Given two orthogonal vectors $A$, $B$, in $\mathbb R^3$ each of length 1. Let $P$ be a vector satisfying the equation $P\times B=A-P$.

Question

Given two orthogonal vectors $A$, $B$, in $\mathbb R^3$ each of length 1. Let $P$ be a vector satisfying the equation $P\times B=A-P$. Prove each of the following statements.

(a) $P$ is orthogonal to $B$ and has length $\frac{1}{2}\sqrt{2}$.

(b) $P, B, P\times B$ form a basis.

(c) $(P\times B)\times B=-P$.

(d) $P=\frac{1}{2}A-\frac{1}{2}(A\times B)$.

I solved the first part of (a). $A \cdot B=(P+(P\times B))\cdot B=P\cdot B=0$. However, I don't know how to find the length of $P$. What I got so far is $||P\times B||=||P||||B||=||P||=||A-P||$.

(b) is trivial since all three vectors are linearly independent. (c) and (d) I haven't been able to get so far.

I would appreciate any solutions or suggestions for the second part of (a) and (c), (d).

Answer 1

Okay !! It looks lengthy. Here are my inputs on this problem:

(a) You have already shown That $P.B = 0$. By the identity we have: $$\implies A\times(P\times B) = A\times(A - P)$$ $$\implies (A \cdot B)P - (A \cdot P) B = A\times A - A \times P$$ $$\implies (A \cdot P) B = A \times P $$ $$\implies || A \cdot P || ||B|| = || A \times P||$$ $$\implies ||A|| ||P|| \cos{\theta} = ||A|| ||P|| \sin{\theta}$$ $$\implies \theta = \frac{\pi}{4}.$$ where $\theta$ is angle between $A$ and $P$. Now again from the relation: $$ P = A - (P\times B) \implies P\cdot P = A\cdot P - (P\times B).A$$ $$\implies ||P||^2 = ||A||\, ||P|| \cos{\frac{\pi}{4}} \implies ||P|| = \frac{1}{\sqrt{2}} = \frac{1}{2}\sqrt{2}.$$ (c) It can be derived just by a vector identity (help): $$(P\times B)\times B = (P\cdot B)B - (B\cdot B)P = -P.$$ (d) $$(P\times B)\times B = -P \implies P = -(A - P)\times B$$ $$\implies P = - A \times B + P \times B = - A \times B + (A - P)$$ $$\implies 2 P = A - A \times B$$ $$ P = \frac{1}{2}A -\frac{1}{2}(A \times B).$$

It is rather long but I hope it will help.

Answer 2

Consider the linear map $L(v):=v\times B+v$ and show that $L$ is bijective. Hence $L(v)=A$ has a unique solution, namely $P$. Now show that $P$ in (d) is the solution. Plug this solution in (c).