Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What are all the squares in upper corners of each vector symbol? –  Thomas E. Jun 8 '12 at 13:37
    
That's your browser. Had it too, now it disappeared. –  Miroslav Cetojevic Jun 12 '12 at 7:53

2 Answers 2

up vote 2 down vote accepted

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.)

share|improve this answer
    
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. –  Miroslav Cetojevic May 25 '12 at 15:13

What exactly is the problem? $-4x + 3y = 0$; $x^2+y^2=1$. This clearly gives $(x,y) = \pm (3/5,4/5)$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.