Vector Algebra
Scalar Triple Product
Grade 12
Question:
<p>The vectors \(u = (al + a_1l_1)\mathbf{i} + (am + a_1m_1)\mathbf{j} + (an + a_1n_1)\mathbf{k}\), \(v = (bl + b_1l_1)\mathbf{i} + (bm + b_1m_1)\mathbf{j} + (bn + b_1n_1)\mathbf{k}\) and \(w = (cl + c_1l_1)\mathbf{i} + (cm + c_1m_1)\mathbf{j} + (cn + c_1n_1)\mathbf{k}\)</p>
<p>(a) form an equilateral triangle</p>
<p>(b) are coplanar</p>
<p>(c) are collinear</p>
<p>(d) are mutually perpendicular</p>
Step-by-Step Solution
Key Concept: The scalar triple product of three vectors equals zero if and only if they are coplanar. The determinant structure allows factorization showing both determinants have a zero column.
Step 1: We compute the scalar triple product: \[[u v w] = \begin{vmatrix} al + a_1l_1 & am + a_1m_1 & an + a_1n_1 \\ bl + b_1l_1 & bm + b_1m_1 & bn + b_1n_1 \\ cl + c_1l_1 & cm + c_1m_1 & cn + c_1n_1 \end{vmatrix}\] Step 2: This can be factored as: \[[u v w] = \begin{vmatrix} a & a_1 & 0 \\ b & b_1 & 0 \\ c & c_1 & 0 \end{vmatrix} \begin{vmatrix} l & l_1 & 0 \\ m & m_1 & 0 \\ n & n_1 & 0 \end{vmatrix} = 0\] Step 3: Since the scalar triple product is zero, the vectors are coplanar.
Correct Answer: B