Question

If the line passing through the origin makes angles ${\theta }_1,{\theta }_2,{\theta }_3$ with the planes $XOY,XOZ$ and $ZOX$ respectively, then prove that ${\cos }^2{\theta }_1+{\cos }^2{\theta }_2+{\cos }^2{\theta }_3=2.$

Answer 1

Direction cosine are given as $\cos {\theta }_1,\cos {\theta }_2,\cos {\theta }_3$

Here $P(x,y,z)$ & $0(0,0,0)$

$OP=\sqrt{(x-0{)}^2+(y-0{)}^2+(z-0{)}^2}$

$OP=\sqrt{x^2+y^2+z^2}$

$r=x^2+y^2+z^2$

According to diagonal

$\cos {\theta }_1=\frac{x}{r}$ $\cos {\theta }_2=\frac{y}{r}$ $\cos {\theta }_3=\frac{z}{r}$

$x=r\cos {\theta }_1$ $y=r\cos {\theta }_2$ $z=r\cos {\theta }_3$

Square & add

$x^2+y^2+z^2=r^2({\cos }^2{\theta }_1+{\cos }^2{\theta }_2+{\cos }^2{\theta }_3)$

$r^2=r^2{\cos }^2{\theta }_1+{\cos }^2{\theta }_2+{\cos }^2{\theta }_3$

So ${\cos }^2{\theta }_1+{\cos }^2{\theta }_2+{\cos }^2{\theta }_3=1$.