Let r- be a vector perpendicular to a- - b- - c- where -a-b-c- - 2- If r- - lb-c- - m c-a- - n a-b- then l - m - n is equal to

The correct option is C 0Given, r #160; a #160; + b #160; + c i.e. #160; [ r a + r b + r c =0..........( 1 ) ...

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Graphical Interpretation of Differentiability

Let r⃗ be a...

Question

Let $\to r$ be a vector perpendicular to $\to a + \to b + \to c$ , where $[\to a \to b \to c] = 2$ . If $\to r = l (\to b \times \to c) + m (\to c \times \to a) + n (\to a \times \to b)$ , then (l + m + n) is equal to

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\to a, \to b, \to c

\to r

(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b)

\to a, \to b, \to c

\to r

[\to b \to c \to r] \to a + [\to c \to a \to r] \to b + [\to a \to b \to r] \to c

\to a, \to b, \to c

\to d

(\to a \times \to b) . (\to c \times \to d) = λ

\to a . \to c = \frac{\sqrt{3}}{2}

\to r

\to a + \to b + \to c

[\to a \to b \to c] = z

\to r = ℓ (\to b \times \to c) + m (\to c \times \to a) + n (\to a \times \to b)

l + m + n

\to a, \to b, \to c

\to d = λ \to a + μ \to b + ν \to c,

λ

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