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Position Vector, a vector that is used to find the relative location of one object to another. In simple words, finding the location of an object relative to another object is done by a position vector. Usually, the position vector starts at the origin/center and ends at any other point in space. So simply, these Vectors are used to find the position of a point w.r.t. its origin. Further in this article, we will discuss position vectors, ways to find them, formulas with solved examples.
Read More: NCERT Solutions For Class 12 Mathematics Chapter 10 Vector Algebra
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Key Terms: Position vector, Vectors, Magnitude, Origins, Three-Dimensional Space, Vector algebra, Two-Dimensional Space, Three-Dimensional Space
Definition of Position Vector
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A position vector is specified as a vector that demonstrates either the position or the location of any given point with respect to any irregular reference point like the origin.
It’s just like a string whose one end is tied to the center and we’re moving the other end to find the relative position of any object.
Read More: Multiplication of a vector by a scalar
The direction of the position vector starts from the origin and points toward the given point.
Now, in a Two-Dimensional Space,
In the cartesian coordinate system, taking O and P(x1, y1) as the origin and another point respectively, then the required position vector which will start from point O to end at point P will be represented as OP.
And in a Three-Dimensional Space
Considering the origin O(0,0,0) and P(x1, y1, z1), then the required position vector will be represented as:
V = x1i + y1j + z1k
Read More: Vector product of two vectorsFinding The Position Vector
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Now, before learning how to find a position vector, let’s cover up some basics.
Let’s imagine that we have given two vectors, A and B and their position vectors are (2, 4) and (3, 5) respectively. Then, the coordinates of the given vectors A and B can be written as:
A = (2,4), B = (3, 5)
Let’s move forward,
For this, consider two points M and N, whose coordinates are M (x1, y1) and N (x2, y2). For finding the position vector of M and N, we will be subtracting their corresponding components and as we discussed, the resultant position vector will be written as:
MN = (x2-x1, y2-y1)
Read More: Types of vectors
The formula of Position Vector
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Now that we’ve learned the way to find a position vector between any two points in the XY plane if we know their positions. So, let’s look up the general formula that can be used to find a position vector between any two points.
For that, consider a point M with coordinates (xk, yk) and another point N with coordinates (xk+1, yk+1), and both are in the XY plane.
Then, the formula to find the position vector AB will be: (xk+1 – xk, yk+1 – yk).
Consider that this position vector AB is of a vector that started from A and ends at B.
Similarly, for the reversed case, if the vector starts from B and ends at A, then the vector position will be: BA = (xk – xk+1, yk – yk+1).
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Things to Remember
- A position vector is specified as a vector that demonstrates either the position or the location of any given point with respect to any irregular reference point like the origin.
- It’s just like a string whose one end is tied to the center and we’re moving the other end to find the relative position of any object.
- In the cartesian coordinate system, taking O and P(x1, y1) as the origin and another point respectively, then the required position vector which will start from point O to end at point P will be represented as OP.
- Considering the origin O(0,0,0) and P(x1, y1, z1), then the required position vector will be represented as:
- V = x1i + y1j + z1k
Read More: Addition of Vectors
Sample Questions
Ques: A = (-4, 6) and B = (5, 12) are the two points given. You have to find the position vector between ‘AB’. Then, determine the magnitude of the vector ‘AB’. (4 marks)
Ans: Two points that are given, lie in the xy-plane, so we can use the method we discussed above to find the position vector AB.
AB = (x2-x1, y2-y1)
Where x1, y1 are the coordinates of given point A and x2, y2 are the coordinates of given point B.
As we have the coordinates of each point, we can easily find the required position vector by putting the values in the above equation, as:
AB = (5-(-4), 12-6)
AB = ((5+ 4), 12-6)
AB = (9, 6)
We have found out that the required position vector AB is a vector that starts at the origin and ends at a point which is 9 units in the right along the x-axis and 6 units upward on the y-axis.
Further, to determine the magnitude of position vector AB.
|AB| = √92 + 62
|AB| = √81 + 36
|AB| = √117
|AB| = 3√13
Ques: The position vector of S1 is given as OS1 = (2, 3), and also of the vector S1S2 = (-3, 6), find the position vector of the point S2 and OS2. (4 marks)
Ans: Let’s start with plotting the vector OS1 which starts from (0, 0) and ends at (2, 3). Next, we will plot the vector OS2, which will start at (0, 0) and end at S2 (x, y).
As we’ve given the position vector S1S2, we will draw it as a vector that starts from S1 and heads to three units left on the x-axis and 3 units upward on the y-axis. Now we’ve got an image, which depicts the formation of a triangle, 0S1S2.
Now, we can use the triangle law (of the vector addition) to find the required position vector OS2:
S1S2 = OS1 + OS2
OS2 = S1S2 – OS1
Now putting the given values in the above equation, we will get,
OS2 = (-3, 6) – (2, 3)
OS2 = (-3, 6) + (-2, -3)
OS2 = (-3-2, 6-3)
OS2 = (-5, 3)
Thus, OS2=(-5, 3) is the required position vector.
Ques: M = (4, m) and Q = (-n, -3) are the two points given, fins the position vector QM. (3 marks)
Ans: The two given points M and N lies in the xy-plane, so we can the formula to find the position vector.
Position Vector = (x2 – x1), (y2 – y1)
Putting the given values in the equation we got,
QM = (-n-4, -3-m).
As the values of m and n are not given, we can not simplify the equation further.
Ques: A point q = (-10, 5, 3) is given, find the position vector of point q. Then, find out the magnitude of the position vector. (3 marks)
Ans: As a point q (-10, 5,3) which lies in xy-plane is given, we can simply write its position vector as:
Q = -10i + 5j -3k.
Next, to find the magnitude of Q, we can solve it as:
|R| = √(-10)2 + (5)2 + (-3)2
|R| = √100 + 25 + 9
|R| = √100 + 25 + 9
|R| = √134
Ques: c = 5i + 6j +3k and d = 2i +5j – 2k are the two given points in the orthogonal system, you have to find the position vector CD. (3 marks)
Ans: As we have two points that lie in the xy-plane, we can use the formula to determine its position vector,
Position Vector = (x2 – x1, y2 – y1, z2 – z1).
Putting given values in the above equation, we get
CD = (2-5, 5-5, -2-3)
CD = (-3, 0, -5)
CD = -3i + 0j -5k
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