Respuesta :
First we need to find the lengths of vector u and v. We can find this by distance formula:
[tex]u= \sqrt{ (9-17)^{2} + (-12-5)^{2} }=18.79 [/tex]
and
[tex]v= \sqrt{ (3-12)^{2} + (-2-4)^{2} }=10.82 [/tex]
We need to find | 3u - 2v |, which is the absolute value of 3u - 2v.
3u = 3 x 18.79 = 56.37
2v = 2 x 10.82 = 21.64
So,
| 3u - 2v | = | 56.37 - 21.64 | = | 34.73 | = 34.73
[tex]u= \sqrt{ (9-17)^{2} + (-12-5)^{2} }=18.79 [/tex]
and
[tex]v= \sqrt{ (3-12)^{2} + (-2-4)^{2} }=10.82 [/tex]
We need to find | 3u - 2v |, which is the absolute value of 3u - 2v.
3u = 3 x 18.79 = 56.37
2v = 2 x 10.82 = 21.64
So,
| 3u - 2v | = | 56.37 - 21.64 | = | 34.73 | = 34.73
First, we find the coordinates of the vectors u and v
[tex]\vec{u}=(9-17; \ \ -12-5)=(-8; \ -17) \\ \\ \vec{v}=(3-12; \ \ -2-4)=(-9; \ -6)[/tex]
Then we find the coordinates of the vector 3u-2v
[tex]3\vec{u}-2\vec{v}=(3(-8)-2(-9); \ 3(-17)-2(-6))=(-6; \ -39)[/tex]
Now we can find the length of the vector 3u-2v
[tex]|3\vec{u}-2\vec{v}|= \sqrt{(-6)^2+(-39)^2} = \sqrt{36+1521}= \sqrt{1557} \approx 39.46 [/tex]
Answer: 39.46 (to the nearest hundredth)
[tex]\vec{u}=(9-17; \ \ -12-5)=(-8; \ -17) \\ \\ \vec{v}=(3-12; \ \ -2-4)=(-9; \ -6)[/tex]
Then we find the coordinates of the vector 3u-2v
[tex]3\vec{u}-2\vec{v}=(3(-8)-2(-9); \ 3(-17)-2(-6))=(-6; \ -39)[/tex]
Now we can find the length of the vector 3u-2v
[tex]|3\vec{u}-2\vec{v}|= \sqrt{(-6)^2+(-39)^2} = \sqrt{36+1521}= \sqrt{1557} \approx 39.46 [/tex]
Answer: 39.46 (to the nearest hundredth)