If xa-b-yb-c-zc-a-r- and -a- b- c-18 then x-y-z-

The correct option is D 8[r⃗.(a⃗+b⃗+c⃗)] r⃗.a⃗=y[a⃗b⃗c⃗] r⃗.b⃗=z[a⃗b⃗c⃗] r⃗.c⃗=x[a⃗b⃗c⃗] Hence r⃗.[a⃗+b⃗+c⃗]=[x+y+z][a⃗b⃗c⃗] r⃗.[a ...

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Byju's Answer

Standard XII

Mathematics

Manipulation of a linear equation

If xa⃗×b⃗+y...

Question

If $x (\to a \times \to b) + y (\to b \times \to c) + z (\to c \times \to a) = \to r$ and $[\to a \to b \to c] = \frac{1}{8}$ then $x + y + z =$

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

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Solution

The correct option is D $8 [\to r . (\to a + \to b + \to c)]$
$\to r . \to a = y [\to a \to b \to c]$
$\to r . \to b = z [\to a \to b \to c]$
$\to r . \to c = x [\to a \to b \to c]$
Hence
$\to r . [\to a + \to b + \to c] = [x + y + z] [\to a \to b \to c]$
$\to r . [\to a + \to b + \to c] = [x + y + z] \frac{1}{8}$
$(8 \to r . [\to a + \to b + \to c]) = x + y + z$

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If x(→a×→b)+y(→b×→c)+z(→c×→a)=→r and [→a →b →c]=18 then x+y+z=

The correct option is D 8[→r.(→a+→b+→c)]→r.→a=y[→a→b→c]→r.→b=z[→a→b→c]→r.→c=x[→a→b→c]Hence→r.[→a+→b+→c]=[x+y+z][→a→b→c]→r.[→a+→b+→c]=[x+y+z]18(8→r.[→a+→b+→c])=x+y+z

If $x (\to a \times \to b) + y (\to b \times \to c) + z (\to c \times \to a) = \to r$ and $[\to a \to b \to c] = \frac{1}{8}$ then $x + y + z =$