If the volume of the tetrahedron, whose vertices are \( A(1, 2, 3), B(-3, -1, 1), C(2, 1, 3) \) and \( D(-1, 2, x) \), is \( \frac{11}{6} \) cubic units, then the value of \( x \) is:

Question

Let \( \vec{a}, \vec{b}, \vec{c} \) be vectors of equal magnitude such that the angle between \( \vec{a} \) and \( \vec{b} \) is \( \alpha \), the angle between \( \vec{b} \) and \( \vec{c} \) is \( \beta \), and the angle between \( \vec{c} \) and \( \vec{a} \) is \( \gamma \). Then the minimum value of \( \cos \alpha + \cos \beta + \cos \gamma \) is:

Answer 1

The volume \( V \) of a tetrahedron with vertices \( A(x_1, y_1, z_1), B(x_2, y_2, z_2), C(x_3, y_3, z_3), \) and \( D(x_4, y_4, z_4) \) is calculated using the formula: \[ V = \frac{1}{6} \left| \text{det} \begin{bmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{bmatrix} \right| \] Given the vertices \( A(1, 2, 3), B(-3, -1, 1), C(2, 1, 3), \) and \( D(-1, 2, x) \), the volume formula becomes: \[ V = \frac{1}{6} \left| \text{det} \begin{bmatrix} 1 & 2 & 3 & 1 \\ -3 & -1 & 1 & 1 \\ 2 & 1 & 3 & 1 \\ -1 & 2 & x & 1 \end{bmatrix} \right| = \frac{11}{6} \] By calculating the determinant of this matrix and solving for \( x \), we find \( x = 3 \).

If the volume of the tetrahedron, whose vertices are \( A(1, 2, 3), B(-3, -1, 1), C(2, 1, 3) \) and \( D(-1, 2, x) \), is \( \frac{11}{6} \) cubic units, then the value of \( x \) is:

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The Correct Option is A

Solution and Explanation

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