ADVANCED MATHEMATICS FORM 6 – VECTOR ANALYSIS- 2

Share this post on:

          USED TO PROVED COMPOUND ANGLE FORMULA OF SINE

          Sin (A + B) = sin A cos B + sin B cos A

          Consider the vector diagram below

          Pg. drawing

          edu.uptymez.com

          edu.uptymez.com = (cos A) edu.uptymez.com + (sin A) edu.uptymez.com + oedu.uptymez.com

          edu.uptymez.com = (cos A) edu.uptymez.com + (sin A) edu.uptymez.com + oedu.uptymez.com

          Hence

          edu.uptymez.com = edu.uptymez.com sin (A + B)

                             But

          edu.uptymez.com = edu.uptymez.com

          edu.uptymez.com = i edu.uptymez.com + kedu.uptymez.com

          edu.uptymez.com = i (o) – j (o) + k (-cos A sin B – sin A cos B)

          edu.uptymez.com = -k [sin A cos B + cos A sin B]

          edu.uptymez.com = edu.uptymez.com2

                   = edu.uptymez.com2

          edu.uptymez.com =  sin A cos B + cos A sin B

                   Also

          edu.uptymez.com= edu.uptymez.com

          edu.uptymez.com = edu.uptymez.com2

          edu.uptymez.com2 = edu.uptymez.com2 B

                          = 1

          Therefore

          Sin A cos B + cos A sin B = (1) (1) sin (A+B)

          Sin (A + B) = Sin A cos B + cos A sin B

                   Proved

  USED TO DETERMINE/ TO PROVE COMPOUND AND FORMULA OF SINE
  Sin (A – B) = sin A cos B – Cos A sin B
   –  Consider the vector diagram below

Pg.6 drawing
edu.uptymez.com

         edu.uptymez.com = (cos A) edu.uptymez.com + (sin A) edu.uptymez.com + oedu.uptymez.com

          edu.uptymez.com = (cos B) edu.uptymez.com + (sin B) edu.uptymez.com + oedu.uptymez.com

          = edu.uptymez.com = edu.uptymez.com sin (A – B)

                    but

          edu.uptymez.com = edu.uptymez.com

          = i edu.uptymez.com – j edu.uptymez.com + k edu.uptymez.com

edu.uptymez.com = edu.uptymez.com2

                   = edu.uptymez.com2

                   = sin A cos B – cos A sin B

                   Also

          edu.uptymez.com = edu.uptymez.com+ 0

          = edu.uptymez.com

                   = edu.uptymez.com

                    = 1

edu.uptymez.com =  edu.uptymez.com

          = edu.uptymez.com

                   = edu.uptymez.com

                    = 1

Therefore

Sin A cos B – cosA sin B = (1) (1) sin (A – B)

Sin (A – B) = sin A cos B – cos A sin B

          Proved

USED TO FIND THE AREA OF THE TRIANGLE

–      Consider the triangle ABC below

Pg. 7 drawing
edu.uptymez.com

Area (A) = ½ x base (b) x height (h)

          A = ½ bh ……i

Also

Sin edu.uptymez.com = edu.uptymez.com

Sin edu.uptymez.com = edu.uptymez.com

h = edu.uptymez.com Sin edu.uptymez.com….ii

And

b = edu.uptymez.com…iii

Substitute …ii and …iii into 1 as follows

A = ½ edu.uptymez.com sin edu.uptymez.com

     A = ½ edu.uptymez.com

USED TO FIND THE AREA OF PARALLELOGRAM

–      Consider the parallelogram below

Pg. 7 drawing

 edu.uptymez.com

          Area = length (l) x height (h)

                   = A = Lh …i

          L = edu.uptymez.com…ii

Also

Sin Q = edu.uptymez.com

edu.uptymez.com

h = edu.uptymez.com Sin Q….ii

Substitute ii and iii into —1 as follows

Area (A) = edu.uptymez.com sin Q

          A = edu.uptymez.com

          Generally

Area (A) = edu.uptymez.com

             = edu.uptymez.com

             = edu.uptymez.com

             = edu.uptymez.com

Where

    edu.uptymez.com = edu.uptymez.com – edu.uptymez.com

    edu.uptymez.com = edu.uptymez.com – edu.uptymez.com

    edu.uptymez.com = edu.uptymez.com – edu.uptymez.com

    edu.uptymez.com = edu.uptymez.com – edu.uptymez.com

USED TO FIND THE VOLUME OF PARALLELOPIPED

–      Consider the diagram below

Pg. 7 drawing
edu.uptymez.com

–      Suppose P Q R and S are the vertices of the parallelepiped, hence the volume (v) of the parallelepiped is given by:

Volume (v) = base area (A) x height (h)

            V = edu.uptymez.com x edu.uptymez.com

             V = edu.uptymez.com

             V = edu.uptymez.com

          V = edu.uptymez.com

Again, for the sides with position vectorsedu.uptymez.com, edu.uptymez.com and edu.uptymez.com

Volume (v) =  edu.uptymez.com x edu.uptymez.com

                           = edu.uptymez.com

                            = edu.uptymez.com

USED TO FIND THE VOLUME OF A TETRAHEDRON

–      Consider the tetrahedron with vertices P, Q, R and S

Pg. 8 drawing

 edu.uptymez.com

Volume (v) = 1/3 x base area x altitude

V = 1/3 x ½ edu.uptymez.com x edu.uptymez.com

           V = 1/6 edu.uptymez.com x edu.uptymez.com

             V = edu.uptymez.com

Therefore

Volume = 1/6 edu.uptymez.com

USED
TO FIND THE VECTOR PERPENDICULAR TO THE PLANE

–      Consider the diagram below

Pg. 8 drawing

 edu.uptymez.com

Where

 edu.uptymez.com = is the vector perpendicular or normal to the plane

 edu.uptymez.com

 edu.uptymez.com = edu.uptymez.com x edu.uptymez.com

Question

26.  Find the area of the triangle ABC whose vertices are A (2, 1, 1) B (3, 2, 1) and C (-2, -4, -1)

27.  The position vector of the points A, B and C are (2, 4, 3), (6, 3, -4) and (7,           5, -5) respectively. Find the angle between  edu.uptymez.com and  edu.uptymez.com and hence the       area of the triangle ABC

28.  Find the area of the parallelogram whose vertices P, Q and R are (2, 1, 1),           (3, 2, 1) (2, 4, 1)

29.  Find the volume of the parallelepiped the edges are A (1, 0, 2) B (2, -1, 3) C (4, 1, 3) and D (1, -1, 1)

30.  Find the volume of the tetrahedron whose sides are  edu.uptymez.com = 2edu.uptymez.com + edu.uptymez.com, edu.uptymez.com = edu.uptymez.com – 3edu.uptymez.com ` + edu.uptymez.com, and edu.uptymez.com edu.uptymez.com – edu.uptymez.com+ 2edu.uptymez.com

Share this post on: