anyone know how to find the magnitude of vectors?
if it is writen in ai+bj form then a and b are the magnitudes.
is they are expressed as rectangular coordinates? then suppose they are in the form x i-cap + y j-cap = vectorA then, \[magnitude = |vectorA| = \sqrt{x^2 + y^2}\]
or you can use distance formula
sqrt(a^2 + b^2)
@timo86m 'a' and 'b' the magnitudes of each rectangular coordinate actually, not the complete vector.
ai + bj + ck+..... magnitude of vector= sqrt of (a^2 + b^2 + c^2 +...)
I was given the initial and terminal point (-8, -12) & (4,1)
so, then your vector is: (-8-4 , -12-1)
magnitudes of the x, j directions yes.
if taking (-8,-12) as A, (4,1) as B. vector AB= \[\left(\begin{matrix}12 \\ 14\end{matrix}\right)\] magnitude= sqrt(12^2 + 14^2)
this is not making sense.
sorry. I'm not very bright
@glgan1 '-12-1 = 13'
okay you can try to draw the 2 points A and B on a graph. just take A and B as 2 different destinations. from A to B, we need to walk 12 steps to the right and 14 steps up. this instruction we can call the vector from A to B. so let say we move from B to A. we need to move 12 steps to the left, 14 steps downwards. so for this case it's called the vector from B to A. so vector A to B is\[\left(\begin{matrix}12 \\ 14\end{matrix}\right)\] vector B to A is\[\left(\begin{matrix}-12 \\ -14\end{matrix}\right)\] vector B to A is just the opposite of vector A to B, thats why we just add negative sign to it. so you may draw on your graph an arrow which represents the vector from A to B. the length of arrow is called the magnitude of the vector, we can say that it is the displacement from A to B if we take both A and B as destinations. so magnitude = sqrt (12^2 + 14^2). it's just the square root of the squares of every elements in the vector.
for magnitude it's still the same if we take vector A to B or vector B to A as (-12)^2 = (12)^2, (-14^2) = (14)^2.
thank you!
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