Question

If the vectors \(2i - j + k\), \(i + 2j - 3k\) and \(3i + aj + 5k\) are coplanar, find the value of \(a\).

• A. \(-2\)
• B. \(-4\)
• C. \(2\)
• D. \(4\)

Hint:

Three vectors are coplanar if the determinant of their components is zero. This represents the scalar triple product condition.

Answer 1

The Correct Option is B

Concept: Three vectors are coplanar if their scalar triple product is zero. \[ \vec{A}\cdot(\vec{B}\times\vec{C}) = 0 \] Equivalently, the determinant formed by their components is zero.

Step 1: Write vectors in component form. \[ \vec{A} = (2,-1,1) \] \[ \vec{B} = (1,2,-3) \] \[ \vec{C} = (3,a,5) \]

Step 2: Form the determinant condition. \[ \begin{vmatrix} 2 & -1 & 1 1 & 2 & -3 3 & a & 5 \end{vmatrix} = 0 \]

Step 3: Expand the determinant. \[ = 2 \begin{vmatrix} 2 & -3 a & 5 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -3 3 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 3 & a \end{vmatrix} \] \[ = 2(10 + 3a) + (5 + 9) + (a - 6) \] \[ = 20 + 6a + 14 + a - 6 \] \[ = 28 + 7a \] Since vectors are coplanar: \[ 28 + 7a = 0 \] \[ a = -4 \] \[ \boxed{a = -4} \]