Answer:
[tex]||3u-2v||=39.4588[/tex]
Step-by-step explanation:
Vector u can also be written with initial point at the origin as:
[tex]u=(9-17,-12-5)\\u=(-8,-17)[/tex]
and vector v in a similar way can be written as:
[tex]v=(3-12,-2-4)\\v=(-9,-6)[/tex]
then the new vector created via the operations: 3u -2v, can be expressed as:
[tex]3u-2v=3\,(-8,-17)-2\,(-9,-6)\\3u-2v=(-6,-39)[/tex]
Now the norm of this vector can be found using the Pythagorean identity:
[tex]||3u-2v||=||(-6,-39||=\sqrt{(-6)^2+(-39)^2} =39.4588[/tex]
