Mathematical Methods for Physics and Engineering - Matematica.NET

7.8 USING VECTORS TO FIND DISTANCESof the sphere. This is easily expressed in vector notation as|r − c| 2 =(r − c) · (r − c) =a 2 , (7.43)where c is the position vector of the centre of the sphere and a is its radius.◮Find the radius ρ of the circle that is the intersection of the plane ˆn · r = p and the sphereof radius a centred on the point with position vector c.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)where r is restricted to lie in the plane and b is the position of the circle’s centre.As b lies on the plane whose normal is ˆn, thevectorb − c must be parallel to ˆn, i.e.b − c = λˆn for some λ. Further, by Pythagoras, we must have ρ 2 + |b − c| 2 = a 2 . Thusλ 2 = a 2 − ρ 2 .Writing b = c + √ a 2 − ρ 2 ˆn and substituting in (7.45) gives(r 2 − 2r · c + √ )a 2 − ρ 2 ˆn + c 2 +2(c · ˆn) √ a 2 − ρ 2 + a 2 − ρ 2 = ρ 2 ,whilst, on expansion, (7.44) becomesr 2 − 2r · c + c 2 = a 2 .Subtracting these last two equations, using ˆn · r = p and simplifying yieldsp − c · ˆn = √ a 2 − ρ 2 .On rearrangement, this gives ρ as √ a 2 − (p − c · ˆn) 2 , which places obvious geometricalconstraints on the values a, c, ˆn and p can take if a real intersection between the sphereand the plane is to occur. ◭7.8 Using vectors to find distancesThis section deals with the practical application of vectors to finding distances.Some of these problems are extremely cumbersome in component form, but theyall reduce to neat solutions when general vectors, with no explicit basis set,are used. These examples show the power of vectors in simplifying geometricalproblems.7.8.1 Distance from a point to a lineFigure 7.14 shows a line having direction b that passes through a point A whoseposition vector is a. To find the minimum distance d of the line from a point Pwhose position vector is p, we must solve the right-angled triangle shown. We seethat d = |p − a| sin θ; so, from the definition of the vector product, it follows thatd = |(p − a) × ˆb|.229