Answer:
Option A,B,C
Explanation:
Plan
(i) Dot product let a and be be two non-zero vectors inclined at an angle θ . Then a.b=|a| |b| cos θ
(ii) The direction of a x b is perpendicular to the plane of a and b
(iii) Vector triple product
a x (b x c)= (a.c) b- (a.b) c
Given that |x|=|y|=|z|= $\sqrt{2}$
and angle between each pair is $\frac{\pi}{3}$
$\therefore$ $ x.y=|x||y|\cos\frac{\pi}{3}$
$=(\sqrt{2})(\sqrt{2})(\frac{1}{2})=1$
Similarly, y.z=z.x=1
also, x . x=|x|2 =2
similarly, y.y=z.z= 2
now, $a \bot $ x and y $\times$ x
Let, a= $\lambda $ [x $\times$ (y $\times$ z)]
a= $\lambda $ [(x.z)y-(x.y)z]
a=$\lambda $ (y-z)
(as x.z =x.y=1)
a.y= $\lambda $ (y-z).y
a.y= $\lambda $(y.y-y.z)=$\lambda $ (2-1)
( as y.y=2=y.z=1)
a.y= $\lambda $
$\Rightarrow$ a= (a.y) (y-z) .......(i)
Also , $b \bot $ y and z $\times$ x
b= $\mu$ { y x (z x x)]
= $\mu$ [(y .x)z-(y.z)x]
= $\mu$ (z-x)
[as y.x=y.z=1)
b.z= $\mu$ (z-x).z
= $\mu$ (z.z -x.z]
= $\mu$ (2-1)= $\mu$
$\Rightarrow$ b= (b.z) (z-x) ...........(ii)
Now using Eqs.( i) and (ii), we get
a.b= (a.y)(y-z).(b.z)(z-x)
= (a.y)(b.z)(y.z-y.x. -z.z+z.x)
=(a,y)(b.z)(1-1-2+1)
a.b= -(a.y)(b.z)
