# Component of a vector perpendicular to another vector.

I have a doubt regarding vectors.

Let's say we have $$2$$ vectors $$\vec{A}$$ & $$\vec{B}$$ in the $$xy$$ plane with unit vectors $$\hat{i}$$ and $$\hat{j}$$. The angle between the $$2$$ is $$\alpha$$. I have to find the component of $$\vec{A}$$ that is perpendicular to $$\vec{B}$$. So one answer is clearly $$\vec{A} \sin\alpha$$?.

Will we also consider $$-\vec{A} \sin\alpha$$? It is perpendicular to $$\vec{B}$$, but is not a component of $$\vec{A}$$ (or is it?).

My approach using an example:

$$\vec{A} = 3\hat{i} + 4\hat{j}$$ $$\vec{B} = \hat{i} + \hat{j}$$ The angle is $$\alpha$$.

Clearly the magnitude of the vector perpendicular to $$\vec{B}$$ is $$\vec{A} \sin\alpha$$. Now about its direction. Let a vector $$\vec{C}$$, in the perpendicular direction be $$x\hat{i} + y\hat{j}$$. Then using dot product of $$\vec{C}$$ and $$\vec{B}$$, we will have $$0$$.

$$\vec{C} \cdot \vec{B} = 0$$ $$(x\hat{i} + y\hat{j})\cdot(\hat{i} + \hat{j}) = 0$$ $$x + y = 0$$ $$x = -y$$.

The vector becomes $$x\hat{i} -y\hat{j}$$ or $$-x\hat{i} + y\hat{j}$$. And so the direction will become $$\frac{1}{\sqrt{2}}(\hat{i} - \hat{j})$$ or $$\frac{1}{\sqrt{2}}(\hat{j} - \hat{i})$$. So do we accept both?

$\newcommand{\Brak}{\langle #1 \rangle}\DeclareMathOperator{\proj}{proj}$If $A$ and $B \neq 0$ are vectors (in an arbitrary inner product space, with the inner product denoted by angle brackets), there exists a unique pair of vectors that are (respectively) parallel to $B$ and orthogonal to $B$, and whose sum is $A$. These vectors are, indeed, given by explicit formulas: $$\proj_{B}(A) = \frac{\Brak{A, B}}{\Brak{B, B}}\, B,\qquad \proj_{B^{\perp}}(A) = A - \proj_{B}(A)$$ The first is sometimes called the component of $A$ along $B$, and the second is the component of $A$ perpendicular/orthogonal to $B$.
The point is, the component of $A$ perpendicular to $B$ is unique (unles you have a definition that explicitly says otherwise) so "no", you need not/should not take both choices of sign.
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