How Do You Find a Vector with a Specific Magnitude in the Opposite Direction?

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Homework Help Overview

The problem involves finding a vector that points in the opposite direction of a given vector <-4,1,2> and has a specific magnitude of 3. The context is within vector mathematics, likely at an introductory level.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the magnitude of the given vector and derive a unit vector in the opposite direction. They express confusion about how to adjust this unit vector to achieve the desired magnitude of 3. Other participants suggest considering the relationship between the magnitude of a vector and its components, and they prompt the original poster to think about scaling the unit vector.

Discussion Status

Participants are engaging in a constructive dialogue, with some providing guidance on how to approach the problem. The original poster acknowledges a realization that the problem may be simpler than initially thought. There is an ongoing exploration of the mathematical concepts involved, but no explicit consensus has been reached.

Contextual Notes

There is a discussion regarding the classification of the problem as precalculus versus calculus, with some participants questioning the appropriateness of the categorization based on their educational experiences. This indicates a potential gap in foundational knowledge related to vectors.

mmont012
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Homework Statement



Find a vector in the direction opposite to <-4,1,2>, that has a magnitude of 3.

Homework Equations


I think that I did the first part of the problem correctly:
<-4,1,2>
magnitude= sqrt[ (-4)^2+1^2+2^2 ]
= sqrt(16+1+4)
= sqrt(21)
(-4/sqrt(21), 1/sqrt(21), 2/sqrt(21) )
to get the opposite direction, I would just change the signs
(4/sqrt(21), -1/sqrt(21), -2/sqrt(21) )

But I am confused at where to go from here. What does it mean "that has a magnitude of 3"? I may have approached this problem incorrectly. Any help would be great. Thank you for stopping by.
 
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You have found a normal vector. The magnitude of a normal vector is ##1##.

Perhaps the answer is more trivial than it first appears, just think:

$$\sqrt{a^2 + b^2 + c^2} = 3$$

Select a vector ##<a, b, c>## such that the above is satisfied. Multiplying the original vector by ##-1## will help. In fact, one is the loneliest one, and the only one you'll ever need for this problem.
 
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As Zondrina has pointed out, you have correctly determined a unit vector in the desired direction. Now, what do you need to multiply this unit vector by to produce a vector having the same direction, but with a magnitude of 3?

Chet
 
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Ah, okay. I think that I was making this problem more difficult than it actual is. Thank you so much for the help!
 
Removed "Calc III" from thread title -- this is really a precalc type of problem. Also moved thread to the precalc forum section.
 
How is this a pre-calc problem if I'm in calc 3 and this is the first time that I've dealt with this? Did my university just not cover this? Or am I behind?
 
mmont012 said:
How is this a pre-calc problem if I'm in calc 3 and this is the first time that I've dealt with this? Did my university just not cover this? Or am I behind?
Vectors in R2 and R3 are often covered in precalculus with trig courses.
 
mmont012 said:
How is this a pre-calc problem if I'm in calc 3 and this is the first time that I've dealt with this? Did my university just not cover this? Or am I behind?
No limits, no derivatives, no integrals = no calculus.

Your calc 3 course may simply be providing a refresher on basic vector math before jumping into vector calculus.
 

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