Component of a vector perpendicular to another vector.

Question

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?

Answer 1

$\newcommand{\Brak}[1]{\langle #1 \rangle}\DeclareMathOperator{\proj}{proj}$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.