Point \(A(-4,2,5)\) and the two vectors, \(\mathbf{R}_{A M}=(20,18-10)\) and \(\mathbf{R}_{A N}=(-10,8,15)\), define a triangle. Find \((a)\) a unit vector perpendicular to the triangle; \((b)\) a unit vector in the plane of the triangle and perpendicular to \(\mathbf{R}_{A N} ;(c)\) a unit vector in the plane of the triangle that bisects the interior angle at \(A\).

We first calculate the cross product of the given vectors: \(\mathbf{N} = \mathbf{R}_{AM} \times \mathbf{R}_{AN}\) \(\mathbf{N} = (20,18,-10) \times (-10,8,15)\) \(\mathbf{N} = \begin{pmatrix} (18)(15) - (-10)(8) \\ (-10)(15) - (20)(15) \\ (20)(8) - (18)(-10) \end{pmatrix}\) \(\mathbf{N} = \begin{pmatrix} 270 + 80 \\ -150 - 300 \\ 160 + 180 \end{pmatrix}\) \(\mathbf{N} = (350, -450, 340)\)

To find the unit vector perpendicular to the triangle, we will normalize the normal vector, \(\mathbf{N}\): \( ||\mathbf{N}|| = \sqrt{(350)^2 + (-450)^2 + (340)^2}\) \( ||\mathbf{N}|| = \sqrt{350000}\) \( \hat{\mathbf{N}} = \frac{\mathbf{N}}{||\mathbf{N}||}\) \( \hat{\mathbf{N}} = \frac{1}{\sqrt{350000}}(350, -450, 340)\)

To find a vector in the plane of the triangle and perpendicular to \(\mathbf{R}_{AN}\), take the cross product of \(\mathbf{R}_{AN}\) and \(\mathbf{N}\): \(\mathbf{P} = \mathbf{R}_{AN} \times \mathbf{N}\) \(\mathbf{P} = (-10,8,15) \times (350,-450,340)\) \(\mathbf{P} = \begin{pmatrix} (8)(340) - (15)(-450) \\ (15)(350) - (-10)(340) \\ (-10)(-450) - (8)(350) \end{pmatrix}\) \(\mathbf{P} = \begin{pmatrix} 2720 + 6750 \\ 5250 + 3400 \\ 4500 - 2800 \end{pmatrix}\) \(\mathbf{P} = (9470, 8650, 1700)\)

To find a unit vector in the triangle plane and perpendicular to \(\mathbf{R}_{AN}\), we will normalize the vector, \(\mathbf{P}\): \( ||\mathbf{P}|| = \sqrt{(9470)^2 + (8650)^2 + (1700)^2}\) \( ||\mathbf{P}|| = \sqrt{194305000}\) \( \hat{\mathbf{P}} = \frac{\mathbf{P}}{||\mathbf{P}||}\) \( \hat{\mathbf{P}} = \frac{1}{\sqrt{194305000}}(9470, 8650, 1700)\)

To find a unit vector in the triangle plane that bisects the interior angle at A, we will add the two given vectors, \(\mathbf{R}_{AM}\) and \(\mathbf{R}_{AN}\), and normalize the sum: \(\mathbf{B} = \mathbf{R}_{AM} + \mathbf{R}_{AN}\) \(\mathbf{B} = (20,18,-10) + (-10,8,15)\) \(\mathbf{B} = (10,26,5)\) \( ||\mathbf{B}|| = \sqrt{(10)^2 + (26)^2 + (5)^2}\) \( ||\mathbf{B}|| = \sqrt{1075}\) \( \hat{\mathbf{B}} = \frac{\mathbf{B}}{||\mathbf{B}||}\) \( \hat{\mathbf{B}} = \frac{1}{\sqrt{1075}}(10, 26, 5)\) In conclusion, the unit vector perpendicular to the triangle is \((\frac{350}{\sqrt{350000}}, -\frac{450}{\sqrt{350000}}, \frac{340}{\sqrt{350000}})\); a unit vector in the plane of the triangle and perpendicular to \(\mathbf{R}_{AN}\) is \((\frac{9470}{\sqrt{194305000}}, \frac{8650}{\sqrt{194305000}}, \frac{1700}{\sqrt{194305000}})\); and a unit vector in the plane of the triangle that bisects the interior angle at A is \((\frac{10}{\sqrt{1075}}, \frac{26}{\sqrt{1075}}, \frac{5}{\sqrt{1075}})\).