Here's the question you clicked on:
KatClaire
Find a unit vector in the same direction as u= [3,0,-7,-4,2] (column)
\[\left(\begin{matrix}3 \\ 0\\-7\\-4\\2\end{matrix}\right)\]
what is the length of the given vector?
suppose you have a vector that is 12 feet in length, how do you find out its "unit" length?
that's all that's given :\
yeah, and to find the length of a vector, you square the parts, add em up, and sqrt it all ....
\[vector:<a,b,c,d,...,k>\]\[distance=\sqrt{a^2+b^2+c^2+d^2+...+k^2}\]
parts: square 3: 9 0: 0 7:49 4:16 2: 4 --- sum: 78 ; so length must be sqrt(78) or did i do something wrong?
I did that and got sqrt 79, yes?
vector = sqrt(78) units to find a vector of 1 unit, we divide both sides by sqrt(78) vector/ sqrt(78) = sqrt(78)/sqrt(78) units vector/ sqrt(78) = 1 unit therefore, a unit vector is created by dividing the components of the vector by its length
18+10=28 .. not 29 :)
sorry my computer like died so I'm on my friends lol so it's 6/sqrt78 @amistre64 then you times that into each unit thingy
each component then gets divided by the vectors length to obtain components of the unit vector. 3/sqrt(78) 0/sqrt(78) - 7/sqrt(78) - 4/sqrt(78) 2/sqrt(78)
another way to notate it is just to scale the original vector by 1/length \[\frac{1}{\sqrt{78}}[3,0,-7,-4,2]\]
oh okay my problem was accepting the fact that my calculator wouldn't put my answers in fractions so I thought I must be wrong hahaha. Thanks!!