WEEK FOUR SS2 FURTHER MATHS SECOND TERM VECTORS IN THREE DIMENSIONS

WEEK FOUR

SS2 FURTHER MATHS SECOND TERM

VECTORS IN THREE DIMENSIONS

CONTENTS

(a) Scalars Product of Vectors in Three Dimensions

(b) Application of Scalar Product

SUB TOPIC: SCALAR PRODUCT OF VECTORS IN THREE DIMENSIONS

The scalar product of three vectors a,b and c is defined as a.(bxc) which is a scalar quantity.

If three vectors a,b and c are given as:

a = a₁i + a₂j + a₃ok

b = b₁i + b₂j + b₃ok

c = c₁i + c₂j + c₃ok

The scalar product is discovered the identical manner because the determinant of a 3×3 matrix.

We denote dot or scalar product of two vectors A and B by A.B. This dot product is outlined because the product of the magnitudes of A and B and the cosine of the angle between them.

A.B =

Scalar product in any other case referred to as dot product (or inside merchandise).

The Scalar product between two perpendicular vectors is zero.

Abstract of the scalar product of unit vectors is offered beneath:

i.i = j.j = ok.ok = 1 right here, θ = 0⁰

i.j = i.ok = j.ok = 0 right here, θ = 90⁰

scalar product is scalar and commutative

A.B = B.A

Word:

1. a.(bxc) =(axb).c
2. Three vectors a,b and c are stated to be coplanar or collinear if their scalar triple product is zero.

Properties of dot product

1. A.B = B.A
2. A.(B+C) = A.B +A.C
3. m(A.B) =(mA).B = (A.B)m the place m is a scalar.
4. i.i = j.j = ok.ok = 1
5. i.j = j.ok = ok.i = 0

If A = a₁i + a₂j + a₃ok and B = b₁i + b₂j + b₃ok

A.B = a₁b₁+ a₂b₂+ a₃b₃

A.A = a₁² + b₂²+ b₃²

B.B = b₁² + b₂²+ b₃²

If A.B = 0 and A and B should not null vectors, then A and B are perpendicular.

Examples

1. Given: a = 2i –j + ok; b = 3i + 2j – ok ; c = i -4j + 3k

present that vectors a,b and c are coplanar.

Resolution:

We have to present {that a}.(bxc) = 0

=2(6-4)+1(9+1)+ (-12-2) = 2 x2 + 10 – 14 = 14 – 14 = 0

Since a.(bxc) = 0, vectors a, b and c are coplanar.

1. If p = 2i + 5j – 3k

q = i + 0j + 5k

r = 3i – 4j + 2k, present if they’re collinear or not.

Resolution:

= 2(0+20) – 5 (2-15) -3 (-4-0) = 2 x20 – 5x – 13 -3x-4

= 40 + 65 + 12 = 117

Since

CLASS ACTIVITY:

1. Given a = i+2j+-3k, b = 2i – j + 2k, c = 3i + j – ok. Discover a.(bxc) = 0
2. P, q and r are three vectors given by 4i – j + 2k, 3i + 2j – 5k and –i+3j + ok respectively. Consider (pxq).r

SUB TOPIC: APPLICATION OF SCALAR PRODUCT

We will start by making use of the scalar product to the strains that type the edges of a triangle.

The appliance shall lead us to determine the cosine rule and the well-known pythagoras’ theorem.

Let

Making use of triangle regulation of vectors

A

C

B

. = (

c² = – 2.

c² = a²+b² – 2abcosθ

The formulation by way of, c² = a²+b² – 2abcosθ is the acquainted cosine rule.

If the vectors are perpendicular, c² = a²+b²

= 0 and the formulation will cut back to c² = a²+b²⁼

That is the well-known Pythagoras’s theorem for a proper angled triangle.

Software of dot product to acquire trigonometric growth.[mediator_tech]

Let O_x and O_y be perpendicular axes with unit vectors instructions OA and OB that are perpendicular to 1 one other. Let OA make angle x⁰ with Ox.

Let unit vector act within the course OC which makes angle y⁰ with OA.

y

B

C

A

x

0

y⁰

x⁰

Resolving alongside Ox and and Oy provides

Resolving alongside Ox and Oy provides

= -isinx + cosx

Resolving alongside OA and OB provides

=

Substituting for

=

… (i)

Resolving alongside Ox and Oy provides

= … (ii)

Since (i) and (ii) characterize the identical vector, corresponding parts have to be equal and so we’ve

Cos(x+y) = cos x cosy – sinxsiny

Sin(x+y) = sin x cosy + cosxsiny

CLASS ACTIVITY:

1. If
2. Show the scalar product to the strains that type the edges of a triangle.

PRACTICE EXERCISE

1. Outline the scalars product of vectors in three dimension.
2. Scalar product in any other case referred to as ____ product
3. Record all of the properties of dot product.
4. Show the purposes of dot product to acquire trigonometric growth.
5. If = 2i + 3j-k, = -i + 2j-4k, discover
6. Show that x = i +2j-3k,

ASSIGNMENT

1. Given {that a} = 3i + 2j + ok and b = 4i – 5j +3k. discover a.(axb); (b) b. (axb)
2. Present if query 2 is collinear.
3. The vertices of a triangle have place vectors 3i – 2j + 7k, 2i + 4j +ok and 5i + 3j – 2k. discover the world of the triangle.
4. Record three properties of scalar product
5. Show the appliance of dot product to acquire trigonometric growth.

KEY WORDS

• VECTORS
• SCALAR
• DOT PRODUCT
• DIMENSION
• PERPENDICULAR
• COLLINEAR
• COPLANAR
• UNIT VECTOR
• DETERMINANT