Forces $P, Q, R$ act along three non-intersecting edges of a cube; which of the following is the central axis.

\frac{- a P - x Q + y P}{R}

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\frac{- a R - y Q + x P}{R}

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\frac{- a R - x P + y Q}{R}

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\frac{a R + x Q + y P}{R}

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Solution

The correct option is A $\frac{- a P - x Q + y P}{R}$
Let the three forces $P, Q, R$ act along. the three non-intersecting edges $B C^{'}, C A^{'}$ and $A B^{'}$ respectively of a cube of side $a$ .
Therefore, we have
(i) the force $P$ acting at the point $B (0, a, 0)$ along $B C^{'}$ whose d.c.'s are $1, 0, 0$ ;
(ii) the force $Q$ acting at the point $C (0. 0, a)$ along $C A^{'}$ whose d.c.'s are $0, 1, 0$ ;
and (iii) the force $R$ acting at the point $A (a, 0, 0)$ along $A B^{'}$ whose d.c.'s are $0, 0, 1$ .
The components $(X_{1}, Y_{1}, Z_{1})$ etc., of these forces parallel to the axes are :
$X_{1} = P, X_{2} = 0, X_{3} = 0$
$Y_{1} = 0, Y_{2} = Q, Y_{3} = 0$
$Z_{1} = 0, Z_{2} = 0, Z_{3} = R$
If these three forces reduce to a single force $R = (X, Y, Z)$ acting at O and a couple $G = (L, M, N)$ , then
$X = 3 \sum r = 1 X_{r} = X_{1} + X_{2} + X_{3} = P, Y = 3 \sum r = 1 Y_{r} = Q, Z = 3 \sum r = 1 Z_{r} = R$
$L = 3 \sum r = 1 (y_{r} Z_{r} - z_{r} Y_{r}) = (y_{1} Z_{1} - z_{1} Y_{1}) + (y_{2} Z_{2} - z_{2} Y_{2}) + (y_{3} Z_{3} - z_{3} Y_{3})$
$= (0 - 0) + (0 - a Q) + (0 - 0) = - a Q$ ,
$M = 3 \sum r = 1 (z_{r} X_{r} - x_{r} Z_{r}) = (z_{1} X_{1} - x_{1} Z_{1}) + (z_{2} X_{2} - x_{2} Z_{2}) + (z_{3} X_{3} - x_{3} Z_{3}) = - a R$
and $N = 3 \sum r = 1 (x_{r} Y_{r} - y_{r} X_{r}) = (x_{1} Y_{1} - y_{1} X_{1}) + (x_{2} Y_{2} - y_{2} X_{2}) + (x_{3} Y_{3} - y_{3} X_{3}) = - a P$ .
$∴$ the equations of the central axis are
$\frac{L - y Z + z Y}{X} = \frac{M - z X + x Z}{Y} = \frac{N - x Y + y X}{Z}$
or $\frac{- a Q - y R + z Q}{P} = \frac{- a R - z P + x R}{Q} = \frac{- a P - x Q + y P}{R}$

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