1991 Paper 2 Q10

Year: 1991
Paper: 2
Question Number: 10

Course: LFM Pure and Mechanics
Section: Vectors

Difficulty: 1600.0 Banger: 1500.0

Problem

A straight stick of length \(h\) stands vertically. On a sunny day, the stick casts a shadow on flat horizontal ground. In cartesian axes based on the centre of the Earth, the position of the Sun may be taken to be \(R(\cos\theta,\sin\theta,0)\) where \(\theta\) varies but \(R\) is constant. The positions of the base and tip of the stick are \(a(0,\cos\phi,\sin\phi)\) and \(b(0,\cos\phi,\sin\phi)\), respectively, where \(b-a=h\). Show that the displacement vector from the base of the stick to the tip of the shadow is \[ Rh(R\cos\phi\sin\theta-b)^{-1}\begin{pmatrix}-\cos\theta\\ -\sin^{2}\phi\sin\theta\\ \cos\phi\sin\phi\sin\theta \end{pmatrix}. \] {[}`Stands vertically' means that the centre of the Earth, the base of the stick and the tip of the stick are collinear, `horizontal' means perpendicular to the stick.

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

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Banger Rating: 1500.0

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Problem source
A straight stick of length $h$ stands vertically. On a sunny day,
the stick casts a shadow on flat horizontal ground. In cartesian axes
based on the centre of the Earth, the position of the Sun may be taken
to be $R(\cos\theta,\sin\theta,0)$ where $\theta$ varies but $R$
is constant. The positions of the base and tip of the stick are $a(0,\cos\phi,\sin\phi)$
and $b(0,\cos\phi,\sin\phi)$, respectively, where $b-a=h$. Show
that the displacement vector from the base of the stick to the tip
of the shadow is 
\[
Rh(R\cos\phi\sin\theta-b)^{-1}\begin{pmatrix}-\cos\theta\\
-\sin^{2}\phi\sin\theta\\
\cos\phi\sin\phi\sin\theta
\end{pmatrix}.
\]
{[}`Stands vertically' means that the centre of the Earth, the base
of the stick and the tip of the stick are collinear, `horizontal'
means perpendicular to the stick.