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

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

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Given vertices of the triangle \triangle ABC  (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Finding each side direction ratios;

\RightarrowDirection ratios of side AB are (-1-3), (1-5),\ and\ (2-(-4)) i.e.,

-4,-4,\ and\ 6.

Therefore its direction cosines values are;

\frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}} or\ \frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}

SImilarly for side BC;

    \RightarrowDirection ratios of side BC are (-5-(-1)), (-5-1),\ and\ (-2-2) i.e.,

-4,-6,\ and\ -4.

Therefore its direction cosines values are;

\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}} or\ \frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}

  \RightarrowDirection ratios of side CA are (-5-3), (-5-5),\ and\ (-2-(-4)) i.e.,

-8,-10,\ and\ 2.

Therefore its direction cosines values are;

\frac{-8}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{-5}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{2}{\sqrt{(-8)^2+(10)^2+(2)^2}} or\ \frac{-8}{2\sqrt{42}},\frac{-10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}\ or\ \frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}

Posted by

Divya Prakash Singh

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