Please show us what you have tried and exactly where you are stuck.Find a vector parallel to AC that passes through (2, −4, 5) and has a magnitude of 8. Vector AC is equal to: (-2, 4, 4)
Please show us what you have tried and exactly where you are stuck.
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What method have you been taught to solve such problems? I would start with calculating the unit vector along the vector AC.
thats not the unit vector thats the magnitude of the parrallel likeI found the unit vector to AC to be 6. With the magnitude in mind i ended with 8/6(-2, 4, 4) which left me with (-8/3, 16/3, 16/3). I am just confused on how to make it pass through (2, -4, 5)
To find the parametric equations of the line passing through the point (-1,2,3) and parallel to the vector <3,0,-1>, we first find the vector equation of the line. Here, r_0=<-1,2,3> and v=<3,0,-1>. Thus, the line has vector equation r=<-1,2,3>+t<3,0,-1>.I found the of the vector AC to be 6. With the magnitude in mind i ended with 8/6(-2, 4, 4) which left me with (-8/3, 16/3, 16/3). I am just confused on how to make it pass through (2, -4, 5)
Find a vector parallel to AC that passes through (2, −4, 5) and has a magnitude of 8
Don't feel bad about your confusion. I taught vector geometry and vector analysis for over forty years but I have no idea what the question means.I found the unit vector to AC to be 6. With the magnitude in mind i ended with 8/6(-2, 4, 4) which left me with (-8/3, 16/3, 16/3). I am just confused on how to make it pass through (2, -4, 5)/QUOTE]
vector AC is equal to: (-2, 4, 4)
Vectors are equivalence classes of objects that have length and direction. As such a vector does not contain a given point. A unit vector has length one. The length \(\|<2,-4,4>\|=\sqrt{4+16+16}=6\). I suspect the question wants the equation of a line.
what do I do with the magnitude of 8? Do i just multiply it to the equation or is that only for unit vectors?To find the parametric equations of the line passing through the point (-1,2,3) and parallel to the vector <3,0,-1>, we first find the vector equation of the line. Here, r_0=<-1,2,3> and v=<3,0,-1>. Thus, the line has vector equation r=<-1,2,3>+t<3,0,-1>.
ref: http://sites.science.oregonstate.ed...estStudyGuides/vcalc/lineplane/lineplane.html
You do nothing. There is no vector of length eight in the question.what do I do with the magnitude of 8? Do i just multiply it to the equation or is that only for unit vectors?
A vector is a direction, passing through a point doesn't apply to it. You can find a vector parallel to AC with magnitude 8.Find a vector parallel to AC that passes through (2, −4, 5) and has a magnitude of 8
vector AC is equal to: (-2, 4, 4)
I have no idea as to what the above could mean.A vector is a direction, passing through a point doesn't apply to it. You can find a vector parallel to AC with magnitude 8.
Are you referring to my post?I have no idea as to what the above could mean.
If each of \(\vec{u}~\&~\vec{v}\) is a vector then \(\vec{u}\) is parallel to \(\vec{v}\) if and only if one is a nonzero scalar multiple of the other.
Suppose that \(P: (a,b,c)\) is a point then \(P+t\vec{u}\) is a line that contains the point \(P\) and is parallel to the vector \(\vec{u}\)
Actually, I was expressing my frustration with the original question.Are you referring to my post?
In mechanics, description of a force vector require:Actually, I was expressing my frustration with the original question.
I found the OP to be such a jumble of wrong headed ideas. Sorry if you took any offence.