Vectors
Vectors
Allen Star Batch
Grade 12
Question:
If three coterminous edges of a tetrahedron are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 4$, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$, $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{4}$ and $\vec{c}$ and $\vec{a}$ is $\frac{\pi}{6}$. The area of the base is $2$ sq. units, then the height of the tetrahedron is:
$3\sqrt{\sqrt{3} - 2}$
$3\sqrt{\sqrt{6} - 2}$
$\frac{3\sqrt{\sqrt{6} - 2}}{2}$
None of these
Step-by-Step Solution
Key Concept: Scalar triple product determinant directly gives tetrahedron volume up to factor of 1/6.
Volume of tetrahedron is $V = \frac{1}{6}|[\vec{abc}]|$ where the scalar triple product is computed via determinant. The calculation shows $|[\vec{abc}]|^2 = 144(\sqrt{6}-2)$, giving $V = \frac{1}{6} \times 12(\sqrt{6}-2) = 2\sqrt{\sqrt{6}-2}$. Height is determined from $V = \frac{1}{3} \times \text{base area} \times \text{height}$.
Correct Answer: 2