A| = magnitude of vector A |B| = magnitude of vector B θ = angle between vector A and vector B (0≤ ≤Ѳ п )

It is also equal to

A * projection of B on A

or

B * projection of A on B

2bc cos as required

Algebraic rules

Suppose u, v and w are vectors (in the plane or in space) with magnitudes |u|, |v| and |w|. Then

The rules are obtained by using the Cartesian representation of the scalar product. We let u = u1i + u2j + u3k, v = v1i + v2j + v3k and w = w1i + w2j + w3k. Then

1. By definition of the scalar product

2. If s is a scalar then

(su) · v = su1v1 + su2v2 + su3v3

= u1sv1 + u2sv2 + u3sv3

= s(u1v1 + u2v2 + u3v3) = s(u · v)

3. For the distributive law note that

u · v + u · w= u1(v1 + w1) + u2(v2 + w2) + u3(v3 + w3)= u1v1 + u1w1 + u2v2 + u2w2 + u3v3 + u3w3

= (u1v1 + u2v1 + u3v3) + (u1w1 + u2w2 + u3w3)= u · (v + w)

4. For the last rule simply note that

Alternatively one could prove the algebraic rules using the Cartesian representation of the scalar product

Conditions for perpendicularity

By definition of the scalar product, u·v = |u||v| cos , where is the angle between u and v. If = 90° then cos = 0. Given two non-zero vectors u and v

If u = u1i + u2j + u3k and v = v1i + v2j + v3k are given in Cartesian form then the above reads

For vectors in the plane drop the third terms.

The scalar product of the standard basis vectors  

By assumption the standard basis vectors i, j and k are mutually perpendicular, and of magnitude one. Hence by definition of the scalar product

1. i · i = j · j = k · k = 1;

2. i · j = i · k = j · k = 0.

The component of a vector parallel to a given vector  

3. By definition of the scalar product

4.5. is the projection of v into the direction of u as shown

below. We call

6.7. the component of v in the direction of u. The geometric

picture is shown below.

8.

Resolution of a vector into components parallel and perpendicular to a given vector

 

9. Given two vectors u and v we want to find the projections of v parallel and perpendicular to the vector v. We already know that the projection of v into the direction of u is

10.11. Looking at the picture below it is evident that the

projection of v into the direction perpendicular to u is n = v - p, that is,

12.

13.

The vector product in terms of magnitudes, angles and the “right-hand rule”

 

The vector product is a product of two vectors in space. The result of the multiplication is a vector, hence the name vector product. Other names for the vector product are cross product or outer product.

If u and v are vectors, their vector product is denoted by u × v. We must assign that product a direction and a magnitude.

We start by specifying the direction. Suppose that u and v are two non-zero and non-parallel vectors in space. When placed tail-to-tail they define a plane. There are two possibilities to choose a unit vector perpendicular to that plane:

We choose n such that the triple u, v, n is right-handed. By this we mean the following. Align a screw-driver along the line perpendicular to the plane defined by u and v. Then turn the screw-driver such that u moves to v through the smaller angle. The direction of n is then given by the direction the

screw m oves.

Note that it is irrelevant which way we align the screw-driver, the direction in which the screw moves will be the same either way.

Having established the direction it remains to define the magnitude of the vector product. We let the magnitude be |u||v| sin n, where is the smaller angle between u and v.

Ѳ is the angle b/w vectors u,v, 0≤ ≤Ѳ п and n is unit vector per. To both u & v.

Then we define

If u and v are parallel or one (or both) of them are zero vectors then we set u×v = 0. Geometrically, the magnitude equals the area of the parallelogram spanned by the vectors u and v when placed tail-to-tail.

Important Note: Unlike in the definition of the scalar product

it is essential that the angle between u to v is the smaller one, that is, the one between 0° and 180°. The vector product in terms of components  

If u = u1i + u2j + u3k and u = v1i + v2j + v3k are two vectors then their vector product is

If you are familiar with determinants of a 3 × 3 matrix, there is an easy way to remember the above formula. Formally,

Vector Cross Product

Both the two operands and the result of the cross product are vectors.

The vector cross product has some useful properties, it produces a vector which is mutually perpendicular to the two vectors being multiplied.

The vector product obeys the following algebraic rules

 

Geometrically speaking, the vector   that results from the cross

product of vectors   and   has a magnitude given by the product

of   and   times the sine of the angle from   to  ,

or  .

The direction is given by the “right hand rule.” If you lay your right hand

palm-up along  , then curl your fingers toward  , the resulting vector   is

in the direction of your thumb. Take a look:

Properties of the cross product:

[ lagrange’s identity]

Proof: |a X b∨¿∨a∨¿ b∨¿¿ sinѲ ∴ |a X b|2 =( |a∨¿b|)2 (1-cos2 ) [Ѳ a . b

=|¿a∨¿ b∨¿¿ cosѲ ] ∴ |a X b∨¿ 2= |¿ a∨¿2¿ a . b¿

¿ b∨¿2|Scalar triple product identities

The triple product can be evaluated using the relation

 

The triple product of vectors a, b, and c is given by

= [a b c ]

 The value of the triple product is equal to the volume of the parallelepiped constructed from the vectors. This can be seen from the figure since

 Magnitude of this projection = a .(b X c )

¿ b X c∨¿¿ [b X c repts. The vector area of

the base of the parallelopiped and the height of the parallelopiped is the projection of a along the normal to the plane containing vectors

b∧¿ c . i.e., along b X c ]∴ volume of parallelepiped = [a b c ] {if a , b , c repts. Three co-terminus edges of a parallelogram }

[a b c ] = 0, if a , b , c are coplanar vectors.

The scalar triple product has the following properties

 

⇨ 1. [

a b c

] = [

b c a¿

= [

c a b¿

If  these vectors are cyclically permuted. The value remains unaltered.

2. the position of dot and cross can be interchanged, provided the cyclic order of vectors remains the same.

3. the value remains the same in mag. but changes the sign, if the cyclic order of vectors is changed.(as shown above)

4. if any two vectors are same and parallel( or collinear),then scalar triple product = 0 [a=¿ λb ]

5. if any three vectors and scalar λ , then [λa b c ¿ = λ[a b c]

Triple vector product:

 The triple vector product has the properties

 

Or

If A,B and C are vectors and x is cross multiplication, then,

A × (B × C) = B (A·C) - C (A·B)

As explained above the dot product gives a scalar, so B and C are scaled by these scalars before they are subtracted.

and also,

(A × B) × C = - C × (A × B)reversing order of multiplication changes sign

= - A (B·C) + B (A·C)using first vector triple product identity

Here are some triple product identities for dot product of cross product:

A · (B × C) = (B × C) · A

A · (B × C) = C · (A × B) = B· (C × A)

A · (B × C) = -A · (C × B)

But ,

The expression (u × v) × w makes sense, but

in general. As a counterexample look at (i × i) × j. We have (i × i) × j = 0 × j = 0 but i × (i × j) = i × k = -j.

This shows that writing u × v × w does not make sense as it is not clear which two vectors to multiply first!

The easiest way to prove the algebraic rules of the vector product is to make use of the Cartesian representation of the vector product. Hence assume that u = u1i + u2j + u3k, v = v1i + v2j + v3k and w = w1i + w2j + w3k.

1. u×v = -(v×u) follows directly from the definition of the vector product;

2. For every scalar s

3. Using the Cartesian representations we get

Condition of parallelism

Two vectors, u and v are parallel if and only if the angle between them is 0° or 180°. Hence, by definition of the vector product we have the following characterisation of parallel vectors

The vector product of the standard basis vectors

Using the definition of the vector product it follows that

1. i × i = j × j = k × k = 0

2. i × j = -(j × i) = k;

3. i × k = -(k × i) = -j;

4. j × k = -(k × j) = i;

Recall that, by convention, the triple i, j, k is always right-handed, so i, k, j is left handed and j, k, i is right handed

Vector products and the area of a triangle

The magnitude of the product u × v is by definition the area of the parallelogram spanned by u and v when placed tail-to-tail. Hence we can use the vector product to compute the area of a triangle formed by three points A, B and C in space.

It follows that the area of the triangle is

. area of parallelogram.

Identities

A×B + B×A = 0anti-commutative, vector cross multiplication is not commutative, changing the order reverses the direction of the resulting vector: A×B = - B×A.

A×(B×C) + B ×(C×A) + C ×(A×B) = 0

Jacobi identity : anti-associative : anti-symmetry So cross multipication is not associative.

A × A = 0 A 'squared' is a zero length vector because sin(0) = 0

A × (B + C) = (A × B) + (A × C)

cross multiplication for vectors is distributive over +

0 × A = 0

Where:

A,B and C = any vectors 0 = zero length vector (all elements are zero) × = vector cross product

For example, the following expression:

A × A × B

We might expect that this is zero, because A × A = 0, but this is not necessarily so because cross multiplication is not associative. We need to specify the order, if it is:

A × (A × B)

then the result is not necessarily zero.

Example19: For any two veceorsa , b , we always have |a. b|≤|a| ¿b| ( Cauchy-Schwartz inequality).

Solution: It is hold trivially, whena =0 or b=0. If a ≠0 or b≠0, then we get

¿ a . b∨ ¿|a|∨b∨¿¿

¿ = |cosѲ| ≤ 1, hence proved.

Example 20: For any two veceorsa , b , we always have |a+b|≤|a|+ ¿b|( triangle inequality).

Solution: t is hold trivially, whena =0 or b=0. If a ≠0 or b≠0, then we get , |a+b|² = (a+b)² = (a+b). (a+b) = |a|²+2 a. b+¿b|²≤|a|²+2 ¿ a. b∨¿+¿b¿|² (∵ x≤|x| for all x∊R)≤ |a|²+2 ¿ a∨¿.| b∨¿+¿b¿|² ( from ex. 19)= (|a|+ ¿b|)²

EXAMPLE 29: Three vectors a , b,c are satisfy the condition a + b+c =0. Evaluate the quantity μ = a . b+b . c +c .a , if ¿ a |=1, | b∨¿=4 and| c∨¿¿ =2.

Solution: a .( a + b+c) =0 ⇨ a . b+c .a = -|a∨²=-1………(i)

Similarily , we get a . b+b . c = -|b∨² =-16…………(ii)

b . c +c .a = -4 ………(iii), by adding, 2μ=-21 ⇨ μ =-21/2EXAMPLE 30: If with reference to the right handed system of mutually per. Unit vectors I,j & k, α = 3i-j, β = 2i+j-3k, then express β

In the form β = β1+ β2 , where β1 , β2 are // & per. To α resp.

Solution: let β1 = λα , λ is scalar, i.e., β1 = 3λi – λj , β2 = β - β1= (2-3λ)I + (1+λ)j - 3k. since β2 is to be per. To α , we should have α.β2 =0

⇨ λ =1/2 ∴ β1=3i2

−1 j2

∧ β2 = (1/2)i+(3/2)j-3k.

NCERT QUESTIONS WITH HINTS

EX.10.3 Question 4 Find the projection vector i+3j+7k on the vector 7i-

j+8k. [ Ans. Projection of a on b = a . b¿ b∨¿¿

⇨ 60

√114 ]

Question 8 Find the magnitude of two vectorsa , b , having the same mag. and such that the angle b/w them is 600 and their scalar product is ½.

Ans. a . b = ¿a∨¿¿ ¿ b∨¿¿ cosѲ ⇨ ½ = ¿a∨¿¿ ¿ b∨¿¿ cos600 = ¿a∨¿ ² ¿ (1/2) (∵ ¿a∨¿=¿¿ ¿ b∨¿¿) ⇨ ¿a∨¿¿ ¿∨b∨¿¿ =1

Question 12 Ifa . a = 0 & a . b = 0, then what can be concluded about the vector b. [Ans. These eqns. For a =0 ⇨ b is any vector.]

Question13 If a , b,c are unit vectors such that a + b+c =0, find the value of a . b+b . c +c . a .

Ans. a + b =-c ……..(i) take dot product with a ⇨ a . ( a + b )=a . (-c ¿¿ ⇨ 1 +a . b+c . a = 0 ……(ii), similarily with b & c , we get 1 +a . b+c . b = 0

…..(iii) & 1 +c . b+c . a = 0 …..(iv) [∵¿a|=¿b|=|c|=¿1], by adding (ii),(iii)&(iv) , we get a . b+b . c +c . a = -3/2

Question 14 If either a =0 or b=0, then a . b = 0. But the converse need not be true. Justify your answer with an example.

Ans. Let a = i-3j+4k, b = 2i-2j+k are non- zero vectors but their dot product is zero.

Question 16 Show that the points A(1,2,7), B(2,6,3),C(3,10,-1)are collinear.

ANS. The position vectors of A,B,C are i-2j+7k, 2i+6j+3k, 3i+10j-k

AB= OB−OA = i+4j-4k, | AB| = √33, |BC |=√33 & | AC |= 2√33

| AB| + |BC |= | AC |= 2√33 ⇨ A,B, C are collinear.

Question 17 show that the vectors 2i-j+k, i-3j-5k & 3i-4j-4k form the vertices of a right angled triangle.

Ans. AB = OB−OA = -i-2j-6k ,| AB|= √41, |BC |=√6 & | AC |= √35

| AB|² = |BC |²+ | AC |² ⇨ ABC IS a right angled triangle.

EX. 10.4 Question 3 If a unit vector a makes angles п/3 with I, п/4 with j & an acute angle Ѳ with k, then find Ѳ and hence, the component of a .

Ans. Let a = a1i+a2j+a3k, ¿ a| =1 , according to ques. (a1i+a2j+a3k).i=|a|.icosп/3 ⇨ a1= ½, similarily a2=1/√2 & a3 = ½ [∵a1²+a2²+a3²=1] , (1/2i+1/√2j+1/2k).k=|1/4+1/2+1/4|cos ⇨Ѳ Ѳ=п/3

Question 6 given that a . b = 0 and a × b = 0. What can you conclude about the vectors a and b ?

Ans. given that a . b = 0 and a × b = 0 ⇨ a =0 or b =0 or a per. To b

And a =0 or b =0 or a // b ⇨ a = 0 b =0 [∵a per. To b & a // b can

Never hold at a time]

Question 8 if either a =0 or b=0, then a × b = 0. is converse true? Justify your answer with an example.

Ans.| a × b|=| a ¿∨¿ b|sinѲn = 0 [∵a =0 ] similarily for b=0

Converse: let a ≠0=i+j+k b≠0=2i+2j+2k, a × b = 0 i.e., a // b so Ѳ=0 but | a × b|=| a ¿∨¿ b|sinѲn = 0 Question 9 find the area of △ with vertices A(1,1,2), B(2,3,5), C(1,5,5).

Ans. Area of △ABC = ½| AB × AC| = ½ i j k1 2 30 4 3

= ½ √61 [ position

vectors of AB and AC are i+j+2k & 4j+3k]

Question 10 find the area of //gm. Whose adjacent sides are determined by the vectors i-j+3k and 2i-7j-k.Find unit vector // to its diagonal [Ans. Area of //gm whose adjacent sides are given | a × b|=

15 √2, where a × b= i j k1 −1 32 −7 1

,unit vector=(a + b)/¿ a + b|

MISC. Question 3 A girl walks 4 km. towards west, then she walks 3 km in a direction 300 east of north and stops. Determine the girl’s displacement from her initial point of departure.

Ans. N A

300 3 km displacement

W B (4 km) O

Let the girl starts from O and the girl walks 4 km towards W. again she walks 3 km along WA.(let WO repts. i and WN repts. j) = WO +WA

=-4i+(3cos600i+3sin600j) ⇨ OA = -4i+3j/2+3√3j/2 = -5i/2+3√3j/2

[WA =WB +BA = 3cos600i+3sin600j ] , so displacement from O to A =

√(−52 ) ²+¿¿)² = √13 km along OA .

Question 4 If a = b + c , then is it true ¿a∨¿¿ = ¿b∨¿ + | c∨¿¿ ? justify your answer.

Ans. ¿a∨¿¿ ² = ¿b + c∨¿ ² ¿ = ¿ + c ¿ .¿ + c) = =| b |²+ | c|²+2|b|| c|cosѲ

When Ѳ=00 ¿a∨¿¿ ² =( ¿b∨¿ + ¿ c∨¿¿ ², if ≠Ѳ 00 , ¿a∨¿¿ ≠ ¿b∨¿ + | c∨¿¿

Question 9 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2a +b) & (a -3b) externally in the ratio 1:2. Also, show that P is the mid point of the line segment RQ.

Ans. R divides PQ externally in the ratio 1:2 ∴ position vector of a

point R is 1. ( a−3 b )−2(2a+ b)1−2

=(3a +5 b) mid point of RQ is (2a +b)= position

vector of P ⇨ P is mid point of RQ and the point R is (3a +5b)

Question 12 let a = i+4j+2k, b = 3i-2j+7k and c = 2i-j+4k. find a vector d which is per. To both a& b , and c . d = 15.

Ans. Let d =d1i+d2j+d3k, since d is per. To a = i+4j+2k, b = 3i-2j+7k, then d1+4d2+2d3=0 & d1-2d2+7d3=0 , by solving d1=32λ, d2=-λ, d3=-14λ (by cross multi.) , put in c . d = 15 ⇨ (2i-j+4k).(32λi -λj -14λk)⇨λ=5/3, then d = 160i/3-5j/3-70k/3

Question 13 the scalar product of the vectors i+j+k with a unit vector Along the sum of vectors 2i+4j-5k and λi+2j+3k is equal to one. Find the value of λ.

Ans. a+ b =t = (2+λ)i+6j-2k , unit vector of t = [(2+λ)i+6j-2k]/¿ t|

(i+j+k). [(2+λ)i+6j-2k]/¿ t|= 1 ⇨ (λ+6)/ ¿ t|= 1, where ¿ t|= √ λ ²+4 λ+44

⇨ λ =1

Question 14 If a , b , c are mutually per. Vectors of equal mag., show that the vectors a +b +c is equally inclined to a , b ¿c .

Ans. a .¿ +b +c) = λ¿a +b +c|cosѲ [ ∵¿a ¿=¿ b ¿=¿c|= λ, a . b =0 ¿ b . c=c.a] ⇨ cosѲ= λ

|a+b +c| , similarily dot product with b , c

Give same angle Ѳ = cos-1¿)

EXTRA QUESTIONS:

Q. 1 If a ¿ b are unit vectors and Ѳ is the angle b/w them, then show that sin(Ѳ/2)= ½ |a −b |.

Ans. |a −b |²= (a −b ). (a −b )= |a∨² −2 a . b+¿ b |²= 2-2cosѲ

[|a∨¿ ¿∨b |=1, A•B = |A| |B| cos(θ)] ⇨ |a −b |²= 4sin2(Ѳ/2)

Q.2 If a × b =c × d , a × c = b × d , show that a - d is // to b - c

Ans. By subtracting above results , we get a ׿ b -c) =¿ ¿ -b¿ × d⇨ a ׿ b -c) = d × ¿ ¿ -c ¿ ⇨ ( a - d ) ׿ b - c )=0 ⇨ a - d is // to b - c

Q.3 If the sum of two unit vectors is a unit vector, show that the mag. of their diff. is √3. [ Hint :use |a + b|² ⇨ 2a. b= -1, put in |a- b|²]

Q.4 If a , b , c are three unit vectors such that a . b = a . c =0 and angle b/w b , c is п/6, prove that a = ∓ 2(b × c ).

Ans. a . b = a . c =0 ⇨ a is per . b , a is per . c ⇨ a is per .¿bothb & c ⇨ a is∥¿ b×c ⇨ a ¿ λ¿) …….(i) , λ is scalar ⇨| a ¿=|λ|∨¿)|= |λ||b|∨c|sinп/6 ⇨ |λ| = 2 ∴ |λ| = ∓2 , put in (i)

Q. 5 If a , b , c are resp. the position vectors of the vertices A,B,C of △ABC, Prove that area of the △ ABC is given by =1/2 |a× b¿+¿)+ c ×a)|

Ans. Area of △ABC = ½ | AB× BC)| = ½ |(b−a) ×(c− b)|∵ AB = OB- OA=¿ b−a , BC = c− b

Q. 6 If |a . b|=12 |a|=10 & |b| = 2, find |a× b|.

Ans. We know that |a× b|² +|a . b|² = |a¿2∨b|² ⇨ 16 [ by lagrange’s identity, |a¿2∨b|²sin² =Ѳ |a¿2∨b|²(1-cos² ) ]Ѳ

Q.7 if a +b +c= 0 , show that a× b=b× c =c ×a.

Ans. a ׿ +c) = a ×(−c ) ⇨ a× b= c ×a , simily take cross product with vector b , we get a× b=b× c, by above results ]

Q.8 ABCD is quad. Such that AB = b , AD = d, AC = m b+p d .Show that the area of the quad. ABCD is ½ |m+p||b× d|

Ans. AB + BD = AD ⇨ BD = AD−AB = d− b ∴ area of quad. ABCD = ½ |AC× BD | = ½ |( m b+p d)× (d− b)| = ½ | m

¿ ¿)- p(d ×b)| = ½ |m+p||b× d| ∵ ¿ ¿)= - (d ×b)]

Q. 9 Let A,B,C &D be any four points in space. Prove that

| AB×CD+ BC× AD + CA× BD| = 4 ( area of △ABC)

Ans. Taking A as orgin , let the position vectors of B,C &D be b , c , d resp. then AB , CA , AD are b ,−c , d resp. & CD =d− c , BC = c− b , BD = d− b ∴ | AB×CD+ BC× AD + CA× BD| =

|b ׿ d− c) + ¿) × d + (−c ) × (d− b )|= 2| AB× AC|

Q.10 Three vectorsa , b , c are coplanar vectors ⇨ ¿ ¿).c =0

Questions on scalar triple product:

1.If a = 2i-3j+4k, b =i+2j-k, c =3i+4j-k then find is a .(b X c) =(a X b) .c ? [ ans. 36=36 ]

2. find the volume of parallelepiped whose edges are a =2i-3j+4k, b =i+2j-k and c = 2i-j+2k [ ans. Use scalar triple product 2 cubic units]

3. Show that three vectors -4i-6j-2k, -i+4j+3k and -8i-j+3k are co-planar. [ find scalar triple product is zero]

4. Show that three vectors i-2j+3k, -2i+3j-4k and i-3j+λk are co-planar. If λ = 5 [ take scalar triple product is zero , put λ=5]

5. Show that four points with position vectors 6i-7j, 16i-19j-4k, 3j-6k and 2i+5j+10k are not co-planar.

[hint: a = AB=10i-12j-4k ,b = AC =-6i+10j-6k, c = AD =-4i+12j+10k

And show scalar triple product =840≠ 0]

6. For any three vectors prove that

[a+b b+ c c+a ]= 2 [a b c]

[hint: [¿+b¿¿ X( b+ c )] .( c+ a) = [aXb+a X c+bX c ¿ .¿) ]

7. If the vectors A =ai+j+k , B = i+bj+k, C =i+j+ck are co-planar, then

11−a

+ 11−b

+ 11−c

=1, where a,b,c ≠ 1.

8. Show that if vectors a , b , c are co-planar, then a+ b , b+c , c+ a are also co-planar.

9.Find the value of λ if the points A(-1,4,-3) , B(3,λ,-5) , C(-3,8,-5), D(-3,2,1) are co-planar. [ ans. Is 2]

10. Let a= 5i-j+7k & b = i-j+mk . find m, such that a +b & a -b are orthogonal. [ ans. Is √(73) ]

*11. Let a, b be two given non-cillinear vectors. Then any vector r coplanar with vector a & b can be uniquely expressed as r = xa + yb ,where x , y are scalars.

Find the value of p which makes the vectors a , b , c co-planar , where

a= 2i-j+k, b = i+2j-3k, c = 3i-pj+5k.

[hint: by using above result , we get x+3y=2, 2x-py=-1 & 5y-3x =1⇨ x=1/2, y=1/2 , p= 4]

12. if a & b are unit vectors and Ѳ is the angle b/w them, then show that sinθ

2 = ½ |a -b |

[hint: ¿ a -b |2 = (a -b).¿ -b ) =¿ a∨¿ 2+ |b |2 -2¿ a ||b|cosѲ =2(1-cosѲ)

As | a |= |b|=1]

13. If two vectors a =i+j+k , b = j-k , find a vector c such that a Xc =b

& a .c=3. [ ans. Is 5/3i+2/3j+2/3k ]

14 . Find angle b/w two vectors with mag. 1 & 2 resp. and when ¿ a Xb

|= √3. [ hint: use sinѲ= ¿ axb∨¿

|a|∨¿ b∨¿¿¿¿ , Ѳ=п/3]

15. If a Xb=c Xd, a Xc= b Xd, show that a –d is // to b -c .

[ hint: by subtracting above given results a X(b -c)=( c -d)Xd ⇨( a -d)X(b -c)=0 ⇨ sinѲ=0, where Ѳ is angle b/w two vectors,a ≠d ;b ≠c]