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(A) Given vectors are $\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\vec{b}=\mu \hat{i}+\hat{j}-\hat{k}$.Since the vectors are orthogonal,their dot

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$\left(\frac{1}{4}, \frac{7}{4}\right)$

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(A) Given vectors are $\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\vec{b}=\mu \hat{i}+\hat{j}-\hat{k}$.Since the vectors are orthogonal,their dot product is zero:$\vec{a} \cdot \vec{b} = (1)(\mu) + (\lambda)(1) + (2)(-1) = 0$$\mu + \lambda - 2 = 0 \Rightarrow \lambda + \mu = 2 \Rightarrow \mu = 2 - \lambda$ (Eq. $1$)Given that $|\vec{a}| = |\vec{b}|$,we square both sides:$|\vec{a}|^2 = |\vec{b}|^2$$1^2 + \lambda^2 + 2^2 = \mu^2 + 1^2 + (-1)^2$$1 + \lambda^2 + 4 = \mu^2 + 1 + 1$$\lambda^2 + 3 = \mu^2$ (Eq. $2$)Substitute $\mu = 2 - \lambda$ into Eq. $2$:$\lambda^2 + 3 = (2 - \lambda)^2$$\lambda^2 + 3 = 4 - 4\lambda + \lambda^2$$3 = 4 - 4\lambda$$4\lambda = 1 \Rightarrow \lambda = \frac{1}{4}$Now,find $\mu$ using $\mu = 2 - \lambda$:$\mu = 2 - \frac{1}{4} = \frac{7}{4}$Thus,$(\lambda, \mu) = \left(\frac{1}{4}, \frac{7}{4}\right)$.

If $\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\vec{b}=\mu \hat{i}+\hat{j}-\hat{k}$ are orthogonal and $|\vec{a}|=|\vec{b}|$,then $(\lambda, \mu) = $

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