If d = lambda (a times b) + mu (b times c) + | vedclass

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(A) Given $d = \lambda (a 	imes b) + \mu (b 	imes c) + 
u (c 	imes a)$ and $[a, b, c] = \frac{1}{8}$.Taking the dot product of $d$ with $c$:$d \cd

Vedclass · Accepted Answer

$8d \cdot (a + b + c)$

Vedclass · Answer

(A) Given $d = \lambda (a 	imes b) + \mu (b 	imes c) + 
u (c 	imes a)$ and $[a, b, c] = \frac{1}{8}$.Taking the dot product of $d$ with $c$:$d \cdot c = \lambda (a 	imes b) \cdot c + \mu (b 	imes c) \cdot c + 
u (c 	imes a) \cdot c$$d \cdot c = \lambda [a, b, c] + 0 + 0 = \lambda \left(\frac{1}{8}ight) \implies \lambda = 8(d \cdot c)$.Similarly,taking the dot product of $d$ with $a$:$d \cdot a = \lambda (a 	imes b) \cdot a + \mu (b 	imes c) \cdot a + 
u (c 	imes a) \cdot a$$d \cdot a = 0 + \mu [b, c, a] + 0 = \mu \left(\frac{1}{8}ight) \implies \mu = 8(d \cdot a)$.Similarly,taking the dot product of $d$ with $b$:$d \cdot b = \lambda (a 	imes b) \cdot b + \mu (b 	imes c) \cdot b + 
u (c 	imes a) \cdot b$$d \cdot b = 0 + 0 + 
u [c, a, b] = 
u \left(\frac{1}{8}ight) \implies 
u = 8(d \cdot b)$.Therefore,$\lambda + \mu + 
u = 8(d \cdot c) + 8(d \cdot a) + 8(d \cdot b) = 8d \cdot (a + b + c)$.

If $d = \lambda (a \times b) + \mu (b \times c) + \nu (c \times a)$ and $[a, b, c] = \frac{1}{8}$,then $\lambda + \mu + \nu$ is equal to

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