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

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Geometric mean of $a$ and $b$

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(A) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{u}, \vec{v}, \vec{w}] = 0$.Given vectors are $\vec{u} = a\hat{i} + a\hat{j} + c\hat{k}$,$\vec{v} = \hat{i} + 0\hat{j} + \hat{k}$,and $\vec{w} = c\hat{i} + c\hat{j} + b\hat{k}$.The scalar triple product is given by the determinant:$\begin{vmatrix} a & a & c \ 1 & 0 & 1 \ c & c & b \end{vmatrix} = 0$Expanding along the second row:$-1(ab - c^2) + 0 - 1(ac - ac) = 0$$-(ab - c^2) = 0$$c^2 - ab = 0$$c^2 = ab$$c = \sqrt{ab}$Thus,$c$ is the geometric mean of $a$ and $b$.

Let $a, b$ and $c$ be distinct non-negative 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 $c = \dots$

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