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Sagot :
### a) (i) Find the coordinates of [tex]\(N\)[/tex] and [tex]\(M\)[/tex]
Coordinates of [tex]\(N\)[/tex]:
The point [tex]\(N\)[/tex] is the midpoint of the line segment [tex]\(OA\)[/tex].
Given:
[tex]\[ O = \binom{0}{0} \][/tex]
[tex]\[ A = \binom{-8}{5} \][/tex]
The midpoint [tex]\(N\)[/tex] is calculated as:
[tex]\[ N = \frac{O + A}{2} = \frac{\binom{0}{0} + \binom{-8}{5}}{2} = \binom{\frac{0 + (-8)}{2}}{\frac{0 + 5}{2}} = \binom{-4}{2.5} \][/tex]
So, the coordinates of [tex]\(N\)[/tex] are [tex]\(\binom{-4}{2.5}\)[/tex].
Coordinates of [tex]\(M\)[/tex]:
The point [tex]\(M\)[/tex] is the midpoint of the line segment [tex]\(AB\)[/tex].
Given:
[tex]\[ A = \binom{-8}{5} \][/tex]
[tex]\[ B = \binom{12}{-5} \][/tex]
The midpoint [tex]\(M\)[/tex] is calculated as:
[tex]\[ M = \frac{A + B}{2} = \frac{\binom{-8}{5} + \binom{12}{-5}}{2} = \binom{\frac{-8 + 12}{2}}{\frac{5 + (-5)}{2}} = \binom{2}{0} \][/tex]
So, the coordinates of [tex]\(M\)[/tex] are [tex]\(\binom{2}{0}\)[/tex].
---
### a) (ii) Find the magnitude of [tex]\(NM\)[/tex]
Vector [tex]\(NM\)[/tex]:
First, we find vector [tex]\(NM\)[/tex]:
[tex]\[ N = \binom{-4}{2.5} \][/tex]
[tex]\[ M = \binom{2}{0} \][/tex]
[tex]\[ NM = M - N = \binom{2 - (-4)}{0 - 2.5} = \binom{6}{-2.5} \][/tex]
Magnitude of [tex]\(NM\)[/tex]:
The magnitude of a vector [tex]\(\binom{a}{b}\)[/tex] is given by [tex]\( \sqrt{a^2 + b^2} \)[/tex]:
[tex]\[ ||NM|| = \sqrt{6^2 + (-2.5)^2} = \sqrt{36 + 6.25} = \sqrt{42.25} = 6.5 \][/tex]
So, the magnitude of [tex]\(NM\)[/tex] is [tex]\(6.5\)[/tex].
---
### b) Express vector [tex]\(NM\)[/tex] in terms of [tex]\(OB\)[/tex]
Vector [tex]\(OB\)[/tex]:
Given:
[tex]\[ O = \binom{0}{0} \][/tex]
[tex]\[ B = \binom{12}{-5} \][/tex]
[tex]\[ OB = B - O = \binom{12}{-5} - \binom{0}{0} = \binom{12}{-5} \][/tex]
We know [tex]\( NM = \binom{6}{-2.5} \)[/tex].
The vector [tex]\(NM\)[/tex] in terms of [tex]\(OB\)[/tex] is:
[tex]\[ NM = \frac{\binom{6}{-2.5}}{|OB|} \cdot OB \][/tex]
where [tex]\( |OB| \)[/tex] is the magnitude of [tex]\(OB\)[/tex]:
[tex]\[ |OB| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \][/tex]
[tex]\[ NM = \frac{\binom{6}{-2.5}}{13} \][/tex]
So, [tex]\( NM = \frac{6}{13} \binom{12}{-5} \)[/tex].
Thus, vector [tex]\(NM\)[/tex] in terms of [tex]\(OB\)[/tex] is [tex]\(\frac{6}{13} OB\)[/tex].
---
### c) Given [tex]\(OP = OM + 2MN\)[/tex], find the coordinates of [tex]\(P'\)[/tex].
Given:
[tex]\[ OP = OM + 2 MN \][/tex]
First, we find [tex]\(OM\)[/tex]:
[tex]\[ OM = M - O = \binom{2}{0} - \binom{0}{0} = \binom{2}{0} \][/tex]
Next, we find [tex]\(MN\)[/tex]:
[tex]\[ MN = N - M = \binom{-4}{2.5} - \binom{2}{0} = \binom{-4 - 2}{2.5 - 0} = \binom{-6}{2.5} \][/tex]
Given:
[tex]\[ OP = OM + 2 MN \][/tex]
Substitute the values:
[tex]\[ OP = \binom{2}{0} + 2 \binom{-6}{2.5} = \binom{2}{0} + \binom{-12}{5} = \binom{2 - 12}{0 + 5} = \binom{-10}{5} \][/tex]
Lastly, apply the translation to find [tex]\(P'\)[/tex]:
[tex]\[ \text{Translation} = \binom{-5}{8} \][/tex]
[tex]\[ P' = OP + \text{Translation} = \binom{-10}{5} + \binom{-5}{8} = \binom{-10 - 5}{5 + 8} = \binom{-15}{13} \][/tex]
So, the coordinates of [tex]\(P'\)[/tex] are [tex]\(\binom{9}{3}\)[/tex].
Coordinates of [tex]\(N\)[/tex]:
The point [tex]\(N\)[/tex] is the midpoint of the line segment [tex]\(OA\)[/tex].
Given:
[tex]\[ O = \binom{0}{0} \][/tex]
[tex]\[ A = \binom{-8}{5} \][/tex]
The midpoint [tex]\(N\)[/tex] is calculated as:
[tex]\[ N = \frac{O + A}{2} = \frac{\binom{0}{0} + \binom{-8}{5}}{2} = \binom{\frac{0 + (-8)}{2}}{\frac{0 + 5}{2}} = \binom{-4}{2.5} \][/tex]
So, the coordinates of [tex]\(N\)[/tex] are [tex]\(\binom{-4}{2.5}\)[/tex].
Coordinates of [tex]\(M\)[/tex]:
The point [tex]\(M\)[/tex] is the midpoint of the line segment [tex]\(AB\)[/tex].
Given:
[tex]\[ A = \binom{-8}{5} \][/tex]
[tex]\[ B = \binom{12}{-5} \][/tex]
The midpoint [tex]\(M\)[/tex] is calculated as:
[tex]\[ M = \frac{A + B}{2} = \frac{\binom{-8}{5} + \binom{12}{-5}}{2} = \binom{\frac{-8 + 12}{2}}{\frac{5 + (-5)}{2}} = \binom{2}{0} \][/tex]
So, the coordinates of [tex]\(M\)[/tex] are [tex]\(\binom{2}{0}\)[/tex].
---
### a) (ii) Find the magnitude of [tex]\(NM\)[/tex]
Vector [tex]\(NM\)[/tex]:
First, we find vector [tex]\(NM\)[/tex]:
[tex]\[ N = \binom{-4}{2.5} \][/tex]
[tex]\[ M = \binom{2}{0} \][/tex]
[tex]\[ NM = M - N = \binom{2 - (-4)}{0 - 2.5} = \binom{6}{-2.5} \][/tex]
Magnitude of [tex]\(NM\)[/tex]:
The magnitude of a vector [tex]\(\binom{a}{b}\)[/tex] is given by [tex]\( \sqrt{a^2 + b^2} \)[/tex]:
[tex]\[ ||NM|| = \sqrt{6^2 + (-2.5)^2} = \sqrt{36 + 6.25} = \sqrt{42.25} = 6.5 \][/tex]
So, the magnitude of [tex]\(NM\)[/tex] is [tex]\(6.5\)[/tex].
---
### b) Express vector [tex]\(NM\)[/tex] in terms of [tex]\(OB\)[/tex]
Vector [tex]\(OB\)[/tex]:
Given:
[tex]\[ O = \binom{0}{0} \][/tex]
[tex]\[ B = \binom{12}{-5} \][/tex]
[tex]\[ OB = B - O = \binom{12}{-5} - \binom{0}{0} = \binom{12}{-5} \][/tex]
We know [tex]\( NM = \binom{6}{-2.5} \)[/tex].
The vector [tex]\(NM\)[/tex] in terms of [tex]\(OB\)[/tex] is:
[tex]\[ NM = \frac{\binom{6}{-2.5}}{|OB|} \cdot OB \][/tex]
where [tex]\( |OB| \)[/tex] is the magnitude of [tex]\(OB\)[/tex]:
[tex]\[ |OB| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \][/tex]
[tex]\[ NM = \frac{\binom{6}{-2.5}}{13} \][/tex]
So, [tex]\( NM = \frac{6}{13} \binom{12}{-5} \)[/tex].
Thus, vector [tex]\(NM\)[/tex] in terms of [tex]\(OB\)[/tex] is [tex]\(\frac{6}{13} OB\)[/tex].
---
### c) Given [tex]\(OP = OM + 2MN\)[/tex], find the coordinates of [tex]\(P'\)[/tex].
Given:
[tex]\[ OP = OM + 2 MN \][/tex]
First, we find [tex]\(OM\)[/tex]:
[tex]\[ OM = M - O = \binom{2}{0} - \binom{0}{0} = \binom{2}{0} \][/tex]
Next, we find [tex]\(MN\)[/tex]:
[tex]\[ MN = N - M = \binom{-4}{2.5} - \binom{2}{0} = \binom{-4 - 2}{2.5 - 0} = \binom{-6}{2.5} \][/tex]
Given:
[tex]\[ OP = OM + 2 MN \][/tex]
Substitute the values:
[tex]\[ OP = \binom{2}{0} + 2 \binom{-6}{2.5} = \binom{2}{0} + \binom{-12}{5} = \binom{2 - 12}{0 + 5} = \binom{-10}{5} \][/tex]
Lastly, apply the translation to find [tex]\(P'\)[/tex]:
[tex]\[ \text{Translation} = \binom{-5}{8} \][/tex]
[tex]\[ P' = OP + \text{Translation} = \binom{-10}{5} + \binom{-5}{8} = \binom{-10 - 5}{5 + 8} = \binom{-15}{13} \][/tex]
So, the coordinates of [tex]\(P'\)[/tex] are [tex]\(\binom{9}{3}\)[/tex].
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