If -a-b-c-1- a-b-b-c-c-a-12 then -a- b-c-

The correct option is A 1√(2) [a, b, c]^2=[ nbsp;a.a nbsp; amp; nbsp;a.b nbsp; amp; nbsp; a.c nbsp;b.a amp; b.b nbsp; amp; ...

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Applications of Dot Product

If |a|=|b|=...

Question

If $∣ ∣ \to a ∣ ∣ = ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣ = 1, \to a . \to b = \to b . \to c = \to c . \to a = \frac{1}{2}$ then $[\to a, \to b, \to c] =$

1

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\frac{1}{2}

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\frac{1}{\sqrt{2}}

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\frac{\sqrt{3}}{2}

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∣ ∣ \to a ∣ ∣ = ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣ = 1

\to a . \to b = \to b . \to c = \to c . \to a = cos θ

θ

1 + (x y) + (x y)^{2} + (x y)^{3} + . . . .

a = 1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . (- 1 < a < 1)

b = 1 + 3 y + 6 y^{2} + 10 y^{3} + . . . (- 1 < b < 1)

{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} = a^{2}

{(x_{2} - x_{3})}^{2} + {(y_{2} - y_{3})}^{2} = b^{2}

{(x_{3} - x_{1})}^{2} + {(y_{3} - y_{1})}^{2} = c^{2}

k ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = (a + b + c) (b + c - a) (c + a - b) \times (a + b - c)

k

a^{2} + b^{2} + c^{2} = - 2

f (x) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} x & (1 + b^{2}) x & (1 + c^{2}) x (1 + a^{2}) x & 1 + b^{2} x & (1 + c^{2}) x (1 + a^{2}) x & (1 + b^{2}) x & 1 + c^{2} x \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣

f (x)

| a | < 1

| b | < 1

s = 1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + . . . i s - .

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