If vec{a}=hat{i}-hat{k}, vec{b}=x hat{i}+hat{ | vedclass

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(A) The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is given by the determinant of the components of the vectors $\vec{a}, \vec{b},$ and $\vec{c

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neither $x$ nor $y$

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(A) The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is given by the determinant of the components of the vectors $\vec{a}, \vec{b},$ and $\vec{c}$.$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 0 & -1 \ x & 1 & 1-x \ y & x & 1+x-y \end{vmatrix}$Applying the column operation $C_3 	o C_3 + C_1$:$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 0 & 0 \ x & 1 & 1 \ y & x & 1+x \end{vmatrix}$Expanding along the first row:$[\vec{a} \vec{b} \vec{c}] = 1 \cdot \begin{vmatrix} 1 & 1 \ x & 1+x \end{vmatrix} - 0 + 0$$[\vec{a} \vec{b} \vec{c}] = (1+x) - x = 1$Since the result is $1$,which is a constant,the value of $[\vec{a} \vec{b} \vec{c}]$ depends on neither $x$ nor $y$.

If $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$,then $[\vec{a} \vec{b} \vec{c}]$ depends on

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