Question
Question: The vector with initial point P (2, -3, 5) and terminal point Q (3, -4, 7) is (a) \(\hat{i}-\hat{j...
The vector with initial point P (2, -3, 5) and terminal point Q (3, -4, 7) is
(a) i^−j^+2k^
(b) 5i^−7j^+12k^
(c) i^+j^−2k^
(d) None of these
Solution
We must first find the position vectors OP and OQ, And then find the relationship between OP, OQ and PQ using the triangle law of addition. Then, with the help of this relation, we can easily find the distance vector PQ.
Complete step by step answer:
Let us assume a point A such that the coordinates of A in a three dimensional coordinate system is (a, b, c). We know that the origin is O (0, 0, 0). We can easily say that the vector starting from the origin O and terminating at point A is called the position vector of point A.
We all know very well that i^ is the unit vector in the direction of positive X-axis, j^ is the unit vector along the direction of positive Y-axis and k^ is the unit vector along the direction of positive Z-axis.
So, we can write the position vector of point A mathematically as
OA=ai^+bj^+ck^.
Similarly, let us assume another point B (d, e, f). The position vector of point B can be written as
OB=di^+ej^+fk^
We have shown the vectors OA, OB and AB in the figure below.
We know that by the triangle law of addition of vectors, we can write
OA+AB=OB.
Thus, we can write
AB=OB−OA
where, AB is the vector with A as starting point and B as terminating point.
Here, in this question, the starting point of this vector is P (2, -3, 5) and terminal point is Q (3, -4, 7). So, we need to replace A with P and B with Q.
So, by using the above concept, we can write
OP=2i^−3j^+5k^OQ=3i^−4j^+7k^
And, using the triangle law of addition, we can write
OP+PQ=OQ
Or, we can write
PQ=OQ−OP
Thus, we get
PQ=(3i^−4j^+7k^)−(2i^−3j^+5k^)
Hence, we now have PQ=i^−j^+2k^.
So, the correct answer is “Option a”.
Note: We know that the subtraction of two vectors OB=di^+ej^+fk^ and OA=ai^+bj^+ck^ is written as AB=(d−a)i^+(e−b)j^+(f−c)k^. WE must remember that the two vectors OA, and OB as shown in the figure, are called position vectors, and the vector AB is called the distance vector.