Resolution of 2 i ^ + 3 j ^ along i ^ + j ^ and i ^ − j ^ I proceeded as below:Let A, a and b be the given vectors in the same order. Let A = λ a + μ b Putting the values and rearranging, 2 i ^ + 3 j ^ = ( λ + μ ) i ^ + ( λ − μ ) j ^ Then, noting that two vectors are equal iff their magnitudes as well as directions are equal, first we equate the directions and taking tan of both sides. 3 2 = λ − μ λ + μ Thus, λ = − 5 μNow, taking magnitudes and squaring, 13 = 2 ( λ 2 + μ 2 )Using our previous equation and doing some work, we get λ = ± 5 2 and μ = ∓ 1 2

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Resolution of   2                    i        ^              +  3                    j        ^             along                     i        ^              +                    j        ^             and                     i        ^              −                    j        ^            I proceeded as below:Let A, a and b be the given vectors in the same order. Let      A    =  λ      a    +  μ      b  Putting the values and rearranging,  2                    i        ^              +  3                    j        ^              =  (  λ  +  μ  )                    i        ^              +  (  λ  −  μ  )                    j        ^            Then, noting that two vectors are equal iff their magnitudes as well as directions are equal, first we equate the directions and taking tan of both sides.      3    2    =            λ      −      μ              λ      +      μ      Thus,   λ  =  −  5  μNow, taking magnitudes and squaring,  13  =  2  (      λ    2    +      μ    2    )Using our previous equation and doing some work, we get  λ  =  ±      5          2             and     μ  =  ∓      1          2

Alison Mcgrath · Accepted Answer

Since   13  =  2  (  26      μ    2    ), I think you miscounted the powers of 2. (The book probably instead worked with the unit vectors (  (                    i        ^              ±                    j        ^              )      /        2  ) The easiest simultaneous solution of   λ  +  μ  =  2  ,    λ  −  μ  =  3 averages the equations to get   λ  =            5      2      , so   μ  =  2  −  λ  =  −            1      2

Miguel Mathis · Accepted Answer

Another way to solve this is using matrix techniques:      [                            2                                      3                      ]    =  λ      [                            1                                      1                      ]    +  μ      [                            1                                      −          1                      ]      =      [                            1                          1                                      1                          −          1                      ]        [                            λ                                      μ                      ]  So take inverse and premultiply and we get       [                            λ                                      μ                      ]    =      [                            5                      /                    2                                      −          1                      /                    2                      ]

Resolve 2 hat(i)+3 hat(j) along hat(i)+hat(j) and hat(i)-hat(j)

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