If vec{a}=2 hat{i}+3 hat{j},vec{b}=3 hat{j}+4 | vedclass

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(D) vector perpendicular to both $\vec{a}$ and $(\vec{b} 	imes \vec{c})$ is given by the cross product $\vec{a} 	imes (\vec{b} 	imes \vec{c})$.Usin

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$-45 \hat{i}+30 \hat{j}+4 \hat{k}$

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(D) vector perpendicular to both $\vec{a}$ and $(\vec{b} 	imes \vec{c})$ is given by the cross product $\vec{a} 	imes (\vec{b} 	imes \vec{c})$.Using the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.First,calculate the dot products:$\vec{a} \cdot \vec{c} = (2 \hat{i} + 3 \hat{j}) \cdot (5 \hat{i} + 4 \hat{k}) = (2)(5) + (3)(0) + (0)(4) = 10$.$\vec{a} \cdot \vec{b} = (2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{j} + 4 \hat{k}) = (2)(0) + (3)(3) + (0)(4) = 9$.Now,substitute these into the formula:$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = 10(3 \hat{j} + 4 \hat{k}) - 9(5 \hat{i} + 4 \hat{k})$$= 30 \hat{j} + 40 \hat{k} - 45 \hat{i} - 36 \hat{k}$$= -45 \hat{i} + 30 \hat{j} + 4 \hat{k}$.

If $\vec{a}=2 \hat{i}+3 \hat{j}$,$\vec{b}=3 \hat{j}+4 \hat{k}$,and $\vec{c}=5 \hat{i}+4 \hat{k}$ are three vectors,then a vector which is perpendicular to $\vec{a}$ and $\vec{b} \times \vec{c}$ is

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