Question

Find the angles of a triangle whose vertices are A (0, −1, −2), B (3, 1, 4) and C (5, 7, 1). 

Answer 1

Given that
\[ \vec{OA} = 0 \hat{i} - 1 \hat{j} - 2 \hat{k} ; \vec{OB} = 3 \hat{i} + 1 \hat{j} + 4 \hat{k} ; \vec{OC} = 5 \hat{i} + 7 \hat{j} + 1 \hat{k} \] 
\[ \vec{AB} = \vec{OB} - \vec{OA} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \Rightarrow \left| \vec{AB} \right| = \sqrt{9 + 4 + 36} = 7\]
\[ \vec{BA} = \vec{OA} - \vec{OB} = - 3 \hat{i} - 2 \hat{j} - 6 \hat{k} \Rightarrow \left| \vec{BA} \right| = \sqrt{9 + 4 + 36} = 7\]
\[ \vec{BC} = \vec{OC} - \vec{OB} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \Rightarrow \left| \vec{BC} \right| = \sqrt{4 + 36 + 9} = 7\]
\[ \vec{CB} = \vec{OB} - \vec{OC} = - 2 \hat{i} - 6 \hat{j} + 3 \hat{k} \Rightarrow \left| \vec{CB} \right| = \sqrt{4 + 36 + 9} = 7\]
\[ \vec{CA} = \vec{OA} - \vec{OC} = - 5 \hat{i} - 8 \hat{j} - 3 \hat{k} \Rightarrow \left| \vec{CA} \right| = \sqrt{25 + 64 + 9} = \sqrt{98} = 7\sqrt{2}\]
\[ \vec{AC} = \vec{OC} - \vec{OA} = 5 \hat{i} + 8 \hat{j} + 3 \hat{k} \Rightarrow \left| \vec{AC} \right| = \sqrt{25 + 64 + 9} = \sqrt{98} = 7\sqrt{2}\]
\[\cos A = \frac{\vec{AB} . \vec{AC}}{\left| \vec{AB} \right|\left| \vec{AC} \right|} = \frac{15 + 16 + 18}{\left( 7 \right)\left( 7\sqrt{2} \right)} = \frac{49}{49\sqrt{2}} = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow A = \cos^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}\]
\[\cos B = \frac{\vec{BA} . \vec{BC}}{\left| \vec{BA} \right|\left| \vec{BC} \right|} = \frac{- 6 - 12 + 18}{\left( 7 \right)\left( 7 \right)} = \frac{0}{49} = 0\]
\[ \Rightarrow B = \cos^{- 1} \left( 0 \right) = \frac{\pi}{2}\]
\[\cos C = \frac{\vec{CB} . \vec{CA}}{\left| \vec{CB} \right|\left| \vec{CA} \right|} = \frac{10 + 48 - 9}{\left( 7 \right)\left( 7\sqrt{2} \right)} = \frac{49}{49\sqrt{2}} = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow C = \cos^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}\]