Question

If the volume of a tetrahedron having \( \vec{i}+2\vec{j}-3\vec{k} \), \( 2\vec{i}+\vec{j}-3\vec{k} \) and \( 3\vec{i}-\vec{j}+p\vec{k} \) as its coterminous edges is 2, then the values of p are the roots of the equation

• A. \( x^2 + 4x - 12 = 0 \)
• B. \( x^2 + 8x + 12 = 0 \)
• C. \( x^2 - 4x - 12 = 0 \)
• D. \( x^2 - 8x + 12 = 0 \)

Hint:

The volume of a parallelepiped with edges \( \vec{a}, \vec{b}, \vec{c} \) is \( |[\vec{a} \vec{b} \vec{c}]| \). The volume of the tetrahedron formed by the same edges is \( \frac{1}{6} \) of this value. Remember the absolute value, as volume cannot be negative.

Answer 1

The Correct Option is A

The volume \(V\) of a tetrahedron with coterminous edges given by vectors \( \vec{a}, \vec{b}, \vec{c} \) is \( V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]| \), where \( [\vec{a} \vec{b} \vec{c}] \) is the scalar triple product.
Given \( V=2 \), we have \( \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]| = 2 \), which implies \( |[\vec{a} \vec{b} \vec{c}]| = 12 \).
The scalar triple product can be calculated as the determinant of the matrix formed by the vectors.
\( [\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 1 & -3 \\ 3 & -1 & p \end{vmatrix} \).
Expanding the determinant along the first row:
\( = 1(1 \cdot p - (-3)(-1)) - 2(2 \cdot p - (-3)(3)) - 3(2(-1) - 1 \cdot 3) \).
\( = 1(p - 3) - 2(2p + 9) - 3(-2 - 3) \).
\( = p - 3 - 4p - 18 + 15 \).
\( = -3p - 6 \).
Now we use the condition \( |[\vec{a} \vec{b} \vec{c}]| = 12 \), so \( |-3p - 6| = 12 \).
This gives two possibilities:
Case 1: \( -3p - 6 = 12 \implies -3p = 18 \implies p = -6 \).
Case 2: \( -3p - 6 = -12 \implies -3p = -6 \implies p = 2 \).
The values of p, which are the roots of the required quadratic equation, are 2 and -6.
To find the equation, we use the fact that an equation with roots \(r_1, r_2\) is \( x^2 - (r_1+r_2)x + r_1r_2 = 0 \).
Sum of roots = \( 2 + (-6) = -4 \).
Product of roots = \( 2 \times (-6) = -12 \).
The equation is \( x^2 - (-4)x + (-12) = 0 \), which simplifies to \( x^2 + 4x - 12 = 0 \).