I am new to the topic of tensors and I was reading the following exercise:
"Pick a point O in Euclidean space, and take any line O1, O2, O3, to be at right angles to each other such to form a 3D Cartesian frame of reference. Let $\alpha_i$ be the angle between a line formed by O and an arbitrary point in space, and the corresponding axis $i=1,2,3$. Interpret geometrically the surface for whose points $\lambda_1= \cos \alpha_1 =$constant. Show that 2 surfaces $\lambda_1=$cons and $\lambda_2=$const intercept in two straight lines provided that $\lambda_1^2+\lambda_2^2 < 1$"
I'm under the impression that the surface is a cone. But I have no idea how to even try to solve the problem. Any input would be kindly appreciated.
If $\vec p=(x,y,z)$ is a point of the surface, then all points $\mu\vec p$ on the line from the origin through $\vec p$ are on the surface, just be cause
$$\lambda_k=\cos(\alpha_k)=\frac{\vec p\cdot \vec o_k}{\|\vec p\|}=\frac{\mu\vec p\cdot o_k}{\|\mu\vec p\|}.$$
Hence the surface is a cone.
Analytically, the equation is
$$\lambda_1=\frac{x_1}{\sqrt{x_1^2+x_2^2+x_3^2}},$$ or $$\left(1-\frac1{\lambda_1^2}\right)x_1^2+x_2^2+x_3^2=0,$$ and similarly for the other axes.
The intersection of one of the cones with the unit sphere is a circle. Then the intersection of two cones goes through the intersection points of the two corresponding circles.
Analytically, you form the system
$$\left(1-\frac1{\lambda_1^2}\right)x_1^2+x_2^2+x_3^2=0,\\ x_1^2+\left(1-\frac1{\lambda_2^2}\right)x_2^2+x_3^2=0,\\ x_1^2+x_2^2+x_3^2=1,$$
which can be solved for $x_1^2,x_2^2,x_3^2$, which gives eight solutions for $x_1,x_2,x_3$. A rigorous sign discussion will result in just two of them.