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Solving Exercise 1: Coordinates of vectors (C7 – Math 10 Kite)
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Solve the exercise Exercise 1, page 65, Math textbook 10 Kite, episode 2
Find the coordinates of the vectors in Figure 16 and represent each of them as two vectors \(\overrightarrow i , \overrightarrow j \)
Solution method
Draw the vectors \(\overrightarrow {OA} = \overrightarrow a ,\overrightarrow {OB} = \overrightarrow b ,\overrightarrow {OC} = \overrightarrow c ,\overrightarrow {OD} = \overrightarrow d \). The coordinates of 4 points A, B, C, D are the coordinates of 4 vectors.
To represent vectors over unit vectors: \(\overrightarrow u {\rm{ }} = {\rm{ }}\left( {a;{\rm{ }}b} \right) \Leftrightarrow \overrightarrow u = a\overrightarrow i + b\overrightarrow j \)
Solution guide
a) Draw the vectors \(\overrightarrow {OA} = \overrightarrow a ,\overrightarrow {OB} = \overrightarrow b ,\overrightarrow {OC} = \overrightarrow c ,\overrightarrow {OD} = \overrightarrow d \)
Based on the figure, we see that the coordinates of 4 points A, B, C, D are:
\(A\left( { – 5; – 3} \right),B\left( {3; – 4} \right),C\left( { – 1;3} \right),D\left( { 2;5} \right)\)
Thus \(\overrightarrow a = \overrightarrow {OA} = \left( { – 5; – 3} \right),\overrightarrow b = \overrightarrow {OB} = \left( {3; – 4} \right) ,\overrightarrow c = \overrightarrow {OC} = \left( { – 1;3} \right),\overrightarrow d = \overrightarrow {OD} = \left( {2;5} \right)\)
b) Since \(\overrightarrow a = \overrightarrow {OA} = \left( { – 5; – 3} \right)\)\(\overrightarrow a = \left( { – 5} \right)\overrightarrow i + \left( { – 3} \right)\overrightarrow j = – 5\overrightarrow i – 3\overrightarrow j \)
Since \(\overrightarrow b = \overrightarrow {OB} = \left( {3; – 4} \right)\) \(\overrightarrow b = 3\overrightarrow i + \left( { – 4} \right)\ overrightarrow j = 3\overrightarrow i – 4\overrightarrow j \)
Since \(\overrightarrow c = \overrightarrow {OC} = \left( { – 1;3} \right)\) \(\overrightarrow c = \left( { – 1} \right)\overrightarrow i + \left ( 3 \right)\overrightarrow j = – \overrightarrow i + 3\overrightarrow j \)
Since \(\overrightarrow d = \overrightarrow {OD} = \left( {2;5} \right)\) \(\overrightarrow d = 2\overrightarrow i + 5\overrightarrow j \)
Solve the exercise Exercise 2, page 65, Math textbook 10 Kite, episode 2
Find the coordinates of the following vectors:
a) \(\overrightarrow a = 3\overrightarrow i \) b) \(\overrightarrow b = – \overrightarrow j \)
c) \(\overrightarrow c = \overrightarrow i – 4\overrightarrow j \) d) \(\overrightarrow d = 0.5\overrightarrow i + \sqrt 6 \overrightarrow j \)
Solution method
Use: \(\overrightarrow u {\rm{ }} = {\rm{ }}\left( {a;{\rm{ }}b} \right) \Leftrightarrow \overrightarrow u = a\overrightarrow i + b \overrightarrow j \)
Solution guide
a) Since \(\overrightarrow a = 3\overrightarrow i \) so \(\overrightarrow a = \left( {3;0} \right)\)
b) Since \(\overrightarrow b = – \overrightarrow j \) so \(\overrightarrow b = \left( {0; – 1} \right)\)
c) Since \(\overrightarrow c = \overrightarrow i – 4\overrightarrow j \) so \(\overrightarrow c = \left( {1; – 4} \right)\)
d) Since \(\overrightarrow d = 0.5\overrightarrow i + \sqrt 6 \overrightarrow j \) so \(\overrightarrow d = \left( {0.5;\sqrt 6 } \right)\)
Solve the exercises Lesson 3, page 65, Math textbook 10 Kite, episode 2
Find pairs of real numbers a and b such that each of the following vector pairs are equal:
a) \(\overrightarrow u = \left( {2a – 1; – 3} \right)\) and \(\overrightarrow v = \left( {3;4b + 1} \right)\)
b) \(\overrightarrow x = \left( {a + b; – 2a + 3b} \right)\) and \(\overrightarrow y = \left( {2a – 3;4b} \right)\)
Solution method
With \(\overrightarrow a = \left( {{x_1};{y_1}} \right)\) and \(\overrightarrow b = \left( {{x_2},{y_2}} \right)\) , we yes: \(\overrightarrow a = \overrightarrow b \Leftrightarrow \left\{ \begin{array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.\ )
Solution guide
a) Let \(\overrightarrow u = \overrightarrow v \Leftrightarrow \left\{ \begin{array}{l}2a – 1 = 3\\ – 3 = 4b + 1\end{array} \right. \Leftrightarrow \ left\{ \begin{array}{l}a = 2\\b = – 1\end{array} \right.\)
So \(\left\{ \begin{array}{l}a = 2\\b = – 1\end{array} \right.\) then \(\overrightarrow u = \overrightarrow v \)
b) \(\overrightarrow x = \overrightarrow y \Leftrightarrow \left\{ \begin{array}{l}a + b = 2a – 3\\ – 2a + 3b = 4b\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = – 2\end{array} \right.\)
So \(\left\{ \begin{array}{l}a = 1\\b = – 2\end{array} \right.\) then \(\overrightarrow x = \overrightarrow y \)
Solution of Exercises Lesson 4, page 66, Math Textbook 10, Kite episode 2
In the coordinate plane Oxy, give three points A(2;3), B(-1; 1), C(3;- 1).
a) Find the coordinates of point M such that \(\overrightarrow {AM{\rm{ }}} = {\rm{ }}\overrightarrow {BC} \) .
b) Find the coordinates of the midpoint N of the line segment AC. Prove \(\overrightarrow {BN} {\rm{ }} = {\rm{ }}\overrightarrow {NM} \) .
Solution method
With \(\overrightarrow a = \left( {{x_1};{y_1}} \right)\) and \(\overrightarrow b = \left( {{x_2},{y_2}} \right)\) , we yes: \(\overrightarrow a = \overrightarrow b \Leftrightarrow \left\{ \begin{array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.\ )
Solution guide
a) Call \(M\left( {a;b} \right) \Rightarrow \overrightarrow {AM} = \left( {a – 2;b – 3} \right)\)
adsense
Vector coordinates \(\overrightarrow {BC} = \left( {4; – 2} \right)\)
Let \(\overrightarrow {AM{\rm{ }}} = {\rm{ }}\overrightarrow {BC} \Leftrightarrow \left\{ \begin{array}{l}a – 2 = 4\\b – 3 = – 2\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a = 6\\b = 1\end{array} \right.\)
So for \(\overrightarrow {AM{\rm{ }}} = {\rm{ }}\overrightarrow {BC} \) then the coordinates of point M are:\(M\left( {6;1} \right) \)
b) Call \(N\left( {x,y} \right) \Rightarrow \overrightarrow {NC} = \left( {3 – x, – 1 – y} \right)\)and \(\overrightarrow {AC) } = \left( {x – 2,y – 3} \right)\)
Since N is the midpoint of AC, \(\overrightarrow {AC} = \overrightarrow {NC} \Leftrightarrow \left\{ \begin{array}{l}x – 2 = 3 – x\\y – 3 = – 1 – y\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}x = \frac{5}{2}\\y = 1\end{array} \right.\) . So \(N\left( {\frac{5}{2},1} \right)\)
We have: \(\overrightarrow {BN} {\rm{ }} = \left( { – \frac{7}{2};0} \right)\) and \(\overrightarrow {NM} = \left( {\frac{{ – 7}}{2};0} \right)\). So \(\overrightarrow {BN} {\rm{ }} = {\rm{ }}\overrightarrow {NM} \)
Solve exercises Exercise 5 page 66 Math textbook 10 Kite episode 2
In the coordinate plane Oxy, give the point M(-1; 3).
a) Find the coordinates of point A symmetrical to point M through the origin O.
b) Find the coordinates of point B symmetrical to point M about the Ox axis.
c) Find the coordinates of point C symmetrical to point M about the axis Oy.
Solution method
Draw a picture and then based on the figure determine the points
Solution guide
a) Based on the figure, we see that \(A\left( {1; – 3} \right)\)
b) Based on the figure, we see that \(B\left( { – 1; – 3} \right)\)
c) Based on the figure, we see that \(C\left( {1;3} \right)\)
Solution of Exercises Lesson 6, page 66, Math Textbook 10, Kite episode 2
In the coordinate plane Oxy, give three noncollinear points A(- 3 ; 1), B(-1; 3), I(4;2). Find the coordinates of two points C and D such that quadrilateral ABCD is a parallelogram with I as the center of symmetry.
Solution method
The center of symmetry of a parallelogram is the midpoint of the two diagonals.
With \(\overrightarrow a = \left( {{x_1};{y_1}} \right)\) and \(\overrightarrow b = \left( {{x_2},{y_2}} \right)\) , we yes: \(\overrightarrow a = \overrightarrow b \Leftrightarrow \left\{ \begin{array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.\ )
Solution guide
Call \(C\left( {a;b} \right),D\left( {m,n} \right) \Rightarrow \overrightarrow {IC} = \left( {a – 4,b – 2} \right )\) and \(\overrightarrow {ID} = \left( {m – 4,n – 2} \right)\)
Since I is the center of parallelogram ABCD, I is the midpoints of AC and BD.
So we have:\(\overrightarrow {AI} = \overrightarrow {IC} \) and \(\overrightarrow {BI} = \overrightarrow {ID} \)
We have: \(\overrightarrow {AI} = \left( {7;1} \right)\) and \(\overrightarrow {BI} = \left( {5; – 1} \right)\)
Do \(\overrightarrow {AI} = \overrightarrow {IC} \Leftrightarrow \left\{ \begin{array}{l}7 = a – 4\\1 = b – 2\end{array} \right. \Leftrightarrow. \left\{ \begin{array}{l}a = 11\\b = 3\end{array} \right.\) .So \(C\left( {11;3} \right)\)
Do \(\overrightarrow {BI} = \overrightarrow {ID} \Leftrightarrow \left\{ \begin{array}{l}5 = m – 4\\ – 1 = n – 2\end{array} \right. \ Leftrightarrow \left\{ \begin{array}{l}m = 9\\n = 1\end{array} \right.\). So \(D\left( {9;1} \right)\)
Solve exercises Exercise 7 page 66 Math textbook 10 Kite episode 2
In the coordinate plane Oxy, let ABC be a triangle. The points M(1;- 2), N(4;- 1) and P(6 ; 2) are the midpoints of the sides BC, CA, AB, respectively. Find the coordinates of the points A, B, C.
Solution method
The moving average is parallel to and one-half of the corresponding bottom edge
With \(\overrightarrow a = \left( {{x_1};{y_1}} \right)\) and \(\overrightarrow b = \left( {{x_2},{y_2}} \right)\) , we yes: \(\overrightarrow a = \overrightarrow b \Leftrightarrow \left\{ \begin{array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.\ )
Solution guide
By product of the mean lines in a triangle we have: \(\overrightarrow {PN} = \overrightarrow {BM} = \overrightarrow {MC} \) and \(\overrightarrow {MP} = \overrightarrow {NA} \)
Call \(A\left( {{a_1},{a_2}} \right),B\left( {{b_1};{b_2}} \right),C\left( {{c_1};{c_2}} \right)\)
We have: \(\overrightarrow {PN} = \left( {2;3} \right)\),\(\overrightarrow {BM} = \left( {1 – {b_1}; – 2 – {b_2}} \right)\), \(\overrightarrow {MC} = \left( {{c_1} – 1;{c_2} + 2} \right)\), \(\overrightarrow {MP} = \left( {5; 4} \right)\), \(\overrightarrow {NA} = \left( {{a_1} – 4;{a_2} + 1} \right)\)
Yes \(\overrightarrow {PN} = \overrightarrow {BM} \Leftrightarrow \left\{ \begin{array}{l}2 = 1 – {b_1}\\3 = – 2 – {b_2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{b_1} = – 1\\{b_2} = – 5\end{array} \right.\) .So \(B\left( { {) – 1; – 5} \right)\)
Yes \(\overrightarrow {PN} = \overrightarrow {MC} \Leftrightarrow \left\{ \begin{array}{l}2 = {c_1} – 1\\3 = {c_2} + 2\end{array} \ right. \Leftrightarrow \left\{ \begin{array}{l}{c_1} = 3\\{c_2} = 1\end{array} \right.\) .So \(C\left( {3;1 } \right)\)
Yes \(\overrightarrow {NA} = \overrightarrow {MP} \Leftrightarrow \left\{ \begin{array}{l}5 = {a_1} – 4\\4 = {a_2} + 1\end{array} \ right.\Leftrightarrow \left\{ \begin{array}{l}{a_1} = 9\\{a_2} = 3\end{array} \right.\) .So \(A\left( {9;3 } \right)\)
