17. Angle between two vectors
18. The scalar or dot product
19. Geometrical interpretation of scalar product
20. Properties of scalar product
Property 1 :
The scalar product of two vectors is commutative
a^{v}.b^{v} = b^{v}.a^{v}
Property 2 : Scalar Product of Collinear Vectors :
(i) When the vectors a^{v} and b^{v} are collinear and are in the same direction, then θ = 0
a^{v}.b^{v} = |a^{v}| |b^{v}| = ab
(i) When the vectors a^{v} and b^{v} are collinear and are in the opposite direction, then θ = π
a^{v}.b^{v} = |a^{v}| |b^{v}|(-1) = -ab
Property 3 : Sign of Dot Product
The dot product a^{v}.b^{v} may be positive or negative or zero.
(i) If the angle between the two vectors is acute (i.e., 0 < θ < 90°) then
cos θ is positive. In this case dot product is positive.
(ii) If the angle between the two vectors is obtuse (i.e., 90 < θ < 180) then
cos θ is negative. In this case dot product is negative.
(iii) If the angle between the two vectors is 90° (i.e., θ = 90°) then
cos θ = cos 90° = 0. In this case dot product is zero.
21. scalar product in terms of components
If a = a1i+a2j+a3k and
b= b1i+b2j+b3k
then a.b = a1b1+a2b2+a3b3
22. Angle between two vectors
If θ is the angle between two vectors,
cos θ = a.b/|a||b|
=> θ = cos^{-1} (a.b/|a||b|)
In component form
If a = a1i+a2j+a3k and
b= b1i+b2j+b3k
θ = cos^{-1}[(a1b1+a2b2+a3b3)/(SQRT(a1²+a2²+a3²)*SQRT(b1²+b2²+b3²))
23. Components of a vector b along and perpendicular to vector a
Component of vector b along vector a == (a.b/|a|²)aComponent of vector b perpendicular to vector a = b- (a.b/|a|²)a
24. Tetrahedron
A tetrahedron is a three dimensional figure formed by four triangles. A tetrahedron in which all edges are equal, is called a regular tetrahedron.
This section illustrates the use of scalar product in proving some relations between edges of tetrahedron.
The relation mentioned are:
1. If two pairs of opposite edges of a tetrahedron are perpendicular, then prove that the opposite edges of the third pair are also perpendicular to each other.
2. Show that, in a tetrahedron, the sum of the squares of two opposite edges is the same for each pair.
3. Any two opposite edges in a regular tetrahedron are penpendicular
25 Application of scalar product in mechanics to find the work done
Word done by a force is a scalar quantity. This can be calculated by finding the scalar product of force vector and displacement vector.
If the force vector is represented by OA and the displacement vector by OB and if the angle between them is θ work done is
W = OA.OB = |OA||OB|cos θ
|OB|cos θ represents the component of OB in the direction of OA
If a number of forces are acting on a particle, then the sum of works done by the separate forces is equal to the work done by the resultant force.
Example: Forces represented by 6i+2j+3k and 3i-2j+6k respectively act on a particle which gets displaced from the point (2,2,-1) to (4,3,1). Find the work done by the forces.
Resultant force F = (6i+2j+3k) + (3i-2j+6k) = (9i+9k)
Displacement d = (4i+3j+k) – (2i+2j-k) = 2i+j+2k
Total work done = (9i+9k). (2i+j+2k)
= 9*2 +9*2 = 18 +18 = 36 units
No comments:
Post a Comment