Introduction to Three Dimensional Geometry

\(\text{Given one linear relation between direction cosines: } 3l + m + 5n = 0\)

\(\Rightarrow m = -3l - 5n\)

\(\text{Second relation: } 6nm - 2nl + 5lm = 0\)

\(\text{Substitute } m = -3l - 5n:\; 6n(-3l - 5n) - 2nl + 5l(-3l - 5n) = 0\)

\(\Rightarrow -18ln - 30n^{2} - 2ln - 15l^{2} - 25ln = 0\)

\(\Rightarrow -15l^{2} - 45ln - 30n^{2} = 0\)

\(\Rightarrow l^{2} + 3ln + 2n^{2} = 0\)

\(\Rightarrow (l + n)(l + 2n) = 0\)

\(\text{Case 1: } l = -n \Rightarrow m = -3(-n) - 5n = -2n\)

\(\Rightarrow \text{direction ratios } \propto (l,m,n) = (-1,-2,1)\)

\(\text{Case 2: } l = -2n \Rightarrow m = -3(-2n) - 5n = n\)

\(\Rightarrow \text{direction ratios } \propto (l,m,n) = (-2,1,1)\)

\(\text{Angle } \theta \text{ between lines with ratios } \vec{a}=(-1,-2,1), \vec{b}=(-2,1,1)\)

\(\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\;\|\vec{b}\|} = \dfrac{(-1)(-2)+(-2)(1)+(1)(1)}{\sqrt{(-1)^2+(-2)^2+1^2}\;\sqrt{(-2)^2+1^2+1^2}}\)

\(\cos\theta = \dfrac{2 - 2 + 1}{\sqrt{6}\;\sqrt{6}} = \dfrac{1}{6}\)

\(\therefore \theta = \cos^{-1}\!\left(\dfrac{1}{6}\right)\)

\(\text{Given one linear relation between direction cosines: } 3l + m + 5n = 0\)

\(\Rightarrow m = -3l - 5n\)

\(\text{Second relation: } 6nm - 2nl + 5lm = 0\)

\(\text{Substitute } m = -3l - 5n:\; 6n(-3l - 5n) - 2nl + 5l(-3l - 5n) = 0\)

\(\Rightarrow -18ln - 30n^{2} - 2ln - 15l^{2} - 25ln = 0\)

\(\Rightarrow -15l^{2} - 45ln - 30n^{2} = 0\)

\(\Rightarrow l^{2} + 3ln + 2n^{2} = 0\)

\(\Rightarrow (l + n)(l + 2n) = 0\)

\(\text{Case 1: } l = -n \Rightarrow m = -3(-n) - 5n = -2n\)

\(\Rightarrow \text{direction ratios } \propto (l,m,n) = (-1,-2,1)\)

\(\text{Case 2: } l = -2n \Rightarrow m = -3(-2n) - 5n = n\)

\(\Rightarrow \text{direction ratios } \propto (l,m,n) = (-2,1,1)\)

\(\text{Angle } \theta \text{ between lines with ratios } \vec{a}=(-1,-2,1), \vec{b}=(-2,1,1)\)

\(\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\;\|\vec{b}\|} = \dfrac{(-1)(-2)+(-2)(1)+(1)(1)}{\sqrt{(-1)^2+(-2)^2+1^2}\;\sqrt{(-2)^2+1^2+1^2}}\)

\(\cos\theta = \dfrac{2 - 2 + 1}{\sqrt{6}\;\sqrt{6}} = \dfrac{1}{6}\)

\(\therefore \theta = \cos^{-1}\!\left(\dfrac{1}{6}\right)\)

Let the line make angles $\alpha,\beta,\gamma$ with the positive $x,y,z$-axes respectively.
Given: $\beta=\dfrac{\alpha}{2}$ and $\gamma=\dfrac{\alpha}{2}$.

For a directed line in 3D, the direction cosines satisfy

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1.$

Put $\beta=\gamma=\alpha/2$

${\cos }^2\alpha +2{\cos }^2\left(\frac{\alpha }{2}\right)=1$

\( \cos ^2 \alpha+\cos \alpha=0 \)
\( \cos \alpha(\cos \alpha+1)=0\)
\( \cos \alpha=0,-1\)
\( \alpha=\frac{\pi}{2}, \pi \)
Now \(\beta=\frac{\alpha}{2} \Rightarrow \frac{\pi}{4}, \frac{\pi}{2} \)
so sum is \(\frac{3 \pi}{4}\)

Let the line make angles $\alpha,\beta,\gamma$ with the positive $x,y,z$-axes respectively.
Given: $\beta=\dfrac{\alpha}{2}$ and $\gamma=\dfrac{\alpha}{2}$.

For a directed line in 3D, the direction cosines satisfy

$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.$$

Substitute $\beta=\gamma=\alpha/2$:

$$\cos^2\alpha+2\cos^2\left(\frac{\alpha}{2}\right)=1.$$

Use $\cos^2\left(\frac{\alpha}{2}\right)=\frac{1+\cos\alpha}{2}$:

$$\begin{gathered} {\cos }^2\alpha +2\cdot \frac{1+\cos \alpha }{2}=1 \\ \Rightarrow {\cos }^2\alpha +1+\cos \alpha =1 \end{gathered}$$

$$\begin{gathered} \Rightarrow {\cos }^2\alpha +\cos \alpha =0 \\ \Rightarrow \cos \alpha (\cos \alpha +1)=0. \end{gathered}$$

So $\cos\alpha=0$ or $\cos\alpha=-1$, giving

$$\alpha=\frac{\pi}{2}\quad\text{or}\quad \alpha=\pi.$$

Hence

$$\beta=\frac{\alpha}{2}=\frac{\pi}{4}\quad\text{or}\quad \frac{\pi}{2}.$$

Sum of all possible $\beta$ values:

$$\frac{\pi }{4}+\frac{\pi }{2}=\frac{3\pi }{4}\mathrm{.}$$