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
v = sqrt(29)/29*(2i+5j)
I don't understand...
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))
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 \[\hat{v}=-\frac{2}{\sqrt{29}}i+\frac{5}{\sqrt{29}}j\]
Wow thank you so much for explaining all that. Really appreciate it
Could you help with my other question??
go ahead and post it (probably best as its own thing though)
Done done and done
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