Parallel and orthogonal

I need math help in this problem: a=(-5, 5 3) b=(-2,-4,-5) (they are vectors)
Decompose the vector b into b=v+w where v is parallel to a and w is orthogonal to a, find v and w

Result

v1 =  2.119
v2 =  -2.119
v3 =  -1.271
w1 =  -4.119
w2 =  -1.881
w3 =  -3.729

Solution:

a=(5,5,3) b=(2,4,5)  b=v+w va=>v=ka  wa=>w.a=0   2=v1+w1 4=v2+w2 5=v3+w3 v1=5 k =2.119 v2=5 k v3=3 k w1 (5)+w2 5+w3 3=0  v1+w1=2 v2+w2=4 v3+w3=5 5k+v1=0 5kv2=0 3kv3=0 5w15w23w3=0  k=25590.423729 v1=125592.118644 v2=125592.118644 v3=75591.271186 w1=243594.118644 w2=111591.881356 w3=220593.728814 a = (-5, 5, 3) \ \\ b = (-2,-4,-5) \ \\ \ \\ b = v+w \ \\ v \parallel a = > v = k a \ \\ \ \\ w \perp a = > w.a = 0 \ \\ \ \\ \ \\ -2 = v_{ 1 }+w_{ 1 } \ \\ -4 = v_{ 2 }+w_{ 2 } \ \\ -5 = v_{ 3 }+w_{ 3 } \ \\ v_{ 1 } = -5 \cdot \ k \ \\ = 2.119 \ \\ v_{ 2 } = 5 \cdot \ k \ \\ v_{ 3 } = 3 \cdot \ k \ \\ w_{ 1 } \cdot \ (-5)+w_{ 2 } \cdot \ 5 + w_{ 3 } \cdot \ 3 = 0 \ \\ \ \\ v_{ 1 }+w_{ 1 } = -2 \ \\ v_{ 2 }+w_{ 2 } = -4 \ \\ v_{ 3 }+w_{ 3 } = -5 \ \\ 5k+v_{ 1 } = 0 \ \\ 5k-v_{ 2 } = 0 \ \\ 3k-v_{ 3 } = 0 \ \\ 5w_{ 1 }-5w_{ 2 }-3w_{ 3 } = 0 \ \\ \ \\ k = \dfrac{ -25 }{ 59 } \doteq -0.423729 \ \\ v_{ 1 } = \dfrac{ 125 }{ 59 } \doteq 2.118644 \ \\ v_{ 2 } = \dfrac{ -125 }{ 59 } \doteq -2.118644 \ \\ v_{ 3 } = \dfrac{ -75 }{ 59 } \doteq -1.271186 \ \\ w_{ 1 } = \dfrac{ -243 }{ 59 } \doteq -4.118644 \ \\ w_{ 2 } = \dfrac{ -111 }{ 59 } \doteq -1.881356 \ \\ w_{ 3 } = \dfrac{ -220 }{ 59 } \doteq -3.728814 \ \\
v2=(2.1186)=125592.1186=2.119v_{ 2 } = (-2.1186) = - \dfrac{ 125 }{ 59 } \doteq -2.1186 = -2.119
v3=(1.2712)=75591.2712=1.271v_{ 3 } = (-1.2712) = - \dfrac{ 75 }{ 59 } \doteq -1.2712 = -1.271
w1=(4.1186)=243594.1186=4.119w_{ 1 } = (-4.1186) = - \dfrac{ 243 }{ 59 } \doteq -4.1186 = -4.119
w2=(1.8814)=111591.8814=1.881w_{ 2 } = (-1.8814) = - \dfrac{ 111 }{ 59 } \doteq -1.8814 = -1.881
w3=(3.7288)=220593.7288=3.729w_{ 3 } = (-3.7288) = - \dfrac{ 220 }{ 59 } \doteq -3.7288 = -3.729



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