bar{a} = hat{i} + hat{j} + hat{k},bar{b} = ha | vedclass

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(B) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are coplanar,their scalar triple product must be zero: $\bar{a} \cdot (\bar{b} 	imes \bar{c})

Vedclass · Accepted Answer

$\frac{-3}{2}$

Vedclass · Answer

(B) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are coplanar,their scalar triple product must be zero: $\bar{a} \cdot (\bar{b} 	imes \bar{c}) = 0$. This is equivalent to the determinant of the components being zero: $\begin{vmatrix} 1 & 1 & 1 \ 1 & -1 & 2 \ x & x-1 & -1 \end{vmatrix} = 0$. Expanding the determinant along the first row: $1((-1)(-1) - (2)(x-1)) - 1((1)(-1) - (2)(x)) + 1((1)(x-1) - (-1)(x)) = 0$. $1(1 - 2x + 2) - 1(-1 - 2x) + 1(x - 1 + x) = 0$. $(3 - 2x) + (1 + 2x) + (2x - 1) = 0$. $3 - 2x + 1 + 2x + 2x - 1 = 0$. $2x + 3 = 0$. $x = \frac{-3}{2}$.

$\bar{a} = \hat{i} + \hat{j} + \hat{k}$,$\bar{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\bar{c} = x\hat{i} + (x-1)\hat{j} - \hat{k}$. If the vector $\bar{c}$ lies in the plane of $\bar{a}$ and $\bar{b}$,then $x=$

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