Help with vector problem please :)

[skitseez]

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)

 

[Deleted member 4993]

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.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

What method have you been taught to solve such problems? I would start with calculating the unit vector along the vector AC.

 

[skitseez]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

What method have you been taught to solve such problems? I would start with calculating the unit vector along the vector AC.

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)

 

[skitseez]

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)

thats not the unit vector thats the magnitude of the parrallel like

 

[Deleted member 4993]

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)

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

 

[pka]

Find a vector parallel to AC that passes through (2, −4, 5) and has a magnitude of 8

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)

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.
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.

 

[skitseez]

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

what do I do with the magnitude of 8? Do i just multiply it to the equation or is that only for unit vectors?

 

[pka]

what do I do with the magnitude of 8? Do i just multiply it to the equation or is that only for unit vectors?

You do nothing. There is no vector of length eight in the question.
Who ever wrote this question has no idea what they are doing.
If I were you I would complain to whom soever is in charge of your course.

 

[lev888]

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)

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.

 

[pka]

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.

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 \(\vec{u}\)

 

[lev888]

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 \(\vec{u}\)

Are you referring to my post?
If we assume that we are asked to find a vector, then yes, we are looking for the scalar multiple that would give us the magnitude 8. Sorry if my post was not clear.
If, on the other hand, we assume that we are asked to find a line passing through a point, then we can do it too, but lines don't have magnitudes.

 

[pka]

Are you referring to my post?

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.

 

[Deleted member 4993]

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.

In mechanics, description of a force vector require:

(i) a magnitude

(ii) a direction and

(iii) a point of application.

This assignment may have been set up loosely to introduce force and moment of a force.