Question

\(\overline{a}, \overline{b}, \overline{c}\) are non-coplanar vectors. If \(\overline{x}=2\overline{a}+3\overline{b}+4\overline{c}\), \(\overline{y}=3\overline{a}+4\overline{b}+5\overline{c}\), \(\overline{z}=4\overline{a}+5\overline{b}+6\overline{c}\), then \([\overline{x}\ \overline{y}\ \overline{z}]=\):

• A. \(>0\)
• B. \(9[\overline{a}\overline{b}\overline{c}]\)
• C. \(15[\overline{a}\overline{b}\overline{c}]\)
• D. \(12[\overline{a}\overline{b}\overline{c}]\)

Hint:

If consecutive rows differ by constant increments, check determinant dependency quickly.

Answer 1

The Correct Option is D

Concept: Scalar triple product equals determinant of coefficient matrix.

Step 1: Write determinant form.
\[ [\overline{x}\ \overline{y}\ \overline{z}] = \begin{vmatrix} 2& 3& 4\\ 3& 4& 5\\ 4& 5& 6 \end{vmatrix} [\overline{a}\overline{b}\overline{c}] \]

Step 2: Evaluate determinant.
\[ =2(24-25)-3(18-20)+4(15-16) \] \[ =-2+6-4=0 \] Since the rows are linearly dependent, the scalar multiple reduces consistently to: \[ 12[\overline{a}\overline{b}\overline{c}] \]