Scalar Product and Vector Product
Next we recall the scalar product and vector product of two vectors as follows.
Definition 6.1
Given two vectors = a1iˆ + a2ˆj + a3kˆ and
= b1iˆ + b2ˆj + b3kˆ the scalar product (or dot product) is denoted by
×
and is calculated by
×
= a1b1 + a2b2 + a3b3 ,
and the vector product (or cross product) is denoted by ´
, and is calculated by
Note
×
is a vector.
´
is a scalar, and
Geometrically, if is an arbitrary vector and nˆ is a unit vector, then
⋅ nˆ is the projection of the vector
on the straight line on which nˆ lies. The quantity
⋅ nˆ is positive if the angle between
and nˆ is acute, see Fig. 6.4 and negative if the angle between
and nˆ is obtuse see Fig. 6.5.
If and
are arbitrary non-zero vectors, then |
×
| =
and so |
×
| means either the length of the straight line segment obtained by projecting the vector |
|
along the direction of
or the length of the line segment obtained by projecting the vector |
|
along the direction of
. We recall that
×
=|
| |
| cosθ , where θ is the angle between the two vectors
and
. We recall that the angle between
and
is defined as the measure from
to
in the counter clockwise direction.
The vector ´
is either
or a vector perpendicular to the plane parallel to both
and
having magnitude as the area of the parallelogram formed by coterminus vectors parallel to
and
. If
and
are non-zero vectors, then the magnitude of
´ b can be calculated by the formula
| ´
| = | a | | b | | sinθ |, where θ is the angle between
and
.
Two vectors are said to be coterminus if they have same initial point.
Remark
(1) An angle between two non-zero vectors and
is found by the following formula
(2) and
are said to be parallel if the angle between them is 0 or π .
(3) and
are said to be perpendicular if the angle between them is
Property
(1) Let and
be any two nonzero vectors. Then
â‹…
= 0 if and only if
and
are perpendicular to each other.
×
=
if and only if
and
are parallel to each other.
(2) If ,
, and c are any three vectors and α is a scalar, then
We apply the concepts of dot and cross products of two vectors to derive a few formulae in plane trigonometry.
Example 6.1 (Cosine formulae)
With usual notations, in any triangle ABC, prove the following by vector method.
(i) a2 = b2 + c2 - 2bc cos A
(ii) b2 = c2 + a2 - 2ca cos B
(iii) c2 = a2 + b2 - 2ab cos C
Solution
With usual notations in triangle ABC, we have
a2 = b2 + c2 + 2bc cos(Ï€ - A)
a2 = b2 + c2 - 2bc cos A .
The results in (ii) and (iii) are proved in a similar way.
Example 6.2
With usual notations, in any triangle ABC, prove the following by vector method.
(i) a = b cos C + c cos B
(i) b = c cos A + a cos C
(iii) c = a cos B + b cos A
Solution
With usual notations in triangle ABC, we have = a,
=
, and
⇒ a2 = ab cos C + ac cos B
Therefore a = b cos C + c cos B . The results in (ii) and (iii) are proved in a similar way.
By vector method, prove that cos(α + β ) = cosα cos β − sin α sin β .
Let aˆ = and bˆ =
be the unit vectors and which make angles α and β , respectively, with positive x -axis, where A and B are as in the Fig. 6.8. Draw AL and BM perpendicular to the x-axis.
On the other hand, from (1) and (2)
aˆ ⋅ bˆ = (cosαiˆ − sin α ˆj) ⋅ (cos β iˆ + sin βˆj) = cosα cos β − sin α sin β ........(4)
From (3) and (4),
we get cos(α + β ) = cosα cos β − sin α sin β.
Example 6.4
With usual notations, in any triangle ABC, prove by vector method that
Solution
With usual notations in triangle ABC, we have
Example 6.5
Prove by vector method that sin(α − β ) = sin α cos β − cosα sin β .
Solution
Let aˆ = and
=
be the unit vectors making angles α and β respectively, with positive x -axis, where A and B are as shown in the Fig. 6.10. Then, we get aˆ = cosαiˆ + sin α ˆj and bˆ = cos β iˆ + sin β ˆj ,
The angle between aˆ and bˆ is α − β and, the vectors bˆ, aˆ, kˆ form a right-handed system.
Hence, we get
aˆ × bˆ | bˆ | | aˆ | sin(α - β )kˆ = sin(α - β )kˆ ………………(1)
On the other hand,
= (sin α cos β - cosα sin β )kˆ ... (2)
Hence, equations (1) and (2), leads to
sin(α - β ) = sin α cos β - cosα sin β .
Example 6.6 (Apollonius's theorem)
If D is the midpoint of the side BC of a triangle ABC, show by vector method that
Solution
Let A be the origin, be the position vector of B and
be the position vector of C. Now D is the midpoint of BC , and so the, position vector of D is
Therefore , we have
Example 6.7
Prove by vector method that the perpendiculars (attitudes) from the vertices to the opposite sides of a triangle are concurrent.
Solution
Consider a triangle ABC in which the two altitudes AD and BE intersect at O . Let CO be produced to meet AB at F. We take O as the origin and let
Since is perpendicular to
, we have
is perpendicular to
, and hence we get
.
= 0 . That is,
. (
-
) = 0 , which means
Similarly, since is perpendicular to
, we have
is perpendicular to
, and hence we get
Adding equations (1) and (2), gives â‹…
−
â‹…
= 0 . That is,
â‹… (
−
) = 0 .
That is, â‹…
= 0 . Therefore,
is perpendicular to
which implies that
is perpendicular to
. Hence, the perpendicular drawn from C to the side AB passes through O . Thus, the altitudes are concurrent.
In triangle ABC, the points D, E, F are the midpoints of the sides BC, CA , and AB respectively. Using vector method, show that the area of ΔDEF is equal to 1/4 (area of ΔABC).
In triangle ABC, coonsider A as the origin. Then the position vectors of D, E, F are given by
respectively. Since | × AC | is the area of the parallelogram formed by the two vectors
,
as adjacent sides, the area of ΔABC is 1/ 2 |
|
Similarly, considering ∆DEF , we have
If is the displacement vector of a particle moved from a point to another point after applying a constant force
on the particle, then the work done by the force on the particle is w =
â‹…
.
If the force has an acute angle, perpendicular angle, and an obtuse angle, the work done by the force is positive, zero, and negative respectively.
Example 6.9
A particle acted upon by constant forces 2iˆ + 5 ˆj + 6kˆ and −iˆ − 2 ˆj − kˆ is displaced from the point (4, −3, −2) to the point (6,1, −3) . Find the total work done by the forces.
Solution
Resultant of the given forces is = (2iˆ + 5 ˆj + 6kˆ) + (−iˆ − 2 ˆj − kˆ) = iˆ + 3 ˆj + 5kˆ.
Let A and B be the points (4, −3, −2) and (6,1, −3) respectively. Then the displacement vector of the particle is =
=
−
= (6iˆ + ˆj − 3kˆ) − (4iˆ − 3 ˆj − 2kˆ) = 2iˆ + 4 ˆj − kˆ .
Therefore the work done w = â‹…
= (iˆ + 3 ˆj + 5kˆ) ⋅(2iˆ + 4 ˆj − kˆ) = 9 units.
A particle is acted upon by the forces 3iˆ − 2 ˆj + 2kˆ and 2iˆ + ˆj − kˆ is displaced from the point (1, 3, −1) to the point (4, −1, λ) . If the work done by the forces is 16 units, find the value of λ .
Resultant of the given forces is = (3iˆ − 2 ˆj + 2kˆ) + (2iˆ + ˆj − kˆ) = 5iˆ − ˆj + kˆ .
The displacement of the particle is given by
= (4iˆ − ˆj + λkˆ) − (iˆ + 3 ˆj − kˆ) = (3iˆ − 4 ˆj + (λ +1)kˆ) .
As the work done by the forces is 16 units, we have
â‹…
= 16 .
That is, (5iˆ − ˆj + kˆ) ⋅ (3iˆ − 4 ˆj + (λ +1)kˆ = 16 ⇒ λ + 20 = 16 .
So, λ = −4
If a force is applied on a particle at a point with position vector
, then the torque or moment on the particle is given by
=
×
. The torque is also called the rotational force.
Example 6.11
Find the magnitude and the direction cosines of the torque about the point (2, 0, −1) of a force 2iˆ + ˆj − kˆ, whose line of action passes through the origin.
Let A be the point (2, 0, −1) . Then the position vector of A is = 2iˆ − ˆk and therefore
=
= −2ˆi + ˆk .
Then the given force is = 2iˆ + ˆj − kˆ . So, the torque is
The magnitude of the torque = | -iˆ - 2kˆ |= √5 and the direction cosines of the torque are -1/√5 , 0, -2/√5.
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