Answer:
Option D
Explanation:
We have , four vectors a,b,c and d such that a.b=0
$|a\times c|=|a||c|,|a\times d|=|a||d|$
From the given condition , we get
$a\perp c$ and $a\perp d$
[$\because$ $|x \times y|=|x||y| \sin theta $ and $\sin \theta =1 \Rightarrow \theta =\frac{\pi}{2}$]
$\Rightarrow$ $\underline{a}||(\underline{c}\times\underline{d})$
$\Rightarrow$ $c \times d =\lambda a$ , where $\lambda$ is a constant.
Now, consider $ [b c d]=b.(c \times d)$
$=b.(\lambda a) =\lambda (b.a)=\lambda (a.b)$
$= \lambda \times 0=0$ [$\because$ a.b=0]
(c) a,b, c are non -coplannar
$\therefore$ $b \times c, c \times a $ and $a \times b$ are also non-coplanar
So, any vector can be expressed as a linear combination of these vectors.
Let , $ d= \lambda (b \times c)+\mu (c \times a)+m(a \times b)$
$a.b=\lambda [a b c]$
$ b.d =\mu ( b c a )$
$c.d=m(c a b)$
$\therefore$ $d=\frac{(a.d)(b\times c)}{[a b c]}+\frac{(b.d)(c \times a)}{[a b c]}+\frac{(c.d)(a\times b)}{[a b c]}$
$\Rightarrow (a.d)(b\times c)+(b.d)(c\times a)+(c-d)(a\times b)=d[a b c]$
$\Rightarrow |(a.d)(b\times c)+(b.d)(c\times a)+(c.d)(a\times b)|=[a b c]$ [$\because$ |d|=1]
