Let \(\vec{a}, \vec{b}, \vec{c} \) be distinct non-negative numbers. If the vectors \(a\hat{i}+a\hat{j}+c\hat{k}\) , \(\hat{i}+\hat{k}\) and \(c\hat{i}+c\hat{j}+b\hat{k}\) lie in a plane, then c is
Given that the vectors are coplanar, the determinant of the matrix formed by the vectors must be zero: \[ \left|\begin{array}{lll} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{array}\right| = 0 \] Using the transformation \( C_1 = C_1 - C_2 \), the determinant becomes: \[ \left|\begin{array}{lll} 0 & a & c \\ 1 & 0 & 1 \\ 0 & c & b \end{array}\right| = 0 \] Expanding this determinant: \[ -1 \cdot (ba - c^2) = 0 \] This simplifies to: \[ c^2 = ba \] Hence, \( b \) is the geometric mean of \( a \) and \( c \).
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