Find a vector of length 11 which is parallel to the line.

[john458776]

This is how I approached it.
Line has a direction vector <3,5,0>
Line has a vector <2,-4,6>)
Magnitude of a direction vector is square root of (34)

The vector will be (11/square root of (34)).<3,5,0>

Which will give me the length of 11.

The approach I used is it correct or I'm still missing something?

I have confident that the approach is good but i doubt it because I wanted to find a vector using a vector on the line instead of using a directional vector.

 

[Dr.Peterson]

Line has a vector <2,-4,6>)

I wanted to find a vector using a vector on the line instead of using a directional vector.

I think when you talk about a "vector on the line", you mean a point on the line (e.g. with position vector <2,-4,6>).

Why would a point on the line be at all relevant to the problem? Isn't the problem all about directions?

Your approach is, indeed, good, except that you haven't yet written the answer in the requested form.

 

[pka]

Line has a direction vector <3,5,0>
Line has a vector <2,-4,6>)

The title of the post is "Find a vector of length 11 which is parallel to the line."
If the direction vector is [imath]<3,5,0>[/imath] then the vector [imath]\dfrac{11}{\sqrt{3^2+5^2}}<3,5,0>[/imath] has length [imath]11[/imath] and is parallel to the given line. Can you explain why?

 

[john458776]

I think when you talk about a "vector on the line", you mean a point on the line (e.g. with position vector <2,-4,6>).

Why would a point on the line be at all relevant to the problem? Isn't the problem all about directions?

Your approach is, indeed, good, except that you haven't yet written the answer in the requested form.

Yes the problem is about directions, now it make sense that the position vector would not be relevant to be used in this case.

 

[john458776]

The title of the post is "Find a vector of length 11 which is parallel to the line."
If the direction vector is [imath]<3,5,0>[/imath] then the vector [imath]\dfrac{11}{\sqrt{3^2+5^2}}<3,5,0>[/imath] has length [imath]11[/imath] and is parallel to the given line. Can you explain why?

I think the vector we are looking for will be parrallel to the directional vector but now we are given that the magnitude of the vector we are looking for is 11, I think they have the same direction.

 

[pka]

I think the vector we are looking for will be parrallel to the directional vector but now we are given that the magnitude of the vector we are looking for is 11, I think they have the same direction.

Any vector which a multiple of the direction vector of a line is parallel to the line.
Given the vector [imath]\vec{v}=<1,-5,-2>[/imath] it has length [imath]\sqrt{1+25+4}=\sqrt{30}[/imath]
Multiply the vector [imath]\vec{v}[/imath] by [imath]\dfrac{1}{\sqrt{30}} [/imath] we get a unit vector.
Now multiply the unit vector by eleven to get a vector that is length eleven parallel to [imath]\vec{v}[/imath]

 

[john458776]

Any vector which a multiple of the direction vector of a line is parallel to the line.
Given the vector [imath]\vec{v}=<1,-5,-2>[/imath] it has length [imath]\sqrt{1+25+4}=\sqrt{30}[/imath]
Multiply the vector [imath]\vec{v}[/imath] by [imath]\dfrac{1}{\sqrt{30}} [/imath] we get a unit vector.
Now multiply the unit vector by eleven to get a vector that is length eleven parallel to [imath]\vec{v}[/imath]

Okay I think I understand a little bit, so even if I use your approach I'm still correct and even if I use my approach I'm still correct right ?

 

[Dr.Peterson]

Okay I think I understand a little bit, so even if I use your approach I'm still correct and even if I use my approach I'm still correct right ?

I don't think your approach is any different! You simply multiplied by an appropriate scalar. The only difference is that you didn't explain why it was correct. (But still, don't forget to write it in the requested form: )

 

[john458776]

I think the vector we are looking for will be parrallel to the directional vector but now we are given that the magnitude of the vector we are looking for is 11, I think they have the same direction.

I don't think your approach is any different! You simply multiplied by an appropriate scalar. The only difference is that you didn't explain why it was correct. (But still, don't forget to write it in the requested form: View attachment 29689)

I think I explained here why I think my approach is correct according to my understanding. Maybe my explanation is lacking something that I haven't known yet.

 

[john458776]

I don't think your approach is any different! You simply multiplied by an appropriate scalar. The only difference is that you didn't explain why it was correct. (But still, don't forget to write it in the requested form: View attachment 29689)

I didnt forget to write it in the requested form, it's just that I dont know how to type math equations here.

 

[Dr.Peterson]

I think I explained here why I think my approach is correct according to my understanding. Maybe my explanation is lacking something that I haven't known yet.

I wasn't saying anything was wrong with your explanation, just trying to guess why pka responded to your initial question in a way that sounded as if you had done something wrong. What you did was correct, and what you said was enough, in my mind.

I didnt forget to write it in the requested form, it's just that I dont know how to type math equations here.

I was just reminding you, in case you missed that point.

You don't need to type anything fancy here; it would be enough to type (33/sqrt(34)) i + (55/sqrt(34)) j, which I presume is your final answer. Or maybe you are required to rationalize denominators.