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

The vertices of △ABC are A(3,5,-4), B(-1,1,2), and C(-5,-5,-2).
The direction ratios of side AB are (-1-3), (91-5), and(2-(-4)) i.e., -4, -4, and 6.
Then,
\(\sqrt{(-4)^2+(-4)^2+(6)^2}\)
=\(\sqrt{16+16+36}\)
=\(\sqrt{68}\)
=\(2\sqrt{17}\)

Therefore, the direction cosines of AB are
\(-\frac{4}{(-4)^2}\)+(-4)²+(6)², \(-\frac{4}{(-4)^2}\)+(-4)²+(6)², \(-\frac{6}{(-4)^2}\)+(-4)²+(6)²
\(-\frac{4}{2\sqrt{17}}\), \(-\frac{4}{2\sqrt{17}}\), \(\frac{6}{2\sqrt{17}}\)
\(-\frac{2}{\sqrt{17}}\), \(-\frac{2}{\sqrt{17}}\), \(\frac{3}{\sqrt{17}}\)

The direction ratios of BC are(-5-(-1)), (-5-1), and (-2-2) i.e.,-4, -6, and -4.

Therefore, the direction cosines of BC are \(-\frac{4}{(-4)^2}\)+(-6)²+(-4)², \(-\frac{6}{(-4)^2}\)+(-6)²+(-4)²,\(-\frac{4}{(-4)^2}\)+(-6)²+(-4)² i.e., \(-\frac{4}{2\sqrt{17}}\), \(-\frac{6}{2\sqrt{17}}\), \(-\frac{4}{2\sqrt{17}}\)
The direction ratios of CA are(-5-3) ,(-5-5), and (-2-(-4)) i.e., -8, -10, and 2.