question archive Question 1: Let vectors: A? =(2,1,−4), B? =(−3,0,1), and C? =(−1,−1,2)
Question 1: Let vectors: A? =(2,1,−4), B? =(−3,0,1), and C? =(−1,−1,2)
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Question 1:
Let vectors: A? =(2,1,−4), B? =(−3,0,1), and C? =(−1,−1,2).
Calculate the following:
(A) What is the value of A? ⋅ B? ?
(B) What is the angle θAB between A? and B? ?
(C) What is the value of 2B? ⋅ 3C? ?
(D) What is the value of 2(B? ⋅ 3C?) ?
Which of the following can be computed? (Choose only 1 of the following)
| A? ⋅B? ⋅C? |
| A? ⋅(B? ⋅C? ) |
| A? ⋅(B? +C? ) |
| 3⋅A? |
FOR PARTS F THROUGH H:
V? 1 and V? 2 are different vectors with lengths V1 and V2 respectively. Find the following:
(F) Express Answer in terms of V? 1, V? 1⋅V? 1 =
(G) If V? 1 and V? 2 are perpendicular, V? 1⋅V? 2 =
(H) Express your answer in terms of V1 and V2. If V? 1 and V? 2 are parallel, V? 1⋅V? 2 =
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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)
