Table of Contents

Graph

Take this generalised funny sphere/circle question that Mr White likes so much apparently: (you can just remove the z values and its the same thing)

Radius unknown

Given that

$$\begin{align} L_{1}:& \ r=\begin{pmatrix} a \\ b \\ c \end{pmatrix} + \lambda \times \begin{pmatrix} d \\ e \\ f \end{pmatrix} \\ S:& \ |r-\begin{pmatrix} g \\ h \\ i \end{pmatrix}|= \alpha \end{align} $$

Find where: i) $L_{1}$ is tangent to $S$. ii) $L_{1}$ intersects $S$ twice. iii) $L_{1}$ doesn't intersect $S$.

Solution (1)

Consider i). This occurs when:

  1. The minimum distance between the line and the sphere centre is exactly the radius of the sphere.
  2. The distance vector between the line and the sphere centre is perpendicular to the direction vector of the line.

Let $\vec{d}$ be the distance vector between the line and the sphere centre. Hence,

$$\begin{align} \vec{d}&=\begin{pmatrix} a \\ b \\ c \end{pmatrix} + \lambda \times \begin{pmatrix} d \\ e \\ f \end{pmatrix} - \begin{pmatrix} g \\ h \\ i \end{pmatrix} \\ &=\begin{pmatrix} a-g+\lambda d \\ b-h+\lambda e \\ c-i+\lambda f \end{pmatrix} \end{align} $$

Consider (1), i.e. $|\vec{d}|=R$. Hence,

$$\begin{align} |\begin{pmatrix} a-g+\lambda d \\ b-h+\lambda e \\ c-i+\lambda f \end{pmatrix}|=\alpha \\ \therefore \sqrt{ (a-g+\lambda d)^{2} + (b-h+\lambda e)^{2} + (c-i+\lambda f)^{2} } = \alpha \end{align} $$

Consider (2), i.e.

$$\vec{d} \cdot \begin{pmatrix} d \\ e \\ f \end{pmatrix} = 0 $$ $$\begin{align} \therefore d(a-g+\lambda d)+e(b-h+\lambda e)+f(c-i+\lambda f)=0 \end{align} $$

Hence, solve lines (6) and (7) simultaneously using CAS, i.e.

$$\begin{cases} \sqrt{ (a-g+\lambda d)^{2} + (b-h+\lambda e)^{2} + (c-i+\lambda f)^{2} } = \alpha \\ -d(a-g+\lambda d)+e(b-h+\lambda e)+f(c-i+\lambda f)=0 \end{cases} \ \ \ \alpha, \lambda $$

And thus: i) When $\alpha = \text{ans}$ (from simultaneous equation) ii) When $\alpha >\text{ans}$ iii) When $0<\alpha<\text{ans}$

Scalar unknown

Given that

$$\begin{align} L_{1}:& \ r=\begin{pmatrix} a \\ \beta \\ c \end{pmatrix} + \lambda \times \begin{pmatrix} d \\ e \\ f \end{pmatrix} \\ S:& \ |r-\begin{pmatrix} g \\ h \\ i \end{pmatrix}|= R \end{align} $$

Find where: i) $L_{1}$ is tangent to $S$. ii) $L_{1}$ intersects $S$ twice. iii) $L_{1}$ doesn't intersect $S$.

Solution (2)

This is exactly the same as above. Same principles, but now we are solving for $\beta$, and thus your answers will be slightly different.

In this case, we will get 2 solutions for $\beta$, which I will call $ans_{1}$ and $ans_{2}$, where $ans_{1} ans_{2}$.

Thus: i) When $\beta =ans_{1}$ or $\beta=ans_{2}$ ii) When $ans{1} < \beta < ans_{2}$ iii) When $\beta < ans_{1}$ or $\beta > ans_{2}$