If r- a-b- - b-c- - c-a- and - a- b- c- - 1 8 - then - - - is equal to

The correct option is A 8r⃗.( a⃗+b⃗+c⃗) Clearly, #160;r.c=λ[ abc ] = 1 8 λ #160;r.a=μ[ abc ] = 1 8 μ #160;r.b=v[ abc ] = 1 8 v #16 ...

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Byju's Answer

Standard XII

Mathematics

Summation Using Sigma

If r⃗=λ a⃗×...

Question

If $\to r = λ (\to a \times \to b) + μ (\to b \times \to c) + ν (\to c \times \to a)$ and $[\to a \to b \to c] = \frac{1}{8},$ then $λ + μ + ν$ is equal to

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

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8 \to r . (\to a + \to b + \to c)

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4 \to r . (\to a + \to b + \to c)

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None of these

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Solution

The correct option is A $8 \to r . (\to a + \to b + \to c)$
Clearly, $r . c = λ [a b c] = \frac{1}{8} λ$
$r . a = μ [a b c] = \frac{1}{8} μ$
$r . b = v [a b c] = \frac{1}{8} v$
$∴ r . (a + b + c) = \frac{1}{8} (λ + μ + v)$
$∴ λ + μ + v = 8 r . (a + b + c)$

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If →r=λ(→a×→b)+μ(→b×→c)+ν(→c×→a) and [→a →b →c]=18, then λ+μ+ν is equal to

The correct option is A 8→r.(→a+→b+→c)Clearly, r.c=λ[abc]=18λr.a=μ[abc]=18μr.b=v[abc]=18v∴r.(a+b+c)=18(λ+μ+v)∴λ+μ+v=8r.(a+b+c)

If $\to r = λ (\to a \times \to b) + μ (\to b \times \to c) + ν (\to c \times \to a)$ and $[\to a \to b \to c] = \frac{1}{8},$ then $λ + μ + ν$ is equal to

The correct option is A $8 \to r . (\to a + \to b + \to c)$
Clearly, $r . c = λ [a b c] = \frac{1}{8} λ$
$r . a = μ [a b c] = \frac{1}{8} μ$
$r . b = v [a b c] = \frac{1}{8} v$
$∴ r . (a + b + c) = \frac{1}{8} (λ + μ + v)$
$∴ λ + μ + v = 8 r . (a + b + c)$