Question 11.14: Leaning rod A rough floor lies in the horizontal plane z = 0...

The external forces acting on the rod are the normal reaction Ni of the wall, the tension force −Q j in the string, the weight force −Mgk, and reaction X of the floor.
The equilibrium equations are therefore

P i -Q j -M g k + X = 0,                        (11.31)

(2 L n ) \times(P i -Q j )+(L n ) \times(-M g k )= 0,                    (11.32)

where we have taken moments about A to eliminate the reaction X. Here, 2L is the length of the rod, and n is the unit vector in the direction \overrightarrow{A B}.
Equation (11.31) serves only to determine the reaction X once P and Q are known. To extract P and Q from the vector equation (11.32), we take components in any two directions other than the n-direction; the easiest choices are the i- and j-directions. On taking the scalar product of equation (11.32) with i, we obtain

\begin{aligned}0 &=2[ n , P i -Q j , i ]+[ n ,-M g k , i ] \\&=2 P[ n , i , i ]-2 Q[ n , j , i ]-M g[ n , k , i ] \\&=0-2 Q n \cdot( j \times i )-M g n \cdot( k \times i ) \\&=2 Q( n \cdot k )-M g( n \cdot j )\end{aligned}

where we have used the notation [u, v, w] to mean the triple scalar product of the vectors u, v and w. Hence

Q=\frac{M g( n \cdot j )}{2( n \cdot k )},

and, by taking the scalar product of equation (11.32) with j and proceed in the same way, we obtain

P=-\frac{M g( n \cdot i )}{2( n \cdot k )}.

Finally, we need to express these answers in terms of the data given in the question. Since the unit vector n is given by

n =\frac{-a i +b j +c k }{2 L},

it follows that the reaction exerted by the wall, and the tension in the string, are

P=\frac{M g a}{2 c}, \quad Q=\frac{M g b}{2 c}.