Question

If \(\alpha,\beta,\gamma\;(\alpha>\beta>\gamma)\) are roots of the equation \[ x^3-x^2-4x+4=0, \] the volume of the parallelepiped whose coterminous edges are \[ \alpha\hat i+\beta\hat j+\gamma\hat k,\; \beta\hat i+\gamma\hat j+\alpha\hat k,\; \gamma\hat i+\alpha\hat j+\beta\hat k \] is

• A. \(\sqrt{13}\)
• B. \(3\)
• C. \(\dfrac{15}{6}\)
• D. \(\dfrac{13}{6}\)

Hint:

The volume of a parallelepiped formed by vectors \(\vec a,\vec b,\vec c\) is \[ \boxed{\left|\vec a\cdot(\vec b\times\vec c)\right|} = \boxed{\left|\det(\vec a,\vec b,\vec c)\right|}. \]

Answer 1

The Correct Option is A

Step 1: Find the roots of the given equation.& nbsp;

Factorizing,

\[ x^3-x^2-4x+4 = (x-2)(x-1)(x+2). \]

Hence,

\[ \alpha=2,\qquad \beta=1,\qquad \gamma=-2. \]

Step 2: Form the determinant.

The three vectors are

\[ (2,1,-2),\qquad (1,-2,2),\qquad (-2,2,1). \]

The required volume is

\[ \left| \begin{vmatrix} 2 & amp; 1 & amp; -2\\ 1 & amp; -2 & amp; 2\\ -2 & amp; 2 & amp; 1 \end{vmatrix} \right|. \]

Step 3: Evaluate the determinant.

Expanding,

\[ \begin{aligned} \Delta & amp;= 2(-2-4)-1(1+4)-2(2-4)\\ & amp;= -12-5+4\\ & amp;= -13. \end{aligned} \]

Hence,

\[ \text{Volume} = \sqrt{|-13|} = \sqrt{13}. \]

Therefore,

\[ \boxed{\sqrt{13}}. \]

Thus, the correct option is

\[ \boxed{(A)}. \]