Ask Uncle Colin: Rotating vectors

Dear Uncle Colin,

I need to find a unit vector in the xy-plane that makes an angle of 45 degrees with the vector $3\mathbf{i}+4\mathbf{j}$. How would you do that?

- Don’t Enjoy Maths Of Integer Vectors Rotating Enough

Hi, DEMOIVRE, and thanks for your message! I can think of several ways to approach this.

Method 1: sneaky complex numbers

Instead of the vector $3\mathbf{i}+4\mathbf{j}$, consider the complex number $3+4\mathrm{i}$.

To rotate the point in the Argand diagram corresponding to $z$ by $\frac{\pi }{4}$ ¹, you multiply by $\frac{1}{\sqrt{2}}(1\pm \mathrm{i})$, with a positive sign for an anti-clockwise rotation.

Ignoring the $\sqrt{2}$ since we’ll be dividing out the modulus shortly, $(3+4\mathrm{i})(1\pm i)=(3\mp 4)+\mathrm{i}(4\pm 3)$, which is either $-1+7\mathrm{i}$ or $7+\mathrm{i}$.

The modulus of these numbers is $5\sqrt{2}$, so the corresponding vectors are $-\frac{1}{5\sqrt{2}}\mathbf{i}+\frac{7}{5\sqrt{2}}\mathbf{j}$ and $\frac{7}{5\sqrt{2}}\mathbf{i}+\frac{1}{5\sqrt{2}}\mathbf{j}$.

Method 2: a dot product

Suppose the required vector is $x\mathbf{i}+y\mathbf{j}$. Then we have $\frac{(3\mathbf{i}+4\mathbf{j})\cdot (x\mathbf{i}+y\mathbf{j})}{5\times |(x\mathbf{i}+y\mathbf{j})|}=\frac{1}{\sqrt{2}}$. (Any vector at 45 degrees to the original one will satisfy this; the $\frac{1}{\sqrt{2}}$ is the cosine of the angle between the vectors).

Let’s make live easy for ourselves and suppose that $|x\mathbf{i}+y\mathbf{j}|=\sqrt{2}$, making our equation $(3\mathbf{i}+4\mathbf{j})(x\mathbf{i}+y\mathbf{j})=5$.

We’ve now got simultaneous equations: $x^2+y^2=2$, from the supposition, and $3x+4y=5$, from the dot product.

Letting $3x=5-4y$, and considering $9x^2+9y^2=18$, we get $(5-4y{)}^2+9y^2=18$, or $25-40y+25y^2=18$

Rearranging, $25y^2-40y+7=0$, which factorises as $(5y-7)(5y-1)$.

This gives $y=\frac{7}{5}$ (with $x=-\frac{1}{5}$) or $y=\frac{1}{5}$ (with $x=\frac{7}{5}$) - and when we normalise, we get the same vector as above.

Method 3: Use the perpendicular

If you construct a perpendicular vector of the same length - either $-4\mathbf{i}+3\mathbf{j}$ or $4\mathbf{i}-3\mathbf{j}$ - and find the midpoint of the ends of the two vectors, the vector this point will be at 45 degrees to the original vector.

The two possible midpoints are at $-\frac{1}{2}\mathbf{i}+\frac{7}{2}\mathbf{j}$ and $\frac{7}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$; both of these, when normalised, give the same answer as method 1.

Method 4: compound angle formula

Let $\theta$ be the angle between the $x$-axis and $3\mathbf{i}+4\mathbf{j}$, so that $\tan (\theta )=\frac{4}{3}$

$\tan (\theta \pm {45}^o)=\frac{\frac{4}{3}\pm 1}{1\mp \frac{4}{3}}$

$\cdots =\frac{4\pm 3}{3\mp 4}$, giving either -7 or $\frac{1}{7}$.

The first corresponds to a vector parallel to $-\mathbf{i}+7\mathbf{j}$ and the second to $7\mathbf{i}+\mathbf{j}$; again, these normalise to the correct answer.

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Hope that’s helpful!

- Uncle Colin

Footnotes:

1. In Argand-world, we do not degree.