Question

If \(O(0,0,0), A(1,2,1), B(2,1,3)\), and \(C(-1,1,2)\) are the vertices of a tetrahedron, then the acute angle between its face \(OAB\) and edge \(BC\) is

• A. \(\cos^{-1} \left( \dfrac{6\sqrt{2}}{5\sqrt{7}} \right)\)
• B. \(\sin^{-1} \left( \dfrac{6\sqrt{2}}{5\sqrt{7}} \right)\)
• C. \(\tan^{-1} \left( \dfrac{6\sqrt{2}}{5\sqrt{7}} \right)\)
• D. \(\dfrac{\pi}{2}\)

Hint:

When the angle is between a vector and a face, find the normal to the face and use the sine of the angle formula: \(\sin \theta = \dfrac{|\vec{n} \cdot \vec{v}|}{\|\vec{n}\| \cdot \|\vec{v}\|}\).

Answer 1

The Correct Option is B

To find the angle between the face \(OAB\) and the edge \(BC\), compute the angle between the vector \(\vec{BC}\) and the normal vector to the plane formed by points \(O, A, B\). Step 1: Construct vectors: \[ \vec{OA} = A - O = (1,2,1), \quad \vec{OB} = B - O = (2,1,3) \] Step 2: Compute the normal vector to triangle OAB: \[ \vec{n} = \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & 1
2 & 1 & 3 \end{vmatrix} = (6 - 1)\hat{i} - (3 - 2)\hat{j} + (1 - 4)\hat{k} = 5\hat{i} - \hat{j} - 3\hat{k} \] Step 3: Vector \(BC\): \[ \vec{BC} = C - B = (-1,1,2) - (2,1,3) = (-3, 0, -1) \] Step 4: Use the angle formula between vector and a plane: \[ \sin \theta = \dfrac{|\vec{n} \cdot \vec{BC}|}{\|\vec{n}\| \cdot \|\vec{BC}\|} \] \[ \vec{n} \cdot \vec{BC} = 5(-3) + (-1)(0) + (-3)(-1) = -15 + 0 + 3 = -12 \Rightarrow |\vec{n} \cdot \vec{BC}| = 12 \] \[ \|\vec{n}\| = \sqrt{25 + 1 + 9} = \sqrt{35}, \quad \|\vec{BC}\| = \sqrt{9 + 0 + 1} = \sqrt{10} \] \[ \Rightarrow \sin \theta = \dfrac{12}{\sqrt{350}} = \dfrac{6\sqrt{2}}{5\sqrt{7}} \Rightarrow \theta = \sin^{-1} \left( \dfrac{6\sqrt{2}}{5\sqrt{7}} \right) \]