Finding unknown given vector a,b and parallel

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The discussion revolves around finding the scalar μ such that the vector combination μc + d is parallel to the vector i + 3j, where c = 3i + 4j and d = i - 2j. The initial attempt led to the equation 3μ + 1 = 3(4μ - 2), resulting in μ = 7/9, but the correct value is μ = -1. Participants clarify that for two vectors to be parallel, one must be a nonzero scalar multiple of the other, leading to the conclusion that the left-hand side should be multiplied by 3 instead of the right-hand side. The confusion arises from the understanding of vector relationships, emphasizing that parallel vectors maintain a specific ratio between their components. The conversation highlights the complexities and common frustrations associated with vector mathematics.
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Homework Statement



Given that c= 3i + 4j and d= i - 2j

find μ when μc + d is parallel to i +3j



Homework Equations





The Attempt at a Solution



3iμ + 4jμ + i + j

i( 3μ + 1) + (4μ -2) j

since it is parallel to i + 3j therefore 3μ+1=3(4μ-2)

giving μ = 7/9

However μ = -1


Additional information:
Apparently (see bold) I am supposed to multiply the LHS by 3 instead of the RHS but by equating I and J vectors and looking at the parallel line it would seem logical to do it the way I have (or am I being retarded?). I can accept that I have the multiply LHS instead of RHS but I don't understand why!
 
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xplosiv3s said:

Homework Statement



Given that c= 3i + 4j and d= i - 2j

find μ when μc + d is parallel to i +3j



Homework Equations





The Attempt at a Solution



3iμ + 4jμ + i + j

i( 3μ + 1) + (4μ -2) j

since it is parallel to i + 3j therefore 3μ+1=3(4μ-2)
Not necessarily. Vectors can be parallel without being equal. Two vectors are parallel if either of them is some nonzero scalar multiple of the other.
xplosiv3s said:
giving μ = 7/9

However μ = -1


Additional information:
Apparently (see bold) I am supposed to multiply the LHS by 3 instead of the RHS but by equating I and J vectors and looking at the parallel line it would seem logical to do it the way I have (or am I being retarded?). I can accept that I have the multiply LHS instead of RHS but I don't understand why!
 
Mark44 said:
nonzero scalar multiple of the other.

So i multiply the other side by 3 because it has to be that? ^
 
xplosiv3s said:
So i multiply the other side by 3 because it has to be that? ^

If a vector is parallel to i+3j, then 3 times the i component is equal to the j component. Since 3*1=3. So 3*(3μ+1)=(4μ-2). I'm not sure why you are doing it the other way around.
 
Dick said:
If a vector is parallel to i+3j, then 3 times the i component is equal to the j component. Since 3*1=3. So 3*(3μ+1)=(4μ-2). I'm not sure why you are doing it the other way around.

Ok thanks! That kinda makes sense!

actually nvm >.>
 
I can understand why everyone hates vectors
 
xplosiv3s said:
I can understand why everyone hates vectors
Vectors are marvelous !
 

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