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Q.

If $a, b c$ are non-coplanar vectors, then find the value of
$\frac{(a + 2 b - c) . (a - b) \times (a - b - c)}{[abc]}$

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a

1

b

2

c

3

d

4

answer is C.

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Detailed Solution

Given:

\frac{(a + 2 b - c) . (a - b) \times (a - b - c)}{[abc]}

[equation I]
Now, as we know that

a, b c

are non-coplanar vectors, therefore, their values are not equal to zero.
Now, let’s first evaluate

(a - b) \times (a - b - c)

term.

\Rightarrow (a - b) \times (a - b - c) = a \times a - a \times b - a \times c - b \times a + b \times b + b \times c

= 0 - a \times b - a \times c - b \times a + 0 + b \times c

[As, cross-product of two similar vectors is equal to zero]

= - a \times b + c \times a + a \times b + b \times c

[We know, in cross-product of vectors,

- b \times a = a \times b

]

= b \times c + c \times a

[Cancelling,

- a \times b

&

\times b

]
Therefore, putting the value in equation I,

\Rightarrow \frac{(a + 2 b - c) . (a - b) \times (a - b - c)}{[abc]} = \frac{(a + 2 b - c) . (b \times c + c \times a)}{[abc]}

= \frac{a . (b \times c) + a . (a \times c) + 2 b . (b \times c) + 2 b . (a \times c) - c . (b \times c) - c . (c \times a)}{[abc]}

= \frac{a . (b \times c) + 0 + 0 + 2 b . (a \times c) - 0 - 0}{[abc]}

[As,

a . (a \times c) = 0

,

2 b . (b \times c) = 0

,

c . (b \times c) = 0

&

c . (c \times a) = 0

]

= \frac{[abc] + 2 [abc]}{[abc]}

[

a . (b \times c) = [abc]

]

= \frac{3 [abc]}{[abc]}

= 3

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