Resolution of Vectors
Resolution of a vector can be done for any finite dimension. But we will discuss only in two and three dimensions. Let us start with two dimension.
Let iˆ and ˆj be the unit vectors along the positive x-axis and the y-axis having initial point at the origin O. Now is the position vector of any point P in the plane. Then can be uniquely written as
= xiˆ + y jˆ for some real numbers x and y. Further
Let (x, y) be the coordinates of the point P. Let L and M be the foots of the perpendiculars drawn from P to the x and y axes. Then
To prove the uniqueness, let x1iˆ + y1 ˆj and x2 iˆ + y2 ˆj be two representations of the same point P. Then
x1iˆ + y1jˆ = x2iˆ + y2jˆ .
This implies that
(x1 − x2 )iˆ − ( y2− y1 ) ˆj = ⇒ x1 − x2 = 0, y2 − y1 = 0.
In other words x1 = x2 and y1 = y2 and hence the uniqueness follows.
In the triangle OLP, OP2 = OL2 + LP2 ;
hence | |= √[ x2 + y2 ].
Observe that if iˆ + jˆ are the unit vectors in the postive directions of x and y axes, then the iˆ and jˆ position vector of the point (6,4) can be written as 6iˆ + 4 jˆ and this is the only way of writing it.
If and are two non-collinear vectors in a plane, then any vector in the plane can be written as the linear combination of and in a unique way. That is, any vector in the plane is of the form l + m for some scalars l and m.
Further if three non collinear vectors are coplanar then any one of the vector can be written as a linear combination of other two. Note that the converse is also true.
If , and are three non-coplanar vectors in the space, then any vector in the space can be written as l + m + n in a unique way for some scalars l, m and n.
Let iˆ and ˆj be the unit vectors in the positive directions of x and y axes respectively. Let be any vector in the plane. Then = xiˆ + y ˆj for some real numbers x and y. Here xiˆ and y ˆj are called the rectangular components of along the x and y axes respectively in two dimension.
What we discussed so far can be discussed in the three dimensional space also.
Let iˆ, ˆj and kˆ be the unit vectors in the direction of postive x, y and z axes respectively having initial point at the origin O. Let be the position vector of any point P in the space. Then can be uniquely written as = xiˆ + y ˆj + zkˆ for some real numbers x, y and z. Further | | = √ [ x 2+ y 2 + z2 ].
Let (x, y, z) be the coordinates of the point P. Let Q be the foot of the perpendicular drawn from P to the xy-plane. Let R and S be the foots of the perpendiculars drawn from Q to the x and y axes respectively. Let = .
Then, OR = x, OS = y, and QP = z. z
Let us find the components of the vector joining the point (x1 , y1 ) to (x2 , y2 ).
Let A and B be the points (x1 , y1 ) and (x2 , y2 ). Let P be the point (x2 − x1 , y2 − y1 ). Then =.
The components of are (x2 − x1 )iˆ and ( y2 − y ) jˆ. Hence the components of in the directions of x and y axes are (x2 − x1 )iˆ ( y2 − y1 ) jˆ.
Similarly if A and B are the points (x1 , y1 , z1 ) and (x2 , y2 , z2 ), then the components of in the directions of x, y and z axes are (x2 − x1)iˆ ( y2 − y1 ) jˆ (z2 − z1 )kˆ .
A vector with three components can be visualised as either a row or column matrix as [x, y, z] or respectively.
Thus any vector = a1iˆ + a2 jˆ + a3kˆ can be obtained from
Hence addition of vectors and multiplication of a vector by a scalar can be defined as follows.
k = ka1iˆ + ka2 ˆj + ka3 kˆ
For k ∈ R, k > 1 yields magnification, 0 < k < 1 yields contraction of a vector and k = 0 yields a zero vector = 0iˆ + 0 ˆj + 0kˆ = .
Using the commutative, associative properties of vector addition and the distributive property of the scalar multiplication we can prove the following.
Find a unit vector along the direction of the vector 5iˆ − 3 ˆj + 4kˆ.
Now we have another unit vector parallel to 5iˆ − 3 ˆj + 4kˆ in the opposite direction. That is,