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

The correct option is A b c×a=( a×b)×(b×c)2 nbsp; nbsp;(given) nbsp; =12( nbsp;a×b)×(b×c) We know that a×b=c,b×c=2a we get nbsp ...

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Multiplication with Vectors

If a×b=c,b×...

Question

If $\to a \times \to b = \to c, \to b \times \to c = 2 \to a$ and $[\to a \to b \to c] = 2$ then $\to c \times \to a =$

\to b

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- \to b

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2 \to b

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\frac{1}{2} \to b

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Solution

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Similar questions

A B C,

cot (\frac{A}{2}) cot (\frac{B}{2}) = c,

cot (\frac{B}{2}) cot (\frac{C}{2}) = a,

cot (\frac{C}{2}) cot (\frac{A}{2}) = b

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

A B C

cot \frac{A}{2} cot \frac{B}{2} = c, cot \frac{B}{2} cot \frac{C}{2} = a and cot \frac{C}{2} cot \frac{A}{2} = b

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

a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} \geq a b c (a + b + c)

∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣

= (a + b + c) (x + a + b + c)^{2}, x \neq 0

a + b + c \neq 0

x

∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣

(a + b + c) (x + a + b + c)^{2}, x \neq 0

a + b + c \neq 0

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