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.