Question 1.5: Three field quantities are given by P = 2ax - az, Q = 2ax - ...

(a)

(P + Q) × (P – Q) = P × (P – Q) + Q × (P – Q)
= P × P – P × Q + Q × P – Q × Q
= 0 + Q × P + Q × P – 0
= 2Q × P

=2\left | \begin{matrix} a_{x} & a_{y} & a_{z} \\ 2 & -1 & 2 \\ 2 & 0 & -1 \end{matrix} \right |

=2\left(1-0\right)a_{x}+2\left(4+2\right)a_{y}+2\left(0+2\right)a_{z}=2a_{x}+12a_{y}+4a_{z}

(b) The only way Q · R × P makes sense is

Q\cdot \left(R\times P\right)=\left(2,-1,2\right)\cdot \left | \begin{matrix} a_{x} & a_{y} & a_{z} \\ 2 & -3 & 1 \\ 2 & 0 & -1 \end{matrix} \right |

=\left(2,-1,2\right)\cdot \left(3,4,6\right) =6-4+12=14

Alternatively:

Q\cdot \left(R\times P\right)=\left | \begin{matrix} 2 & -1 & 2 \\ 2 & -3 & 1 \\ 2 & 0 & -1 \end{matrix} \right |

To find the determinant of a 3 × 3 matrix, we repeat the first two rows and cross multiply; when the cross multiplication is from right to left, the result should be negated as shown diagrammatically here. This technique of finding a determinant applies only to a 3 × 3 matrix. Hence from figure,

Q\cdot \left(R\times P\right)=+6+0-2+12-0-2=14

as obtained before.

(c) From eq. (1.29)

A · (B × C) = B · (C × A) = C · (A × B)

P · (Q × R) = Q · (R × P) = 14

or

P · (Q × R) = (2, 0, -1) · (5, 2, -4)
= 10 + 0 + 4 = 14

(d)

\sin \theta_{QR}=\frac{\left|Q\times R\right| }{\left|Q\right| \left|R\right| } =\frac{\left|\left(5,2,-4\right) \right| }{\left|\left(2,-1,2\right) \right|\left|\left(2,-3,1\right) \right| }

=\frac{\sqrt{45} }{3\sqrt{14} } =\frac{\sqrt{5} }{\sqrt{14} } =0.5976

(e)

P × (Q × R) = (2, 0, -1) × (5, 2, -4)
= (2, 3, 4)

Alternatively, using the bac-cab rule,

P × (Q × R) = Q(P · R) – R(P · Q)
= (2, -1, 2) (4 + 0 – 1) – (2, -3, 1) (4 + 0 – 2)
= (2, 3, 4)

(f ) A unit vector perpendicular to both Q and R is given by

a=\frac{\pm Q\times R}{\left|Q\times R\right| } =\frac{\pm \left(5,2,-4\right) }{\sqrt{45} } =\pm \left(0.745,0.298,-0.596\right)

Note that |a| = 1, a · Q = 0 = a · R, Any of these can be used to check a.

(g) The component of P along Q is

P_{Q}=\left|P\right| \cos \theta_{PQ}a_{Q}

=\left(P \cdot a_{Q}\right)a_{Q}=\frac{\left(P\cdot Q\right)Q }{\left|Q\right|^{2} }

=\frac{\left(4+0-2\right)\left(2,-1,2\right) }{\left(4+1+4\right) }=\frac{2}{9}\left(2,-1,2\right)

=0.4444a_{x}-0.2222a_{y}+0.4444a