Let $\vec{A}$ and $\vec{B}$ be two vectors that are perpendicular to each other. Find the flaw in the logic.

Question

Take an arbitrary vector $\vec{K}$ in any direction between $\vec{A}$ and $\vec{B}$. We can easily show that $\vec{A}$ has a component along $\vec{K}$, i.e. $\vec{A_{K}}$. Now $\vec{A_{K}}$ has a component along $\vec{B}$. Hence $\vec{A}$ has a component along $\vec{B}$. But $\vec{A}$ and $\vec{B}$ are perpendicular. Find the error in the logic.

Answer 1

Without loss of generality, we can take $A$ and $B$ to be unit vectors that are orthogonal to each other, and also without loss of generality we can take

$ A = (0, 1) $

$ B = (1, 0) $

Also, let's take $K$ as follows

$ K = (\cos \theta, \sin \theta ) $ where $ 0 \lt \theta \lt 90^\circ $

Now all our vectors are unit vectors. To show the flaw in the logic, I need the orthogonal vector to $K$. Let's call it $M$, then

$ M = (- \sin \theta , \cos \theta ) $

Decomposing $A$ onto $K$ and $M$ , we find that

$ A_K = \dfrac{(A \cdot K)}{(K \cdot K)} K = \sin \theta ( \cos \theta, \sin \theta) $

Similarly

$ A_M = \dfrac{(A \cdot M) }{ (M \cdot M) } M = \cos \theta (- \sin \theta, \cos \theta ) $

So that

$ A = A_K + A_M $

Now if we project $A$ onto $B$, the results will be the zero vector because $(A \cdot B) = 0 $. Now comes the point,

Take the projection of $A_K$ onto $B$, you get

$ (A_K)_B = \dfrac{A_K \cdot B }{B \cdot B} B = \sin \theta \cos \theta (1, 0) $

To complete the story of the projection of $A$ onto $B$, we need also to find the projection of $A_M$ on to $B$, and this is given by

$ (A_M)_B = \dfrac{A_M \cdot B}{B \cdot B } B = - \sin \theta \cos \theta (1, 0)$

Therefore,

$ (A_K)_B + (A_M)_B = (0, 0) $ the zero vector (as expected)

In summary, to compare the projection of a vector $A$ onto another vector $B$ to the projection of $A$ onto a third vector $K$ , we need an additional vector $M$ so that $A$ can be fully decomposed as a linear combination of its projections onto $K$ and $M$. And then, the projection of $A$ onto $B$ has to be compared to the vectorial sum of its projections $A_K$ and $A_M$ onto $B$.

Answer 2

The error in the logic is assuming that because $\vec{A_k}$ has a component along $\vec{B}$, $\vec{A}$ also has a component along $\vec{B}$.

$\vec{A_K}$ has an arbitrary angle between $\vec{A}$ and $\vec{B}$, and cannot be used to prove that $\vec{A}$ and $\vec{B}$ are perpendicular.

Answer 3

The error in your logic is that projecting along two different vectors in succesion is not in general a projection. Define $$P_K: \mathbb{R}^2 \rightarrow \text{span K} \\ v \rightarrow \langle v,K \rangle K$$ and $$P_B: \text{span K}\rightarrow \text{span B} \\ v \rightarrow \langle v,B \rangle B$$ Then applying $(P_B \circ P_K)(A)$ gives us $$P_B(\langle A,K\rangle K)=\langle A,K\rangle \langle K,B\rangle B$$

This of course corresponds not to a general projection along $B$ which is $$P_B ^{\text{general}}: \mathbb{R}^2 \rightarrow \text{span B} \\ v \rightarrow \langle v,B \rangle B$$

Answer 4

$A$ does have a component along $B$: the vector $0$. Now, what you probably meant is that $A$ has a non-zero component along $K$: $A_K$, and $A_K$ has a non-zero component along $B$. IF having a non-zero component were a transitive relation, you'll have that $A$ has a non-zero component along $B$, a contradiction. That simply means that "having a non-zero component" is not transitive.