Given that,
|x|=|y|=|z|=√2 and
θ=60∘ Thus,
x⋅y=|x||y|cosθ=(√2)(√2)cos60∘
=2×=1 Similarly
y⋅z=z⋅x=1 and
x⋅x=|x||x|cos0∘=(√2)(√2)×1=2
y⋅y=2z⋅z=2 Now,
a=x×(y×z)=(x⋅z)y−(x⋅y)z=1⋅y−1⋅z
a=y−z....(i)
Also,
b=y×(z×x)=(y⋅z)z−(y⋅z)x=1⋅z−1⋅x
b=z−x....(ii)
Adding Eqs. (i) and (ii), we get
a+b=(y−z)+(z−x) y−x=a+b...(iii)
Again
c=x×y Taking cross product with
x on both sides
x×c=x×(x×y)=(x⋅y)x−(x⋅x)y x×c=(1)x−(2)y⇒x×c=x−2y....(iv)
Finally,
c=x×y Taking cross product with
y on both sides
y×c=y×(x×y) y×c=(y⋅y)x−(y⋅x)y=(2)x−(l)y
y×c=2x−y...(v)
Subtracting Eq. (v) from Eq. (iv), we get
x×c−y×c=(+x−2y)−(2x−y)
(x−y)×c=−x−y or
x+y=(y−x)×c From Eq. (iii),
y−x=a+b ∴x+y=(a+b)×c....(vi)
Subtracting Eq. (iii) from Eq. (vi), we get
(x+y)−(y−x)=(a+b)×c−(a+b)
2x=(a+b)×c−(a+b) or
x=[(a+b)×c−(a+b)]