If d = ,(a b) + ,(b c) + ,(c a)and [a ,b ,c] = 1 8 , then + + is … - Vidyadip

MCQ

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

✓
$8d\,.\,(a + b + c)$
B
$8d\, \times \,(a + b + c)$
C
$\frac{{d\,}}{8}.\,(a + b + c)$
D
$\frac{{d\,}}{8} \times \,(a + b + c)$

✓

Answer

Correct option: A.

$8d\,.\,(a + b + c)$

a
(a) $d\,.\,c = \lambda (a \times b)\,.\,c + \mu (b \times c)\,.\,c + \nu (c \times a)\,.c$

$ = \lambda \,[a\,b\,c] + 0 + 0 = \lambda \,[a\,b\,c] = \frac{\lambda }{8}$

Hence $\lambda = 8(d\,.\,c),$ $\mu = 8(d\,.\,a)$ and $\nu = 8(d\,.\,b)$

Therefore, $\lambda + \mu + \nu = 8d\,.\,c + 8d.\,a + 8d\,.\,b$

$ = 8d\,.\,(a + b + c).$

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