Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 11

Table of contents
- a 11
  - Solution a
- b 11
  - Solution b

a

Solution

Expand the vector triple product in Cartesian components. Its first component is

[a \times (b \times c)]_{x} = a_{y} (b_{x} c_{y} - b_{y} c_{x}) - a_{z} (b_{z} c_{x} - b_{x} c_{z}) = b_{x} (a_{y} c_{y} + a_{z} c_{z}) - c_{x} (a_{y} b_{y} + a_{z} b_{z}) = b_{x} (a \cdot c) - c_{x} (a \cdot b),

where adding

a_{x} b_{x} c_{x}

to both products introduces cancelling terms. Likewise the other two components are

[a \times (b \times c)]_{y} [a \times (b \times c)]_{z} = a_{z} (b_{y} c_{z} - b_{z} c_{y}) - a_{x} (b_{x} c_{y} - b_{y} c_{x}) = b_{y} (a \cdot c) - c_{y} (a \cdot b), = a_{x} (b_{z} c_{x} - b_{x} c_{z}) - a_{y} (b_{y} c_{z} - b_{z} c_{y}) = b_{z} (a \cdot c) - c_{z} (a \cdot b) .

Together these prove

a \times (b \times c) = (a \cdot c) b - (a \cdot b) c .

Now use the cyclic property of the scalar triple product and the just-proved identity:

(a \times b) \cdot (c \times d) = a \cdot [b \times (c \times d)] = a \cdot [(b \cdot d) c - (b \cdot c) d] .

This proves Lagrange's identity in the requested form,

(a \times b) \cdot (c \times d) = (a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c) .

b

Solution

The sphere of centre

a

p

∣ r - a ∣^{2} = p^{2} .

Let

d = ∣ b - a ∣

be the distance between the centres. A nondegenerate circle of intersection of two spheres exists exactly when

∣ p - q ∣ < d < p + q .

The upper inequality excludes disjoint or externally tangent spheres. The lower excludes strict containment and internal tangency. It also excludes coincident centres, which give either no intersection or an entire common sphere rather than one circle. These inequalities are sufficient because a triangle with side lengths

p, q, d

then exists with nonzero area, and rotating its third vertex about the line of centres traces a circle.

For a point on both spheres, expand their equations:

r^{2} - 2 a \cdot r + a^{2} = p^{2}, r^{2} - 2 b \cdot r + b^{2} = q^{2} .

Subtracting cancels

r^{2}

and gives the plane

2 (b - a) \cdot r = p^{2} - q^{2} + b^{2} - a^{2} .

Its normal is parallel to the line of centres. Under the strict inequalities, the plane cuts either sphere in a genuine circle.

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