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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^\hat{i}-\hat{j}+2\hat{k}
(b) 5i^7j^+12k^5\hat{i}-7\hat{j}+12\hat{k}
(c) i^+j^2k^\hat{i}+\hat{j}-2\hat{k}
(d) None of these

Explanation

Solution

We must first find the position vectors OP and OQ\overrightarrow{OP}\text{ and }\overrightarrow{OQ}, And then find the relationship between OP, OQ and PQ\overrightarrow{OP},\text{ }\overrightarrow{OQ}\text{ and }\overrightarrow{PQ} using the triangle law of addition. Then, with the help of this relation, we can easily find the distance vector PQ\overrightarrow{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^\hat{i} is the unit vector in the direction of positive X-axis, j^\hat{j} is the unit vector along the direction of positive Y-axis and k^\hat{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^\overrightarrow{OA}=a\hat{i}+b\hat{j}+c\hat{k}.
Similarly, let us assume another point B (d, e, f). The position vector of point B can be written as
OB=di^+ej^+fk^\overrightarrow{OB}=d\hat{i}+e\hat{j}+f\hat{k}
We have shown the vectors OA, OB and AB\overrightarrow{OA},\text{ }\overrightarrow{OB}\text{ and }\overrightarrow{AB} in the figure below.
We know that by the triangle law of addition of vectors, we can write
OA+AB=OB\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}.
Thus, we can write
AB=OBOA\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}
where, AB\overrightarrow{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^ \begin{aligned} & \overrightarrow{OP}=2\hat{i}-3\hat{j}+5\hat{k} \\\ & \overrightarrow{OQ}=3\hat{i}-4\hat{j}+7\hat{k} \\\ \end{aligned}
And, using the triangle law of addition, we can write
OP+PQ=OQ\overrightarrow{OP}+\overrightarrow{PQ}=\overrightarrow{OQ}
Or, we can write
PQ=OQOP\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}
Thus, we get
PQ=(3i^4j^+7k^)(2i^3j^+5k^)\overrightarrow{PQ}=\left( 3\hat{i}-4\hat{j}+7\hat{k} \right)-\left( 2\hat{i}-3\hat{j}+5\hat{k} \right)
Hence, we now have PQ=i^j^+2k^\overrightarrow{PQ}=\hat{i}-\hat{j}+2\hat{k}.
So, the correct answer is “Option a”.

Note: We know that the subtraction of two vectors OB=di^+ej^+fk^\overrightarrow{OB}=d\hat{i}+e\hat{j}+f\hat{k} and OA=ai^+bj^+ck^\overrightarrow{OA}=a\hat{i}+b\hat{j}+c\hat{k} is written as AB=(da)i^+(eb)j^+(fc)k^\overrightarrow{AB}=\left( d-a \right)\hat{i}+\left( e-b \right)\hat{j}+\left( f-c \right)\hat{k}. WE must remember that the two vectors OA, and OB\overrightarrow{OA},\text{ and }\overrightarrow{OB} as shown in the figure, are called position vectors, and the vector AB\overrightarrow{AB} is called the distance vector.