Vectors
Vectors
Allen Star Batch
Grade 12
Question:
Let $ABCD$ be a tetrahedron in which position vectors of $A, B, C$ and $D$ are $\vec{i} + \vec{j} + \vec{k}, 2\vec{i} + 2\vec{j} + 2\vec{k}, 3\vec{i} + 2\vec{j} + \vec{k}$ and $2\vec{i} + 3\vec{j} + 2\vec{k}$. If $ABC$ be the base of tetrahedron then height of tetrahedron is:
$\frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{\sqrt{5}}$
$\frac{1}{\sqrt{3}}\sqrt{2}$
None of these
Step-by-Step Solution
Key Concept: Tetrahedron height uses the magnitude of scalar triple product divided by base area.
The cross product $\vec{AB} \times \vec{AC} = (\vec{i} + \vec{k}) \times (2\vec{i} + \vec{j}) = \vec{k} + 2\vec{j} - \vec{i}$. The height of the tetrahedron is $h = \frac{|\vec{AD} \cdot (\vec{AB} \times \vec{AC})|}{6|\vec{AB} \times \vec{AC}|} = \frac{|(\vec{i} + 2\vec{j} + \vec{k}) \cdot (-\vec{i} + 2\vec{j} + \vec{k})|}{6\sqrt{6}} = \frac{4}{6\sqrt{6}} = \frac{\sqrt{6}}{9}$.
Correct Answer: 3