Question

The four points whose position vectors are given by \( 2\bar{a}+3\bar{b}-\bar{c} \), \( \bar{a}-2\bar{b}+3\bar{c} \), \( 3\bar{a}+4\bar{b}-2\bar{c} \) and \( \bar{a}-6\bar{b}+6\bar{c} \) are

• A. Collinear
• B. Coplanar
• C. Vertices of a square
• D. Vertices of a rectangle

Hint:

Zero scalar triple product of three vectors formed from four points implies the points lie on the same plane.

Answer 1

The Correct Option is B

Step 1: Check Coplanarity:

Let the points be P, Q, R, S with coordinates from coefficients: \( P(2, 3, -1) \), \( Q(1, -2, 3) \), \( R(3, 4, -2) \), \( S(1, -6, 6) \). Vectors: \( \vec{PQ} = (-1, -5, 4) \)
\( \vec{PR} = (1, 1, -1) \)
\( \vec{PS} = (-1, -9, 7) \)
Step 2: Scalar Triple Product:

\[ \Delta = \begin{vmatrix} -1 & -5 & 4 \\1 & 1 & -1 \\ -1 & -9 & 7 \end{vmatrix} \] \[ \Delta = -1(7 - 9) - (-5)(7 - 1) + 4(-9 - (-1)) \] \[ \Delta = -1(-2) + 5(6) + 4(-8) = 2 + 30 - 32 = 0 \] Since \( \Delta = 0 \), vectors are coplanar.
Step 3: Final Answer:

The points are Coplanar.