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If the vectors $$\vec{a}=i-j+2k; \vec{b}=2i+4j+k$$ and $$\vec{c}=2013\lambda i+j-\mu k$$ are mutually orthogonal then $$(\lambda, \mu)$$=...


A
(32013;2)
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B
(32013;2)
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C
(32013;2)
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D
(32013;2)
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Solution

The correct option is A $$(\frac{-3}{2013};-2)$$

$$\vec{a}=i-\hat{j}+2k$$
$$\vec{b}=2i-4j+k $$
$$\vec{c}=2013 \lambda+j-\mu k $$
$$\therefore\vec{a}$$ & $$ \vec{c} $$ orthogonal 
$$\Rightarrow \vec{a} .\vec{c}=0$$
$$\Rightarrow 2013\lambda -1-2\mu =0………1$$
Also , $$\vec{b}$$ & $$\vec{c}$$ are orthogonal
$$\Rightarrow \vec{b} , \vec{c} =0$$
$$\Rightarrow 4026\lambda +4-\mu =0……2$$
Now solving these equation, we get
equation 1 $$\times 2$$ equation 2
$$4026\lambda -2-4\mu =0\\ 4026\lambda +4-\mu =0\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad -6-3\mu =0\\ \quad \quad \quad \quad \quad \quad 3u=-6 \Rightarrow\boxed { \mu =-2 } $$
Also ,$$2013\lambda -1+4=0$$
$$\lambda=\dfrac{-3}{2013}= \dfrac{-1}{671} \quad \therefore \left(\dfrac{-3}{2013},-2\right)$$
$$\Rightarrow$$ option $$\left(A\right)$$ is correct


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