Question

If the points A, B, C, D with position vectors \(\vec{i} + \vec{j} - \vec{k}, -\vec{i} + 2\vec{k}, \vec{i} - 2\vec{j} + \vec{k}, 2\vec{i} + \vec{j} + \vec{k}\) form a tetrahedron, then angle between faces ABC and ABD is

• A. \(\cos^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)\)
• B. \(\cos^{-1}\left(\dfrac{-4}{5}\right)\)
• C. \(\cos^{-1}\left(\dfrac{3}{5}\right)\)
• D. \(\cos^{-1}\left(\dfrac{\sqrt{29}}{\sqrt{33}}\right)\)

Hint:

Use cross products to get normals to planes and dot product to find angle between them.

Answer 1

The Correct Option is A

Find normals to faces ABC and ABD using cross product: \[ \text{Normal to ABC: } \vec{AB} \times \vec{AC}, \quad \text{Normal to ABD: } \vec{AB} \times \vec{AD} \] Use dot product between normals to find angle: \[ \cos \theta = \dfrac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}||\vec{n_2}|} \] Substitute and simplify to get \(\cos^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)\).