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I want a simple and intuitive way of solving for $t$ given that $\mathbf{i},\, \mathbf{j},\, \mathbf{k}$ are the unit vectors in $\mathbf{R}^3$. $$ 5 = (2 \mathbf{i} - 3 \mathbf{k} + (\mathbf{i} - \mathbf{j} + \mathbf{k})\, t)\cdot(2 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}) $$ I want to point out that I often get equations provided like this and I am not looking to solve it. I do understand that $(4-12+t\,(2-3-4))=5$ provides me with the answer 13/5. Now I am simply interested in solving this with as little as possible transformation of the expressions.

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  • $\begingroup$ a quick simple and intuitive way would do! $\endgroup$ – ALEXANDER May 5 '15 at 10:13
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i = {1, 0, 0}; 
j = {0, 1, 0};
k = {0, 0, 1}; 
    Solve[5 == (2 i - 3 k + (i - j + k) t).(2 i + 3 j + 4 k), t]

(* {{t -> 13/3}} *)

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  • $\begingroup$ How does mathematica know to match terms when using the dot. I thought that it was the case that i needed to use {} to be able to use the dot product $\endgroup$ – ALEXANDER May 5 '15 at 10:24
  • $\begingroup$ Mathematica substitutes the (list) values for $i$, $j$, and $k$ when executing Solve. $\endgroup$ – David G. Stork May 5 '15 at 10:27
  • $\begingroup$ One could also do {i, j, k} = IdentityMatrix[3];, or use UnitVector[] $\endgroup$ – J. M.'s technical difficulties May 5 '15 at 11:45
  • $\begingroup$ @Guesswhoitis. That's what I did in the answer of which this is a duplicate... $\endgroup$ – Jens May 5 '15 at 17:19

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