Question

Given the vectors v=-2i+5j and w=3i+4j, determine 1. 1/2v 2. w-v 3. length of w 4. the unit vector for v

Answer 1

v = sqrt(29)/29*(2i+5j)

Answer 2

I don't understand...

Answer 3

1. (1/2)V=(1/2)*(-2i+5j)=-i+(5/2)j

2. w-v=3i+4j-(-2i+5j)=5i-j

3. length of w=mod(w)=\[=\sqrt{3^2+4^2}\]
= 5

4. unit vector for v= v/(mod v)=(-2i+5j)/(sqrt(29))

Answer 4

1. Take half of the components of v. Basically, take half of -2 (from -2i) and take half of 5 (from 5j) to get \[\frac{1}{2}v=-i+\frac{5}{2}j\]


2. Here, you subtract the corresponding components of v from w. So subtract -2 (from -2i in vector v) from 3 (from 3i in vector w) to get 3-(-2) = 3+2 = 5. Do the same for the j components to get 4-5 = -1. So \[v-w = (3i+4j)-(-2i+5j) = (3i-(-2i))+(4j-5j)=5i-j\]


3. The length of vector w is \[||w||=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\]

So the length of the vector w is 5 units.


4. The unit vector of v is simply the vector v divide by it's length. So first find the length of v

\[||v||=\sqrt{(-2)^2+5^2}=\sqrt{4+25}=\sqrt{29}\]


and now divide the components of the vector v by this length to get


\[\hat{v}=\frac{v}{||v||}=\frac{-2i+5j}{\sqrt{29}}=-\frac{2}{\sqrt{29}}i+\frac{5}{\sqrt{29}}j\]


So the unit vector of v is \[\hat{v}=-\frac{2}{\sqrt{29}}i+\frac{5}{\sqrt{29}}j\]

Answer 5

Wow thank you so much for explaining all that. Really appreciate it

Answer 6

Could you help with my other question??

Answer 7

go ahead and post it (probably best as its own thing though)

Answer 8

Done done and done