Question

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

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

Hint:

Three vectors are coplanar if the determinant formed by their components equals zero. \[ \vec{A}\cdot(\vec{B}\times\vec{C})=0 \] This is a standard shortcut used in vector algebra problems.

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 must be zero. \[ \begin{vmatrix} a_1 & a_2 & a_3 b_1 & b_2 & b_3 c_1 & c_2 & c_3 \end{vmatrix}=0 \]

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

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

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