x, y, z are three vectors each of magnitude s | vedclass

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(A) Given that, $|x|=|y|=|z|=\sqrt{2}$ and $	heta=60^{\circ}$.Thus, $x \cdot y = |x||y| \cos 60^{\circ} = (\sqrt{2})(\sqrt{2}) 	imes \frac{1}{2} = 1

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$\frac{1}{2}[(a+b) 	imes c-(a+b)]$

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(A) Given that, $|x|=|y|=|z|=\sqrt{2}$ and $	heta=60^{\circ}$.Thus, $x \cdot y = |x||y| \cos 60^{\circ} = (\sqrt{2})(\sqrt{2}) 	imes \frac{1}{2} = 1$.Similarly, $y \cdot z = z \cdot x = 1$ and $x \cdot x = y \cdot y = z \cdot z = |x|^2 = 2$.Now, $a = x 	imes (y 	imes z) = (x \cdot z)y - (x \cdot y)z = 1 \cdot y - 1 \cdot z = y - z$ $\dots(i)$.Also, $b = y 	imes (z 	imes x) = (y \cdot x)z - (y \cdot z)x = 1 \cdot z - 1 \cdot x = z - x$ $\dots(ii)$.Adding $(i)$ and $(ii)$, we get $a + b = y - x$, so $y - x = a + b$ $\dots(iii)$.Given $c = x 	imes y$. Taking cross product with $x$ and $y$ respectively:$x 	imes c = x 	imes (x 	imes y) = (x \cdot y)x - (x \cdot x)y = x - 2y$ $\dots(iv)$.$y 	imes c = y 	imes (x 	imes y) = (y \cdot y)x - (y \cdot x)y = 2x - y$ $\dots(v)$.Subtracting $(v)$ from $(iv)$, we get $(x - y) 	imes c = (x - 2y) - (2x - y) = -x - y$, which implies $x + y = (y - x) 	imes c$.Substituting $y - x = a + b$ from $(iii)$, we get $x + y = (a + b) 	imes c$ $\dots(vi)$.Subtracting $(iii)$ from $(vi)$, we get $(x + y) - (y - x) = (a + b) 	imes c - (a + b)$.$2x = (a + b) 	imes c - (a + b)$.Therefore, $x = \frac{1}{2}[(a + b) 	imes c - (a + b)]$.

$x, y, z$ are three vectors each of magnitude $\sqrt{2}$ and each making an angle $60^{\circ}$ with one another. If $a=x \times(y \times z), b=y \times(z \times x)$, $c=x \times y$, then $x=$

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