Question

Question 1:

Let vectors: A⃗ =(2,1,−4), B⃗ =(−3,0,1), and C⃗ =(−1,−1,2).

Calculate the following:

(A) What is the value of  A⃗ ⋅ B⃗ ?

(B) What is the angle θAB between A⃗ and B⃗ ?

(C) What is the value of 2B⃗ ⋅ 3C⃗ ?

(D) What is the value of 2(B⃗ ⋅ 3C⃗) ?

Which of the following can be computed? (Choose only 1 of the following)

A⃗ ⋅B⃗ ⋅C⃗  

A⃗ ⋅(B⃗ ⋅C⃗ )

A⃗ ⋅(B⃗ +C⃗ )

3⋅A⃗

FOR PARTS F THROUGH H:

V⃗ 1 and V⃗ 2 are different vectors with lengths V1 and V2 respectively. Find the following:

(F) Express Answer in terms of V⃗ 1,   V⃗ 1⋅V⃗ 1 =

(G) If V⃗ 1 and V⃗ 2 are perpendicular, V⃗ 1⋅V⃗ 2 =

(H) Express your answer in terms of V1 and V2.  If V⃗ 1 and V⃗ 2 are parallel,   V⃗ 1⋅V⃗ 2 =

Answer 1

Given,

A = (2,1,−4), B = (−3,0,1), and C = (−1,−1,2)

in vectr form

A = 2i + j - 4 k

B = -3 i + 0 j + k

C = -i - j + 2 k

A) A. B = (2i + j - 4 k).(-3i + 0j + k) = 2 x -3 + 1 x 0 - 4 x 1 = -6 - 4 = -10

Hence, A.B = -10

B)cos(theta) = A.B/(AB)

A = sqrt (2^2 + 1^2 + 4^2) = sqrt (21)

B = sqrt (3^2 + 0^2 + 1^2) = sqrt (10)

cos(theta) = -10/14.49 = -0.6901

theta = cos^-1(-0.6901) = 133.64 deg

Hence, theta = 133.64 deg

C)2B = (-6i + 0j + 2k)

3C = (-3i - 3j + 6k)

2B.3C = (-6i + 0j + 2k).(-3i - 3j + 6k) = -6 x -3 + 0 x -3 + 2 x 6 = 18 + 12 = 30

2B.3C = 30

D)2(B.3C) = 2[(-3i + 0j + k).(-3i - 3j + 6k)] = 2 (-3 x -3 + 0 + 1 x 6) = 2 x (9 + 6) = 30

Hence, 2(B.3C) = 30

F)V1.V1 = (V1)^2

2)V1.V2 = V1 V2 cos(theta)

when perpendicular theta= 90 deg ; cos90 = 0

V1.V2 = 0

H)when parallel, theta = 0

V1.V2 = (V1) (V2)