If r-b-c-c-a-a-b and -abc-2 - then -

The correct option is B 12r(a+b+c) We have nbsp; r=αb×c+βc×a+γa×b nbsp; ⇒r.a=αb×c.a+βc×a.a+γa.×b.a nbsp; r.a=αb×c.a+0+0 r.a=αb×c. ...

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Condition for Two Lines to be on the Same Plane

If r=αb×c+β...

Question

If $\to r = α \to b \times \to c + β \to c \times \to a + γ \to a \times \to b$ and $[\to a \to b \to c] = 2$ , then $α + β + γ =$

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

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\frac{1}{2} \to r (\to a + \to b + \to c)

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

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4

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Solution

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Similar questions

r = α b \times c + β c \times a + γ a \times b

[a b c] = 2

α + β + γ

a, b, c

r = α b \times c + β c \times a + γ a \times b, [a b c] = 2

r

a + b + c

\frac{π}{4}

| r | = \sqrt{2}

α + β + γ

a sin α + b sin β + c sin γ = 0

= a cos α + b cos β + c cos γ

a, b, c, \in R

- π < α, β, γ \leq π .

A = e^{i α}, B = e^{i β}, C = e^{i γ}

\sqrt{- 1} = i

∴ a A + b B + c C = 0

\frac{a}{A} + \frac{b}{B} + \frac{c}{C} = 0

(a A)^{3} + (b B)^{3} + (c C)^{3} = 3 a b c A B C

{(\frac{a}{A})}^{3} + {(\frac{b}{B})}^{3} + {(\frac{c}{C})}^{3} = \frac{3 a b c}{A B C}

sin α + 2 sin β + 3 sin γ = 0,

α + β - γ

\to a, \to b, \to c

\to P = \to a + \to b + \to c

\to Q = - \to 4 a + \to 3 b + \to 4 c

\to R = \to a + α \to b + β \to c

α

a = cos α + i sin α, b = cos β + i sin β, c = cos γ + i sin γ

\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = - 1

cos (β - γ) + cos (γ - α) + cos (α - β) =

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