The vectors x= (-4, p, -2) and y= (-2, 3, 6) are such that cos-1 (4/21) = θ, where θ is the angle between x and y. Determine the value(s) of p.

Question

The vectors X= (-4, p, -2) and Y= (-2, 3, 6) are such that cos^-1 (4/21) = θ, where θ is the angle between X and Y. Determine the value(s) of p.

Answer 1

The inner product divided by the product of the magnitudes equals the cosine of the angle between them.

Thus, (-4,p,-2) . (-2, 3, 6) = 8 + 3p - 12 = 3p - 4

|-4, p, -2| * |-2,3,6| = sqrt(4^2+p^2+(-2)^2)*sqrt((-2)^2+3^2+6^2) = 7sqrt(20+p^2)

Thus,

(3p - 4)/7sqrt(20+p^2) = 4/21

28 sqrt(20+p^2) = 21(3p-4)

4 sqrt(20+p^2) = 3(3p-4)

Squaring both sides,

16(20+p^2) = 9(3p-4)^2

320+16p^2 = 81p^2 - 216p + 144

65p^2 - 216p - 176 = 0

(p - 4)(65p + 44) = 0

p = 4, - 44/65

As we squared, though, we may have introduced an extra root, so we substitute back into

(3p - 4)/7sqrt(20+p^2)

when p = 4, (3*4-4)/7sqrt(20+4^2) = 8/(7*6) = 4/21

when p=-44/65, (3*-44/65-4)/7sqrt(20+(-44/65)^2) =

-392/65/(7/65*sqrt(86436)) =

-392/(7*294) =

-98*4/(7*98*3) =

-4/21

This is, of course, not a solution, but the negative of it (introduced by squaring, as we expected)

p = 4