Question

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, −4), (−1, 1, 2) and (−5, −5, −2).

Answer 1

\[\text {The vertices of ∆ ABC are }A \left( 3, 5, - 4 \right), B \left( - 1, 1, 2 \right)\text{ and } C \left( - 5, - 5, - 2 \right) .\]

\[\text{The direction ratios of AB are} \left( - 1 - 3 \right), \left( 1 - 5 \right), \left[ 2 - \left( - 4 \right) \right], \text{i . e} . - 4, - 4, 6 . \]

\[\text{Therefore, the direction cosines of AB are}\]

\[\frac{- 4}{\sqrt{\left( - 4 \right)^2 + \left( - 4 \right)^2 + \left( 6 \right)^2}}, \frac{- 4}{\sqrt{\left( - 4 \right)^2 + \left( - 4 \right)^2 + \left( 6 \right)^2}}, \frac{6}{\sqrt{\left( - 4 \right)^2 + \left( - 4 \right)^2 + \left( 6 \right)^2}}\]

\[ = \frac{- 4}{2\sqrt{17}}, \frac{- 4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}} \]

\[ = \frac{2}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{- 3}{\sqrt{17}}\]

\[\text{The direction ratios of BC are} \left[ - 5 - \left( - 1 \right) \right], \left( - 5 - 1 \right), \left( - 2 - 2 \right), \text{i . e} . - 4, - 6, - 4 . \]

\[\text{Therefore, the direction cosines of BC are}\]

\[\frac{- 4}{\sqrt{\left( - 4 \right)^2 + \left( - 6 \right)^2 + \left( - 4 \right)^2}}, \frac{- 6}{\sqrt{\left( - 4 \right)^2 + \left( - 6 \right)^2 + \left( - 4 \right)^2}}, \frac{- 4}{\sqrt{\left( - 4 \right)^2 + \left( - 6 \right)^2 + \left( - 4 \right)^2}}\]

\[ = \frac{- 4}{2\sqrt{17}}, \frac{- 6}{2\sqrt{17}}, \frac{- 4}{2\sqrt{17}} \]

\[ = \frac{2}{\sqrt{17}}, \frac{3}{\sqrt{7}}, \frac{2}{\sqrt{17}}\]

\[\text{The direction ratios of CA are} \left[ 3 - \left( - 5 \right) \right], \left[ 5 - \left( - 5 \right) \right], \left[ - 4 - \left( - 2 \right) \right],\text{ i . e} . 8, 10, - 2 . \]

\[\text{Therefore, the direction cosines of CA are}\]

\[\frac{8}{\sqrt{\left( 8 \right)^2 + \left( 10 \right)^2 + \left( - 2 \right)^2}}, \frac{10}{\sqrt{\left( 8 \right)^2 + \left( 10 \right)^2 + \left( - 2 \right)^2}}, \frac{- 2}{\sqrt{\left( 8 \right)^2 + \left( 10 \right)^2 + \left( - 2 \right)^2}}\]

\[ = \frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{- 2}{2\sqrt{42}} \]

\[ = \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{- 1}{\sqrt{42}}\]