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

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(B) Given vectors are $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$,$\vec{b}=\hat{i}+3 \hat{j}-\hat{k}$,and $\vec{c}=3 \hat{i}-\hat{j}-2 \hat{k}$.First,we cal

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$2025$

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(B) Given vectors are $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$,$\vec{b}=\hat{i}+3 \hat{j}-\hat{k}$,and $\vec{c}=3 \hat{i}-\hat{j}-2 \hat{k}$.First,we calculate the dot products:$\vec{a} \cdot \vec{a} = (2)^2 + (1)^2 + (3)^2 = 4 + 1 + 9 = 14$$\vec{b} \cdot \vec{b} = (1)^2 + (3)^2 + (-1)^2 = 1 + 9 + 1 = 11$$\vec{c} \cdot \vec{c} = (3)^2 + (-1)^2 + (-2)^2 = 9 + 1 + 4 = 14$$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} = (2)(1) + (1)(3) + (3)(-1) = 2 + 3 - 3 = 2$$\vec{a} \cdot \vec{c} = \vec{c} \cdot \vec{a} = (2)(3) + (1)(-1) + (3)(-2) = 6 - 1 - 6 = -1$$\vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{b} = (1)(3) + (3)(-1) + (-1)(-2) = 3 - 3 + 2 = 2$Now,substitute these values into the determinant:$\Delta = \left|\begin{array}{ccc} 14 & 2 & -1 \ 2 & 11 & 2 \ -1 & 2 & 14 \end{array}ight|$Expanding along the first row:$\Delta = 14(11 	imes 14 - 2 	imes 2) - 2(2 	imes 14 - 2 	imes (-1)) - 1(2 	imes 2 - 11 	imes (-1))$$\Delta = 14(154 - 4) - 2(28 + 2) - 1(4 + 11)$$\Delta = 14(150) - 2(30) - 1(15)$$\Delta = 2100 - 60 - 15 = 2025$.

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