Let vec{a} = hat{i} - hat{k},vec{b} = xhat{i} | vedclass

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

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neither $x$ nor $y$

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(D) The scalar triple product 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 	imes \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 a constant $1$,it does not depend on $x$ or $y$.

Let $\vec{a} = \hat{i} - \hat{k}$,$\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$,and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$. Then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ depends on:

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