Let a=2 hat{i}-3 hat{j}+4 hat{k}, b=7 hat{i}+ | vedclass

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(B) Given vectors are $a=2 \hat{i}-3 \hat{j}+4 \hat{k}$, $b=7 \hat{i}+2 \hat{j}-3 \hat{k}$, and $c=\hat{i}+\hat{j}+\hat{k}$.Since $x$ is perpendicular

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$\hat{i}+34 \hat{j}+25 \hat{k}$

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(B) Given vectors are $a=2 \hat{i}-3 \hat{j}+4 \hat{k}$, $b=7 \hat{i}+2 \hat{j}-3 \hat{k}$, and $c=\hat{i}+\hat{j}+\hat{k}$.Since $x$ is perpendicular to both $a$ and $b$, $x$ must be parallel to the cross product $a 	imes b$.Thus, $x = \lambda(a 	imes b)$.Calculating the cross product $a 	imes b$:$a 	imes b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 4 \ 7 & 2 & -3 \end{vmatrix} = \hat{i}(9-8) - \hat{j}(-6-28) + \hat{k}(4+21) = \hat{i} + 34 \hat{j} + 25 \hat{k}$.So, $x = \lambda(\hat{i} + 34 \hat{j} + 25 \hat{k})$.Given $x \cdot c = 60$, we substitute $x$ and $c$:$(\lambda \hat{i} + 34 \lambda \hat{j} + 25 \lambda \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 60$.$\lambda + 34 \lambda + 25 \lambda = 60$.$60 \lambda = 60$, which gives $\lambda = 1$.Therefore, $x = \hat{i} + 34 \hat{j} + 25 \hat{k}$.

Let $a=2 \hat{i}-3 \hat{j}+4 \hat{k}$, $b=7 \hat{i}+2 \hat{j}-3 \hat{k}$, and $c=\hat{i}+\hat{j}+\hat{k}$. The vector $x$ such that $x \cdot c=60$ and $x$ is perpendicular to both $a$ and $b$ is:

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