use of triple scalar product to prove an equality

Question

$a,b,c$ are unit vectors mutually inclined at an angle $\theta$. Given that $$(a\times b) + (b\times c)=pa+qb+rc$$ where $p,q,r$ are constants and $a,b,c$ represent the above mentioned vectors. Prove that $$r^2 + p^2 + \frac{q^2}{\cos\theta}=2.$$

Answer 1

Let $\, \mathbf{v}=\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c} = p\mathbf{a}+q\mathbf{b}+r\mathbf{c}$

$\mathbf{a} \cdot \mathbf{v}$ gives

$$\mathbf{a} \cdot \mathbf{b} \times \mathbf{c} = p+(q+r)\cos \theta \tag{1}$$

$\mathbf{b} \cdot \mathbf{v}$ gives

$$0 = q+(r+p)\cos \theta \tag{2}$$

$\mathbf{c} \cdot \mathbf{v}$ gives

$$ \mathbf{c} \cdot \mathbf{a} \times \mathbf{b} = r+(p+q)\cos \theta \tag{3}$$

$(1)-(3)$, $$0=p(1-\cos \theta)+r(\cos \theta-1)$$

For $\cos \theta \ne 1$, $$p=r \tag{4}$$

Substitute $(4)$ in $(2)$,

$$q=-2p\cos \theta \tag{5}$$

$\mathbf{v} \cdot \mathbf{v}$ gives

$$ (\mathbf{a}\times \mathbf{b})^2+(\mathbf{b}\times \mathbf{c})^2+2 \begin{vmatrix} \mathbf{a} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{b} \\ \mathbf{a} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{c} \end{vmatrix} = p^2+q^2+r^2+2(pq+qr+rp)\cos \theta $$

\begin{align*} 2\sin^2 \theta+2(\cos^2 \theta-\cos \theta) &= (p^2+r^2)+q^2+2rp\cos \theta+2q(r+p)\cos \theta \\ 2(1-\cos \theta) &= 2p^2+4p^2\cos^2 \theta+2p^2\cos \theta-8p^2\cos^2 \theta \\ &= 2p^2(1+\cos \theta-2\cos^2 \theta) \\ &= 2p^2(1-\cos \theta)(1+2\cos \theta) \\ p^2 &= \frac{1}{1+2\cos \theta} \\ p^2+r^2+\frac{q^2}{\cos \theta} &= \frac{2}{1+2\cos \theta}+\frac{\cos \theta}{1+2\cos \theta} \\ &= 2 \end{align*} provided $\, -\dfrac{1}{2} < \cos \theta <1$