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Question: Find the terminal point of the vector \[\overrightarrow {{\text{PQ}}} \] whose initial point is \[P(...

Find the terminal point of the vector PQ→\overrightarrow {{\text{PQ}}} whose initial point is P(1,2)P(1,2) and whose components along x-axis and y-axis are −2 - 2 and 33 respectively.

Explanation

Solution

Hint : We are asked to find the terminal or end point of the given vector. The x-component and y-component is given, using those from the vector. The initial point is also given use this point and the vector you formed using the given components to find the end point.

Complete step-by-step answer :
Given a vector PQ→\overrightarrow {{\text{PQ}}}
Initial point of the vector is P(1,2)P(1,2)
x-component of the vector is x=−2x = - 2
y-component of the vector is y=3y = 3
Therefore, the vector PQ→\overrightarrow {{\text{PQ}}} can be written as, PQ→=−2i^+3j^\overrightarrow {{\text{PQ}}} = - 2\widehat {\text{i}} + 3\widehat {\text{j}} (i)
Let the terminal point be Q(x,y)Q(x,y)
Let us draw a diagram to understand the problem properly,

Now, observing the figure we can write the vector PQ→\overrightarrow {{\text{PQ}}} in terms of their initial and final position vectors as,
PQ→=Q→−P→\overrightarrow {{\text{PQ}}} = \overrightarrow {\text{Q}} - \overrightarrow {\text{P}} (ii)
The initial point P(1,2)P(1,2) can be written in vector form as,
P→=1i^+2j^\overrightarrow {\text{P}} = 1\widehat {\text{i}} + 2\widehat {\text{j}} (iii)
The final point Q(x,y)Q(x,y) can be written in vector form as,
Q→=xi^+yj^\overrightarrow {\text{Q}} = x\widehat {\text{i}} + y\widehat {\text{j}} (iv)
Now, putting the values from equation (iii) and (iv) in equation (ii), we get

\Rightarrow \overrightarrow {{\text{PQ}}} = (x - 1)\widehat {\text{i}} + (y - 2)\widehat {\text{j}} $$ Now, putting the value of $$\overrightarrow {{\text{PQ}}} $$ from equation (i), we get $$ - 2\widehat {\text{i}} + 3\widehat {\text{j}} = (x - 1)\widehat {\text{i}} - (y - 2)\widehat {\text{j}}$$ Now, equating x-component we get, $$- 2 = x - 1 \\\ \Rightarrow x = - 1 $$ Equating y-component we get, $$3 = y - 2 \\\ \Rightarrow y = 5 $$ Therefore, the terminal point is $$Q( - 1,5)$$ **So, the correct answer is “ $$Q( - 1,5)$$ ”.** **Note** : A vector has a starting point and an ending or terminal point. The vector can be formed by subtracting the terminal position vector from the initial position vector. A vector has both magnitude and direction. Suppose we have a vector $$\overrightarrow {\text{A}} = a\widehat i + b\widehat j$$ then its magnitude can be written as, $${\text{A}} = \sqrt {{a^2} + {b^2}} $$ . Always remember these basic concepts of vectors as you may have to use these in solving many problems.