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Define a vector of length 1 orthogonal to $\vec{v} = (-4 \qquad 3)^t$

I'm looking for the solution in terms of $\vec{a} = \binom{x}{y}$.

How do I go about it? I'm familiar with addition, subtraction and multiplication of vectors and scalars.

I tried to use the formula

$\vec{a} * \vec{b} = ||\vec{a}|| * ||\vec{b}|| * \cos{(\alpha)}$

but since the unknown vector is orthogonal to $\vec{v}$, $\cos{(90°)}$ becomes $0$, thus everything becomes $0$. I'm at a loss here.

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  • $\begingroup$ What are all the squares in upper corners of each vector symbol? $\endgroup$
    – T. Eskin
    Jun 8 '12 at 13:37
  • $\begingroup$ That's your browser. Had it too, now it disappeared. $\endgroup$ Jun 12 '12 at 7:53
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You could first find a vector that is simply orthogonal to $(-4,3)$. To produce one for a vector with only two components, you could simply interchange the components and switch the sign of one component.

So $(3,4)$ is orthogonal to $(-4,3)$ (check: $3(-4)+4\cdot3=0$). Alternatively, set all but one component of your soon to be orthogonal vector to an (almost) arbitrary value, and solve for the last component.

OK, so $(3,4)$ is orthogonal to $(-4,3)$, but it is not a unit vector. To get a unit vector, divide $(3,4)$ by its norm. This will produce a vector of norm one that will still be orthogonal to $(-4,3)$. Here, the norm of $(3,4)$ is $\sqrt{3^2+4^2}=5$. The desired vector is ${1\over5}(3,4)=(3/5,4/5)$. (Note the negative of this vector will also work.)

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  • $\begingroup$ Ah yes, I had wondered how to find an orthogonal vector in the first place. Didn't realize that I could just mirror it to the x-axis. $\endgroup$ May 25 '12 at 15:13
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What exactly is the problem? $-4x + 3y = 0$; $x^2+y^2=1$. This clearly gives $(x,y) = \pm (3/5,4/5)$

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