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Vector Algebra 
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#1
Jan2708, 11:47 PM

P: 3

1. The problem statement, all variables and given/known data
I have trouble solving this following problem: Find a unit vector parallel to this equation 2xz=4. The answer is supposed to come out as (2x+4z)/(square root of 20). It would be great if someone can show me how to get this answer. 2. Relevant equations 3. The attempt at a solution 


#2
Jan2808, 06:01 AM

P: 13

2xz=4
represents the family of points in the form (x, 0, 2x4) and the points (0,0,4) and (1,0,2) are part of that family, with them we can form the vector (1,0,2) So we get the family of vectores parallel to the line by x.(1,0,2) Notice from your solution its easy to see that they used: x = 2 Now we just need to divide (2, 0, 4) by its norm which is sqrt(2^2 + 4^2) 


#3
Jan2808, 06:13 AM

Math
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PF Gold
P: 39,338

First, the problem, as you stated it, does not make sense! An "equation" is not a geometric object and you cannot talk about something being parallel to an "equation"!
I suspect that you mean "Find a unit vector parallel to the graph of the equation 2xz= 4". Of course, then, the answer cannot be "[itex](2x+4z)/\sqrt{20}[/itex] because that is an "expression", not a vector. Perhaps you mean [itex](2\vec{i}+ 4\vec{k})/\sqrt{20}[/itex]. Further, if we are to interpret this in 3 dimensions (implied by using x and z rather than x and y), then the set of points satisfying 2x z= 4 is a plane and there are and infinite number of vectors parallel to that plane. If, instead, you mean a unit vector, in the xzplane, parallel to the line given by 2x z= 4, a simple way to do this is to note that if x= 0, z= 4 and if z= 0, x= 2. That is, the line goes through (0, 4) and (2, 0). A vector from (0, 4) to (2, 0) is [itex]2\vec{i}+ 4\vec{k}[/itex]. That vector has length [itex]\sqrt{2^2+ 4^2}= \sqrt{20}[/itex]. A unit vector pointing in the same direction as that line (in the xzplane) is [tex](2\vec{i}+ 4\vec{k})/\sqrt{20}[/itex] 


#4
Jan2808, 06:37 AM

P: 13

Vector Algebra
y it could be in [tex]\ R^2[/tex] my bad



#5
Jan2808, 10:58 AM

P: 3

Thank you for the help



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