Question

Prove that the points \[\hat{i} - \hat{j} , 4 \hat{i} + 3 \hat{j} + \hat{k} \text{ and }2 \hat{i} - 4 \hat{j} + 5 \hat{k}\] are the vertices of a right-angled triangle.

Answer 1

Given the points \[\hat{i} - \hat{j} , 4 \hat{i} + 3 \hat{j} + \hat{k}\] and\[2 \hat{i} - 4 \hat{j} + 5 \hat{k}\]  Are A, B and C respectively.
Then,
\[\overrightarrow{AB} = 4 \hat{i} + 3 \hat{j} + \hat{k} - \hat{i} + \hat{j} = 3 \hat{i} + 4 \hat{j} + \hat{k} . \]
\[ \overrightarrow{BC} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} - 4 \hat{i} - 3 \hat{j} - \hat{k} = - 2 \hat{i} - 7 \hat{j} + 4 \hat{k} . \]
\[ \overrightarrow{CA} = \hat{i} - \hat{j} - 2 \hat{i} + 4 \hat{j} - 5 \hat{k} = - \hat{i} + 3 \hat{j} - 5 \hat{k} .\]
\[\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = 3 \hat{i} + 4 \hat{j} + \hat{k} - 2 \hat{i} - 7 \hat{j} + 4 \hat{k} - \hat{i} + 3 \hat{j} - 5 \hat{k} = \vec{0} .\]
The given points forms a vertices of a triangle.
Now,
\[\left| \overrightarrow{AB} \right| = \sqrt{9 + 16 + 1} = \sqrt{26} . \]
\[\left| \overrightarrow {BC} \right| = \sqrt{4 + 49 + 16} = \sqrt{69} . \]
\[\left| \overrightarrow{CA} \right| = \sqrt{1 + 9 + 25} = \sqrt{35} .\]
\[\left| \overrightarrow{AB} \right|^2 + \left| \overrightarrow{CA} \right|^2 = 26 + 35 = 61\] ≠ \[\left| \overrightarrow{BC} \right|^2\]
The given triangle is not right-angled.