Vector Algebra
Coplanar Vectors
Grade 12
Question:
<p>Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be distinct non-negative numbers. If the vectors \(a\vec{i} + a\vec{j} + c\vec{k}\), \(\vec{i} + \vec{k}\) and \(c\vec{i} + c\vec{j} + b\vec{k}\) lie in a plane, then \(c\) is</p>
<p>(a) HM of a and b</p>
<p>(b) 0</p>
<p>(c) AM of a and b</p>
<p>(d) GM of a and b</p>
Step-by-Step Solution
Key Concept: Coplanarity of three vectors requires their scalar triple product to be zero, which leads to a relationship between the coefficients.
Step 1: Three vectors are coplanar if their scalar triple product equals zero. Step 2: Set up the determinant: \[\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0\] Step 3: Apply \(C_1 \to C_1 - C_2\): \[\begin{vmatrix} 0 & a & c \\ 1 & 0 & 1 \\ 0 & c & b \end{vmatrix} = 0\] Step 4: Expand along first column: \[-1(ab - c^2) = 0\] Step 5: Therefore: \[c^2 = ab\] \[c = \sqrt{ab}\] ∴ \(c\) is the Geometric Mean of \(a\) and \(b\). Answer is (d).
Correct Answer: D