Question

Vector u has its initial point at (17, 5) and its terminal point at (9, -12). Vector v has its initial point at (12, 4) and its terminal point at (3, -2). Find ||3u − 2v||. Round your answer to the nearest hundredth.

Answer 1

First, we find the coordinates of the vectors u and v

$\vec{u}=(9-17; \ \ -12-5)=(-8; \ -17) \\ \\ \vec{v}=(3-12; \ \ -2-4)=(-9; \ -6)$

Then we find the coordinates of the vector 3u-2v

$3\vec{u}-2\vec{v}=(3(-8)-2(-9); \ 3(-17)-2(-6))=(-6; \ -39)$

Now we can find the length of the vector 3u-2v

$|3\vec{u}-2\vec{v}|= \sqrt{(-6)^2+(-39)^2} = \sqrt{36+1521}= \sqrt{1557} \approx 39.46$

Answer: 39.46 (to the nearest hundredth)

Answer 2

First we need to find the lengths of vector u and v. We can find this by distance formula:

$u= \sqrt{ (9-17)^{2} + (-12-5)^{2} }=18.79$

and

$v= \sqrt{ (3-12)^{2} + (-2-4)^{2} }=10.82$

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