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### What is the angle between the position vectors a and b, where a = (6i - j + 3k) and b = (-4i + 2j + 10k)?

cos(AOB) = a.b/(|a| x |b|)

a.b = (6i - j + 3k).(-4i + 2j + 10k)

= (6 x -4) + (-1 x 2) + (3 x 10)

= -24 + -2 + 30

= 4

(|a|)^2 = 6^2 + (-1)^2 + 3^2

= 36 + 1 + 9 = 46

-> |a| = 46^(1/2)

(|b|)^2 = (-4)^2 + 2^2 + 10^2

= 16 + 4 + 100 = 120

-> |b| = 120^(1/2) = 2 x 30^(1/2)

cos(AOB) = a.b/(|a| x |b|)

= 4/(46^(1/2) x 2 x 30^(1/2))

= 4/(4 x 345^(1/2))

= 1/(345^(1/2))

AOB = cos^-1 (1/(345^(1/)))

= 86.9 degrees(3 significant figures)

2 years ago

Answered by David, an A Level Maths tutor with MyTutor

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