If r - - b - c - - c - a - - a - b and -abc- - 2- then - - - - - is equal to

The correct option is B 12 r. (a + b + c) r.a = α (a. b × c) + β (a. c × a) + γ (a. a × b) = α [abc] + 0 + 0 Similarly, r.b = β [abc] and ...

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Byju's Answer

Standard IX

Chemistry

Properties of Alpha, Beta and Gamma Radiations

If r = α b ...

Question

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

r . [b \times c + c \times a + a \times b]

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

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2 r . (a + b + c)

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4

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Solution

The correct option is B $\frac{1}{2} r . (a + b + c)$
$r . a = α (a . b \times c) + β (a . c \times a) + γ (a . a \times b)$
$= α [a b c] + 0 + 0$
Similarly, $r . b = β [a b c]$ and $r . c = γ [a b c]$
$∴ \frac{1}{2} r . (a + b + c) = \frac{1}{2} (r . a + r . b + r . c)$
$= \frac{1}{2} (α + β + γ) [a b c]$
$= \frac{1}{2} (α + β + γ) \times 2 = α + β + γ$
Hence, $α + β + γ = \frac{1}{2} r \cdot (a + b + c)$

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Similar questions

Q. 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 =

Q. Let

a, b, c

be unit vectors with

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

and the angle between

r

and

a + b + c

\frac{π}{4}

with

| r | = \sqrt{2}

, then the maximum value of

α + β + γ

Q. If

a = c o s α + i s i n α, b = c o s β + i s i n β, c = c o s γ + i s i n γ

and

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

, then

c o s (β - γ) + c o s (γ - α) + c o s (α - β)

is equal to

Q. If

α, β, γ

be the three angles given by

cos α = \frac{a}{b + c}, cos β = \frac{b}{c + a}

and

cos γ = \frac{c}{a + b}

where

a, b, c

are the sides of a

△ A B C

, then

tan \frac{α}{2} tan \frac{β}{2} tan \frac{γ}{2}

is equal to

a sin α + b sin β + c sin γ = 0

= a cos α + b cos β + c cos γ

where

a, b, c, \in R

and

- π < α, β, γ \leq π .

Then assume

A = e^{i α}, B = e^{i β}, C = e^{i γ}

where

\sqrt{- 1} = i

∴ a A + b B + c C = 0

and

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

, then we get

(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,

then value of

α + β - γ

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If r=αb×c+βc×a+γa×b and [abc]=2, then α+β+γ is equal to

The correct option is B 12r.(a+b+c)r.a=α(a.b×c)+β(a.c×a)+γ(a.a×b)=α[abc]+0+0Similarly, r.b=β[abc] and r.c=γ[abc]∴12r.(a+b+c)=12(r.a+r.b+r.c)=12(α+β+γ)[abc]=12(α+β+γ)×2=α+β+γHence, α+β+γ=12r⋅(a+b+c)

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