Year: 1998
Paper: 3
Question Number: 8
Course: UFM Pure
Section: Vectors
No solution available for this problem.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
\begin{questionparts}
%\item[(i)] Consider the sphere of radius $a$ and centre the origin.
%Show that the line through the point with position vector
%${\bf b}$ and parallel to a unit
%vector ${\bf m}$ intersects the sphere at two points if
%$$
%a^2 > {\bf b}.{\bf b} -({\bf b}.{\bf m})^2 \,.
%$$
%What is the corresponding condition for there to be precisely one
%point of intersection?
%If this point has position vector ${\bf p}$, show that the line
%is perpendicular to ${\bf p}$.
\item Show that the line ${\bf r} ={\bf b} + \lambda {\bf m}$,
where $\bf m$ is a unit vector,
intersects the sphere ${\bf r}\cdot {\bf r} = a^2$ at two points if
$$
a^2 > {\bf b}\cdot{\bf b} -({\bf b}\cdot{\bf m})^2 \,.
$$
Write down the corresponding condition for there to be precisely one
point of intersection.
If this point has position vector ${\bf p}$, show that ${\bf m}\cdot{\bf p}=0$.
\item
Now consider a second sphere of radius $a$
and a plane perpendicular to a unit vector~${\bf n}$.
The centre of the sphere
has position vector ${\bf d}$
and the
minimum distance from the origin to the plane is $l$. What is the
condition for the plane to be tangential to this second sphere?
\item
Show that the first and second spheres intersect at right angles
({\em i.e.\ }the two radii to each point of
intersection are perpendicular) if
$$
{\bf d}\cdot{\bf d} = 2 a^2 \,.
$$
\end{questionparts}