If x, y and z are non-zero real numbers and v | vedclass

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(C) Given vectors are $\vec{a}=x \hat{i}+2 \hat{j}$,$\vec{b}=y \hat{j}+3 \hat{k}$,and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.First,calculate the cros

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$9$

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(C) Given vectors are $\vec{a}=x \hat{i}+2 \hat{j}$,$\vec{b}=y \hat{j}+3 \hat{k}$,and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ x & 2 & 0 \ 0 & y & 3 \end{vmatrix} = \hat{i}(6-0) - \hat{j}(3x-0) + \hat{k}(xy-0) = 6 \hat{i} - 3x \hat{j} + xy \hat{k}$.Given $\vec{a} 	imes \vec{b} = z \hat{i} - 3 \hat{j} + \hat{k}$.Comparing coefficients,we get $z=6$,$3x=3 \Rightarrow x=1$,and $xy=1 \Rightarrow y=1$.Now,the scalar triple product is $[\vec{a} \vec{b} \vec{c}] = (\vec{a} 	imes \vec{b}) \cdot \vec{c}$.$[\vec{a} \vec{b} \vec{c}] = (6 \hat{i} - 3 \hat{j} + \hat{k}) \cdot (x \hat{i} + y \hat{j} + z \hat{k})$.Substituting $x=1, y=1, z=6$:$[\vec{a} \vec{b} \vec{c}] = (6 \hat{i} - 3 \hat{j} + \hat{k}) \cdot (1 \hat{i} + 1 \hat{j} + 6 \hat{k}) = (6)(1) + (-3)(1) + (1)(6) = 6 - 3 + 6 = 9$.

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