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I've lately been facing lot of trouble in solving vector equations.Like the one below :

Let $a,b,c$ be non-coplanar unit vectors,equally inclined to one another at an angle $\theta$.If $a×b+b×c=pa+qb+rc$,find the scalars p,q and r in terms of $\theta$.

What would be the shortest method to solve this problem? In my book they took nearly 2 pages!But I guess there might be a shorter method to solve such type of problems.What say ?

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    $\begingroup$ How does your book solve this? My first instinct would be to take the dot product of that equation with $a,b$, and $c$. Using $u \cdot (u \times v) = 0 \quad \forall u,v \in \Bbb{R}$, $a \cdot a = b \cdot b = c \cdot c$, and $a \cdot b = b \cdot c = a \cdot c = \cos\theta$, we get three equations in $p,q,r$ and trigonometric functions of $\theta$. The only hard/boring part is computing $a \cdot (b \times c)$ (cfr. triple product and this formula). $\endgroup$ – A.P. Dec 25 '15 at 23:56
  • $\begingroup$ @A.P. I think 3 dot products won't suffice.In my book they used cross product and lot of (which seem unnecessary) geometric manipulations.Phew! Well even I was thinking of applying dot product :-) $\endgroup$ – user220382 Dec 26 '15 at 0:03
  • $\begingroup$ As long as the resulting linear equations are independent there won't be any problem. The determinant of the corresponding matrix is $1 + 2 \cos^3\theta - 3 \cos^2 \theta$, which is $0$ only at $2k\pi$ and $\pm \frac{2}{3}\pi + 2k\pi$ for $k \in \Bbb{Z}$. The first case is easy and the other two can be considered separately. $\endgroup$ – A.P. Dec 26 '15 at 0:16
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First of all, observe that we can safely assume $0 < \theta < \frac{2}{3} \pi$. Indeed, it makes sense to consider $\theta$ to be the smallest (positive) angle between, say, $a$ and $b$. Furthermore, clearly for $\theta = 0$ we have $a = b = c$, while for $\theta = \frac{2}{3} \pi$ we have $a,b,c$ coplanar.

Now, recall that $$ a \cdot (b \times c) = c \cdot (a\times b) $$ coincides with the volume of the parallelepiped with sides defined by $a,b,c$, i.e. $$ V = \sqrt{1 + 2 \cos^3\theta - 3 \cos^2\theta}. $$

Since we're after a representation of a vector in the $\Bbb{R}^3$-basis $\{a,b,c\}$, let's consider the system of linear equations obtained by taking the dot product of $a \times b + b \times c = pa + qb + rc$ by $a,b$, and $c$, respectively: $$ \begin{cases} a \cdot (a \times b) + a \cdot (b \times c) = p(a \cdot a) + q(a \cdot b) + r(a \cdot c) \\ b \cdot (a \times b) + b \cdot (b \times c) = p(b \cdot a) + q(b \cdot b) + r(b \cdot c) \\ c \cdot (a \times b) + c \cdot (b \times c) = p(c \cdot a) + q(c \cdot b) + r(c \cdot c). \end{cases} $$ Recalling that $u \cdot (u \times v) = 0$ for every $u,v \in \Bbb{R}^3$, that $u \cdot u = 1$ for every unit vector $u$, and that $a \cdot b = b \cdot c = a \cdot c = \cos\theta$, this becomes $$ \begin{cases} p + q \cos\theta + r \cos\theta = V \\ p \cos\theta + q + r \cos\theta = 0 \\ p \cos\theta + q \cos\theta + r = V \end{cases} \tag{1} \label{eq:1} $$ or, in matrix form $$ A \mathbf{p} = \begin{pmatrix} 1 & \cos\theta & \cos\theta \\ \cos\theta & 1 & \cos\theta \\ \cos\theta & \cos\theta & 1 \end{pmatrix} \begin{pmatrix} p \\ q \\ r \end{pmatrix} = \begin{pmatrix} V \\ 0 \\ V \end{pmatrix}. $$ Interestingly, for $0 < \theta < \frac{2}{3} \pi$ we have that $\det A = V^2 > 0$ (try to see why), so $\eqref{eq:1}$ has exactly one solution, which is $$ \begin{pmatrix} \dfrac{1-\cos\theta}{V},\; \dfrac{2 (\cos\theta-1) \cos\theta}{V},\; \dfrac{1-\cos\theta}{V} \end{pmatrix}^T. $$


† Since doing this by hand is a bit tedious, here's how to compute it in Mathematica

m = {{1, Cos[t], Cos[t]}, {Cos[t], 1, Cos[t]}, {Cos[t], Cos[t], 1}};
v = Sqrt[Det[m]];
p = Assuming[0 <= t <= 2/3 Pi, LinearSolve[m, {v, 0, v}]] // Simplify
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  • $\begingroup$ Nice...but in the ending I was hoping for a manual method...as I might need to solve it manually during exams :-P.Otherwise I loved that parallelopiped thing :-D $\endgroup$ – user220382 Dec 26 '15 at 15:47
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    $\begingroup$ Oh, $3 \times 3$ linear systems are quite manageable by hand, it's just that I don't trust myself with most non-straightforward computations. $\endgroup$ – A.P. Dec 26 '15 at 16:13

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