Answer
Verified
434.1k+ views
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 \[\overrightarrow {{\text{PQ}}} \]
Initial point of the vector is \[P(1,2)\]
x-component of the vector is \[x = - 2\]
y-component of the vector is \[y = 3\]
Therefore, the vector \[\overrightarrow {{\text{PQ}}} \] can be written as, \[\overrightarrow {{\text{PQ}}} = - 2\widehat {\text{i}} + 3\widehat {\text{j}}\] (i)
Let the terminal point be \[Q(x,y)\]
Let us draw a diagram to understand the problem properly,
Now, observing the figure we can write the vector \[\overrightarrow {{\text{PQ}}} \] in terms of their initial and final position vectors as,
\[\overrightarrow {{\text{PQ}}} = \overrightarrow {\text{Q}} - \overrightarrow {\text{P}} \] (ii)
The initial point \[P(1,2)\] can be written in vector form as,
\[\overrightarrow {\text{P}} = 1\widehat {\text{i}} + 2\widehat {\text{j}}\] (iii)
The final point \[Q(x,y)\] can be written in vector form as,
\[\overrightarrow {\text{Q}} = x\widehat {\text{i}} + y\widehat {\text{j}}\] (iv)
Now, putting the values from equation (iii) and (iv) in equation (ii), we get
\[\overrightarrow {{\text{PQ}}} = \left( {x\widehat {\text{i}} + y\widehat {\text{j}}} \right) - \left( {1\widehat {\text{i}} + 2\widehat {\text{j}}} \right) \\
\Rightarrow \overrightarrow {{\text{PQ}}} = (x - 1)\widehat {\text{i}} + (y - 2)\widehat {\text{j}} \]
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.
Complete step-by-step answer:
Given a vector \[\overrightarrow {{\text{PQ}}} \]
Initial point of the vector is \[P(1,2)\]
x-component of the vector is \[x = - 2\]
y-component of the vector is \[y = 3\]
Therefore, the vector \[\overrightarrow {{\text{PQ}}} \] can be written as, \[\overrightarrow {{\text{PQ}}} = - 2\widehat {\text{i}} + 3\widehat {\text{j}}\] (i)
Let the terminal point be \[Q(x,y)\]
Let us draw a diagram to understand the problem properly,
Now, observing the figure we can write the vector \[\overrightarrow {{\text{PQ}}} \] in terms of their initial and final position vectors as,
\[\overrightarrow {{\text{PQ}}} = \overrightarrow {\text{Q}} - \overrightarrow {\text{P}} \] (ii)
The initial point \[P(1,2)\] can be written in vector form as,
\[\overrightarrow {\text{P}} = 1\widehat {\text{i}} + 2\widehat {\text{j}}\] (iii)
The final point \[Q(x,y)\] can be written in vector form as,
\[\overrightarrow {\text{Q}} = x\widehat {\text{i}} + y\widehat {\text{j}}\] (iv)
Now, putting the values from equation (iii) and (iv) in equation (ii), we get
\[\overrightarrow {{\text{PQ}}} = \left( {x\widehat {\text{i}} + y\widehat {\text{j}}} \right) - \left( {1\widehat {\text{i}} + 2\widehat {\text{j}}} \right) \\
\Rightarrow \overrightarrow {{\text{PQ}}} = (x - 1)\widehat {\text{i}} + (y - 2)\widehat {\text{j}} \]
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.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
A rainbow has circular shape because A The earth is class 11 physics CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Find the value of the expression given below sin 30circ class 11 maths CBSE
What is the length of the alimentary canal in human class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What is BLO What is the full form of BLO class 8 social science CBSE