STEP III 1998 question 8 solution - The Student Room
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STEP III 1998 question 8 solution

TODO: write up the question and add solutions for parts (ii) and (iii)

(i) TODO


(ii) By definition of a plane perpendicular to unit vector \mathbf{n} with minimum distance l from the origin, the points \mathbf{r} in the plane satisfy \mathbf{r} \cdot \mathbf{n}  = l

The points \mathbf{p} of the second sphere satisfy (\mathbf{p}-\mathbf{d})\cdot(\mathbf{p}-\mathbf{d}) = a^2

For the the plane to be tangential to the second sphere, it must meet the radius vector (from the centre of the second sphere to the point of contact) at a right angle.

Let the point of contact be \mathbf{s} The radius vector is (\mathbf{s}-\mathbf{d}) which has length a and is parallel to \mathbf{n}

Since the scalar product of two parallel vectors is the product of their magnitudes:

(\mathbf{s}-\mathbf{d}) \cdot \mathbf{n} = a

(\mathbf{s} \cdot \mathbf{n}) - (\mathbf{d} \cdot \mathbf{n}) = a

Since, by definition, \mathbf{s} is in the plane:

\mathbf{s} \cdot \mathbf{n}  = l Therefore:

l - (\mathbf{d} \cdot \mathbf{n}) = a

This is the condition required.


(iii) TODO

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