Finding vectors in $\mathbb R^n$ with Euclidean norm 1
I have a couple of questions here which ask:
Find two vectors in $\mathbb R^2$ with Euclidean norm 1, whose Euclidean inner product with (3, -1) is zero.
and
Show that there are infinitely many vectors in $\mathbb R^3$ with Euclidean norm 1 whose Euclidean inner product with (1,-3, 5) is zero.
Can anyone explain how to solve these types of problems in general? My textbook doesn't show the solution for these types.
$\endgroup$4 Answers
$\begingroup$Hint: You are looking for a vector $(x,y)\in \mathbb R^2$ such that $\|(x,y)\|=1$ and $(x,y)\cdot (3,-1)=0$. The latter condition reads: $3x-y=0$. Now, the unity norm condition reads $x^2+y^2=1$. So, you two equations that you need to solve. A similar approach to problem 2 will furnish a solution. Note that drawing things may help you visualize what the algebra is encoding.
$\endgroup$ $\begingroup$Let's assume that a vector $v=(a,b)$ satisfies the first criterium, that is, $$3a+(-1)b=0\,.$$ That means that $b=3a$. So we have $v=(a,3a)$, and we also want $|v|=1$. Calculating length using coordinates is always done by the Pythagorean theorem, so that we want $$1=|v|^2=a^2+(3a)^2$$ Solve this for $a$, you will get two solutions.
If you draw it in coordinate system, you will see that these are exactly the vectors that are orthogonal to the given $(3,-1)$.
For the second question, if you draw it, the solution will be a whole circle on the plane that is orthogonal to the given vector.
$\endgroup$ $\begingroup$Hint: Start with the zero inner product condition, i.e. first find all vectors with inner product with $(3,-1)$ zero.
$\endgroup$ $\begingroup$You're being asked about vectors perpendicular ("with innner product zero") to a given vector. In 2d, all such vectors form a line, but clearly only the unit vector of that line (and its additive inverse) has magnitude 1.
In 3d, however, there is a whole plane perpendicular to a given vector, and the set of vectors in that plane with unit magnitude forms a circle.
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