Let vec{a}=3 hat{i}+hat{j} and vec{b}=hat{i}+ | vedclass

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(A) Given $\vec{a} = 3 \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 2 \hat{j} + \hat{k}$.Using the vector triple product formula $\vec{a} 	imes (\vec{

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$-5$

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(A) Given $\vec{a} = 3 \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 2 \hat{j} + \hat{k}$.Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.Comparing this with the given equation $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \vec{b} + \lambda \vec{c}$,we get:$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{b} + \lambda \vec{c}$.Since $\vec{b}$ and $\vec{c}$ are non-parallel,we can equate the coefficients of $\vec{b}$ and $\vec{c}$:$\vec{a} \cdot \vec{c} = 1$ and $-(\vec{a} \cdot \vec{b}) = \lambda$.Now,calculate $\vec{a} \cdot \vec{b}$:$\vec{a} \cdot \vec{b} = (3 \hat{i} + \hat{j}) \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = (3)(1) + (1)(2) + (0)(1) = 3 + 2 = 5$.Therefore,$\lambda = -(\vec{a} \cdot \vec{b}) = -5$.

Let $\vec{a}=3 \hat{i}+\hat{j}$ and $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$. Let $\vec{c}$ be a vector satisfying $\vec{a} \times(\vec{b} \times \vec{c})=\vec{b}+\lambda \vec{c}$. If $\vec{b}$ and $\vec{c}$ are non-parallel,then the value of $\lambda$ is.

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