Let a-2-3k- b-2-2k- and c-2-k- and a-b-b-c-c-a-0- then the value of 6- is

Given : {(a×b)×(b×c)}×(c×a)=0 ⇒{[a b c]b [a b b]c}×(c×a)=0 ⇒ [a b c]((a·b)c (b·c)a)=0 ∵ nbsp;[a b b]=0 ⇒ nbsp; [a b c]=0 OR a·b=b·c=0 ⇒α amp;2 amp; 3 1 ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

Let a=αî+2ĵ-3...

Question

Let $\to a = α^i + 2^j - 3^k, \to b =^i + 2 α^j - 2^k$ and $\to c = 2^i - α^j +^k$ and ${(\to a \times \to b) \times (\to b \times \to c)} \times (\to c \times \to a) = \to 0,$ then the value of $6 α$ is

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Solution

Given :
${(\to a \times \to b) \times (\to b \times \to c)} \times (\to c \times \to a) = \to 0 \Rightarrow {[\to a \to b \to c] \to b - [\to a \to b \to b] \to c} \times (\to c \times \to a) = \to 0 \Rightarrow [\to a \to b \to c] ((\to a \cdot \to b) \to c - (\to b \cdot \to c) \to a) = \to 0 ∵ [\to a \to b \to b] = 0 \Rightarrow [\to a \to b \to c] = 0$ OR $\to a \cdot \to b = \to b \cdot \to c = 0 \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} α & 2 & - 3 1 & 2 α & - 2 2 & - α & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
OR $α + 4 α + 6 = 2 - 2 α^{2} - 2 = 0 \Rightarrow 10 - 15 α = 0$
OR $α = 0 = - \frac{6}{5} (contradictory,not possible)$
$\Rightarrow α = \frac{10}{15} = \frac{2}{3}$

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Work Done as Dot Product

Standard XII Physics

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Let →a=α^i+2^j−3^k,→b=^i+2α^j−2^k and →c=2^i−α^j+^k and {(→a×→b)×(→b×→c)}×(→c×→a)=→0, then the value of 6α is

Let $\to a = α^i + 2^j - 3^k, \to b =^i + 2 α^j - 2^k$ and $\to c = 2^i - α^j +^k$ and ${(\to a \times \to b) \times (\to b \times \to c)} \times (\to c \times \to a) = \to 0,$ then the value of $6 α$ is