Answer
(a)The proof is shown below.
(b)$(A\times B)\cdot C=72.2$
Work Step by Step
We can express the three vectors $A,B$ and $C$ in terms of their components.
$A=A_{x}i+A_{y}j+A_{z}k$
$B=B_{x}i+B_{y}j+B_{z}k$
$C=C_{x}i+C_{y}j+C_{z}k$
(a) Now, $A\cdot(B\times C)=(A_{x}i+A_{y}j+A_{z}k)\cdot[(B_{y}C_{z}-B_{z}C_{y})i+(B_{z}C_{x}-B_{x}C_{z})j+(B_{x}C_{y}-B_{y}C_{x})k]$
$=A_{x}(B_{y}C_{z}−B_{z}C_{y})+A_{y}(B_{z}C_{x}-B_{x}C_{z})+A_{z}(B_{x}C_{y}-B_{y}C_{x})$->(1)
Again, $(A\times B).C=[(A_{y}B_{z}-A_{z}B_{y})i+(A_{z}B_{x}-A_{x}B_{z})j+(A_{x}B_{y}-A_{y}B_{x})k]\cdot(C_{x}i+C_{y}j+C_{z}k)$
$=(A_{y}B_{z}-A_{z}B_{y})C_{x}+(A_{z}B_{x}-A_{x}B_{z})C_{y}+(A_{x}B_{y}-A_{y}B_{x})C_{z}$
$=A_{x}(B_{y}C_{z}−B_{z}C_{y})+A_{y}(B_{z}C_{x}-B_{x}C_{z})+A_{z}(B_{x}C_{y}-B_{y}C_{x})$->(2)
On comparing equation (1) and (2) ,we find that
$A\cdot(B\times C)=(A\times B)\cdot C$
(b)Given:
$|A|=5.00$ and $\theta_{A}=26.0^{\circ}$
$|B|=4.00$ and $\theta_{B}=63.0^{\circ}$
$|C|=6.00$ in the (+)ve z-direction and $A$ and $B$ are in the xy-plane
Now,$|(A\times B)|=|A||B|sin\theta=5.00\times4.00sin(37.0^{\circ})=12.04$ , where $\theta=63.0^{\circ}-26.0^{\circ}=37.0^{\circ}$ is the angle between $A$ and $B$ from $A$ to $B$
By the right hand rule $(A\times B)$ is in the (+)ve z direction.
Thus,
$(A\times B).C=12.04\times6.00=72.2$
