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(A) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{u} \vec{v} \vec{w}] = 0$.The given vecto

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$\sqrt{ab}$

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(A) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{u} \vec{v} \vec{w}] = 0$.The given vectors are $\vec{u} = a \hat{i} + a \hat{j} + c \hat{k}$,$\vec{v} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k}$,and $\vec{w} = c \hat{i} + c \hat{j} + b \hat{k}$.The condition for coplanarity is given by the determinant:$\left|\begin{array}{lll}a & a & c \ 1 & 0 & 1 \ c & c & b\end{array}ight| = 0$Expanding the determinant along the second row:$-1 \cdot \left|\begin{array}{ll}a & c \ c & b\end{array}ight| + 0 \cdot \left|\begin{array}{ll}a & c \ c & b\end{array}ight| - 1 \cdot \left|\begin{array}{ll}a & a \ c & c\end{array}ight| = 0$$-1(ab - c^2) - 1(ac - ac) = 0$$-(ab - c^2) - 0 = 0$$c^2 - ab = 0$$c^2 = ab$Since $a, b, c$ are positive numbers,$c = \sqrt{ab}$.

Let $a, b$ and $c$ be distinct positive numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}$,$\hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ are coplanar,then $c$ is equal to:

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