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Mathematics 15 Online
OpenStudy (hang254):

What is the magnitude and direction of GH if G(3, -2) and H(6, 4)? magnitude: 6.71 units; direction 63.43° magnitude: 6.71 units: direction: 26.57° magnitude: 3.61 units; direction: 26.57° magnitude: 3.61 units: direction: 63.43°

OpenStudy (turingtest):

first find the resultant vector what is it?

OpenStudy (hang254):

im not sure what that is

OpenStudy (turingtest):

it's the same thing you just did; subtract the final coordinates from the initial ones

OpenStudy (turingtest):

v=<xh-xg,yh-yg>

OpenStudy (hang254):

oh right, i thought this was solved a different way.

OpenStudy (turingtest):

we will have to do more, but the first step is finding the vector components

OpenStudy (hang254):

ok, the components are <3,6>

OpenStudy (turingtest):

ok, so the magnitude (length) for a vector is found using the pythagorean theorem on its components

OpenStudy (hang254):

could i use the distance formula?

OpenStudy (turingtest):

|dw:1343580631354:dw|\[\vec v=\langle x,y\rangle\]the magnitude is\[\|\vec v\|=\sqrt{x^2+y^2}\]

OpenStudy (turingtest):

if that is what you mean by the distance formula (hopefully you can see how they are related)

OpenStudy (turingtest):

and for the angle we use a little trigonometry I presume you have taken or are taking trig, right?

OpenStudy (hang254):

this is actually geometry

OpenStudy (waleed_imtiaz):

how to calculate the angle ?

OpenStudy (turingtest):

I know to calculate the angle with trig|dw:1343580920560:dw|\[\tan\theta=\frac yx\implies\theta=\tan^{-1}\frac yx\]but with pure geometry I guess we have to do some reasoning about the fact that the y component is twice as long as the x component

OpenStudy (hang254):

which trig function would we use for this?

OpenStudy (turingtest):

tangent, (which leads to inverse tangent) as I wrote above

OpenStudy (hang254):

ok, i got the magnitude. 6.71

OpenStudy (turingtest):

|dw:1343581180531:dw|

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