Vector Algebra
Coplanarity of Vectors
Grade None
Question:
<p>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 \(c\hat{i} + c\hat{j} + b\hat{k}\) lie in a plane, then \(c\) is</p>
<p>the Geometric Mean of \(a\) and \(b\).</p>
<p>the Arithmetic Mean of \(a\) and \(b\).</p>
<p>equal to zero.</p>
<p>the Harmonic Mean of \(a\) and \(b\).</p>
Step-by-Step Solution
Key Concept: Three vectors are coplanar if and only if their scalar triple product equals zero. Set up the determinant of the three vectors and solve for c using the condition that the determinant = 0.
Step 1: Three vectors are coplanar when their scalar triple product = 0. Form the determinant: $\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$ Step 2: Expand along row 2 (contains zeros): $-1\begin{vmatrix} a & c \\ c & b \end{vmatrix} + 1\begin{vmatrix} a & a \\ c & c \end{vmatrix} = 0$ Step 3: Evaluate the 2×2 determinants: $-(ab - c^2) + (ac - ac) = 0$ $-ab + c^2 = 0$ $c^2 = ab$ Step 4: Since a, b, c are distinct non-negative numbers and $c^2 = ab$, we have $c = \sqrt{ab}$ (the geometric mean of a and b). ∴ Answer: A
Correct Answer: A