Question

Three vectors are given as A=<3.5, -5.3, 1.1> B=<4.5, 7.5, -7> C=<-3.9, 4.8, -6>

a) Determine the angle between A and B.

b) Determine the angle between A and C

c) Compute the dot product of B and C

d) Determine the size of the area associated with vectors B and C.

e) Determine the quantity (B (dot)(A(cross)C))

Answer 1

The given vectors in coordinate form will be:

A = 3.5 i - 5.3 j + 1.1 k

B = 4.5 i + 7.5 j - 7 k

C = -3.9 i + 4.8 j -6 k

a)Let theta be the angle between A and B

We know that angle between two vectors is given by:

cos(theta) = (a . b)/a b

A = sqrt (3.5^2 + -5.3^2 + 1.1^2) = 6.45

B = sqrt (4.5^2 + 7.5^2 + -7^2) = 11.20

A.B = ( 3.5 i - 5.3 j + 1.1 k ).(4.5 i + 7.5 j - 7 k) = 3.5 x 4.5 - 5.3 x 7.5 - 1.1 x 7 = 15.75 - 39.75 = -31.7

cos(theta) = -31.7/(6.45 x 11.20) = 116.27 deg

Hence, theta = 116.27 deg

b) A = 6.45

C = sqrt (-3.9^2 + 4.8^2 + -6^2) = 8.62

A.C = ( 3.5 i - 5.3 j + 1.1 k ).(-3.9 i + 4.8 j - 6k) = 3.5 x -3.9 + -5.3 x 4.8 + 1.1 x -6 = -45.69

cos(alpha) = A.C/AC

alpha = cos^-1(-45.69/6.45 x 8.62) = 145.26 deg

Hence, alpha = 145.26 deg

C)B.C = (4.5 i + 7.5 j - 7 k).(-3.9 i + 4.8 j - 6k) = 4.5 x -3.9 + 7.5 x 4.8 + -7 x -6 = 60.45

Hence, B.C = 60.45 units

D)The size of area will be equal to the cross product, so

BxC = (4.5 i + 7.5 j - 7 k) x (-3.9 i + 4.8 j - 6k) = i (7.5 x -6 - (-7) x 4.8) - j (4.5 x -6 - (-7) x 3.9) + k (4.5 x 4.8 - 7.5 x -3.9)

BxC = i (-45 + 33.6) -j(-27 -27.3) + k(21.6 + 29.25) = (-11.4 i + 54.3 j + 50.85 k)

BxC = (-11.4 i + 54.3 j + 50.85 k)

A = lBxCl = sqrt (-11.4^2 + 54.3^2 + 50.85^2) = 75.26
units^2

Hence, A = 75.26 units^2