Vectors for a Given Angle and Vector

As a high school math teacher, I often have to come up with my own problems to give on worksheets and tests.  Many times I will take an existing problem from the textbook and change the numbers, but the trick is to change the numbers in such a way so that the level of difficulty of the problem is maintained.  To accomplish this, I often have to work the problems backwards from the answer to get to the question.  For example, a typical factoring problem would be to factor x² + 5x + 4.  To solve this, you find two numbers that multiply to 4 and add up to 5, which are 1 and 4, and so the solution is (x + 1)(x + 4).  If I wanted a problem similar to this I can’t just randomly change the numbers in x² + 5x + 4, because the solution might have a fraction or radical or both, which would make the problem more difficult.  However, I can randomly change a number in the solution and then work backwards to get a new problem of the same difficulty.  So if I wanted the solution to be (x + 1)(x + 5) instead, I would work backwards by multiplying it out and get x² + 6x + 5 as the new problem.

Working backwards may be relatively simple for factoring, but for other problems, such as finding the angle between two vectors, it can be more difficult.  A typical problem in trigonometry would be to find the angle between the vectors (4, 3) and (3, -4).  Using tan q = ^{m1 – m2}/_{1 + m1m2}, with m₁ = ³/₄ and m₂ = ^-4/₃, q = tan^-1(^25/12/₀) = 90°.  

But what if you want the solution to be a different angle, like 45°?  What other vector makes a 45° angle with (4, 3)?  Asking the question backwards like this does not make for a typical trigonometry problem.

To find vectors for a given angle and vector, we need rearrange the equation tan q = ^{m1 – m2}/_{1 + m1m2}.  Multiplying both sides by 1 + m₁m₂, we get tan q (1 + m₁m₂) = m₁ – m₂.  Distributing, we get tan q + tan q m₁m₂ = m₁ – m₂.  Rearranging terms, we get m₂ + tan q = m₁ – tan q m₁m₂.  Factoring out m₁ we get m₂ + tan q = m₁ (1 – tan q m₂), which means m₁ = ^{m2 + tan q}/_{1 – tanq m2}.  A similar argument rearranges tan q = ^{m1 – m2}/_{1 + m1m2} to m₂ = ^{m1 – tan q}/_{1 + tanq m1}.  Since the vectors can be interchanged without changing the angle between them, we can combine both formulas to make m₁ = ^{m2 ± tan q}/_{1 ∓ tan q m2}.  Therefore, to find a missing vector of a given angle and another vector, we can use the formula m₁ = ^{m2 ± tan q}/_{1 ∓ tan q m2}, where q is the angle and m₁ and m₂ are the slopes of the two vectors.  This means that the other vector that makes a 45° angle with (4, 3) will have a slope of ^{m2 ± tan q}/_{1 ∓ tan q m2} = ^{3/4 ± tan 45°}/_{1 ∓ tan 45° 3/4} = 7 or ^-1/₇.  A few vectors with these slopes can include either (1, 7) or (7, -1).