Line and plane intersecting with spherical section

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Given the line $r:\ x=1+2t,\ y=2+3t,\ z=-1-t$ and the plane $\pi:\ x-2y+3z+4=0$.

(a) Verify that $r$ intersects $\pi$.

(b) Find the point of intersection.

(c) Find the angle that $r$ makes with $\pi$.

(d) Compute the distance of $P(3,1,2)$ from $\pi$.

(e) Write the parametric equations of the line through $P$ perpendicular to $\pi$.

(f) Find the radius of the circle obtained by cutting the sphere $x^2+y^2+z^2+x+y+z-16=0$ with $\pi$.

(g) Find the volume and surface area of that sphere.

Solution

(a) Substituting the parametric equations of $r$ into the equation of $\pi$: $(1+2t)-2(2+3t)+3(-1-t)+4=-2-7t=0\Rightarrow t=-\tfrac27$ (unique solution), so $r$ meets $\pi$ in exactly one point.

(b) Substituting $t=-\tfrac27$ into the equations of $r$ gives $I=\left(\tfrac37,\tfrac87,-\tfrac57\right)\approx(0.43,\,1.14,\,-0.71)$.

(c) The direction vector of $r$ is $\vec d=(2,3,-1)$ and the normal vector of $\pi$ is $\vec n=(1,-2,3)$: $\sin\theta=\dfrac{|\vec d\cdot\vec n|}{|\vec d|\,|\vec n|}=\dfrac{7}{14}=\tfrac12\Rightarrow\theta=30°$.

(d) $d(P,\pi)=\dfrac{|3-2+6+4|}{\sqrt{14}}=\dfrac{11}{\sqrt{14}}\approx2.940$.

(e) The line through $P$ perpendicular to $\pi$ has direction the normal vector $\vec n=(1,-2,3)$: $x=3+t,\ y=1-2t,\ z=2+3t$.

(f) The sphere has centre $C=\left(-\tfrac12,-\tfrac12,-\tfrac12\right)$ and radius $R=\sqrt{16.75}\approx4.093$. The distance from the centre to the plane is $d(C,\pi)=\dfrac{3}{\sqrt{14}}\approx0.802$, so the radius of the section is $\rho_{\text{sez}}=\sqrt{R^2-d^2}\approx4.013$.

(g) Volume and surface area of the sphere: $V=\tfrac43\pi R^3\approx287.15$ and $S=4\pi R^2\approx210.49$. $\boxed{t=-\tfrac27;\ I\left(\tfrac37,\tfrac87,-\tfrac57\right);\ \theta=30°;\ d_P\approx2.94;\ \rho_{\text{sez}}\approx4.01;\ V\approx287.15;\ S\approx210.49}$