If overline{a}=hat{i}-hat{k},overline{b}=x ha | vedclass

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(C) The scalar triple product $\overline{a} \cdot(\overline{b} 	imes \overline{c})$ is given by the determinant of the components of the vectors $\ov

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neither $x$ nor $y$

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

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

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