Mathematical Methods for Physics and Engineering : A Comprehensive Guide

7.8 USING VECTORS TO FIND DISTANCES

of the sphere. This is easily expressed in vector notation as

|r−c|^2 =(r−c)·(r−c)=a^2 , (7.43)

wherecis the position vector of the centre of the sphere andais its radius.

Find the radiusρof the circle that is the intersection of the planenˆ·r=pand the sphere

of radiusacentred on the point with position vectorc.

The equation of the sphere is

|r−c|^2 =a^2 , (7.44)

and that of the circle of intersection is

|r−b|^2 =ρ^2 , (7.45)

whereris restricted to lie in the plane andbis the position of the circle’s centre.
Asblies on the plane whose normal isnˆ,thevectorb−cmust be parallel toˆn,i.e.
b−c=λnˆfor someλ. Further, by Pythagoras, we must haveρ^2 +|b−c|^2 =a^2. Thus
λ^2 =a^2 −ρ^2.
Writingb=c+

√

a^2 −ρ^2 ˆnand substituting in (7.45) gives

r^2 − 2 r·

(

c+

√

a^2 −ρ^2 nˆ

)

+c^2 +2(c·nˆ)

√

a^2 −ρ^2 +a^2 −ρ^2 =ρ^2 ,

whilst, on expansion, (7.44) becomes

r^2 − 2 r·c+c^2 =a^2.

Subtracting these last two equations, usingnˆ·r=pand simplifying yields

p−c·nˆ=

√

a^2 −ρ^2.

On rearrangement, this givesρas

√

a^2 −(p−c·nˆ)^2 , which places obvious geometrical
constraints on the valuesa,c,nˆandpcan take if a real intersection between the sphere
and the plane is to occur.

7.8 Using vectors to find distances

This section deals with the practical application of vectors to finding distances.

Some of these problems are extremely cumbersome in component form, but they

all reduce to neat solutions when general vectors, with no explicit basis set,

are used. These examples show the power of vectors in simplifying geometrical

problems.

7.8.1 Distance from a point to a line

Figure 7.14 shows a line having directionbthat passes through a pointAwhose

position vector isa. To find theminimum distancedof the line from a pointP

whose position vector isp, we must solve the right-angled triangle shown. We see

thatd=|p−a|sinθ; so, from the definition of the vector product, it follows that

d=|(p−a)×bˆ|.