Question

Question: Use the result that for A? =Axi^+Ayj^+Azk^ and B? =Bxi^+Byj^+Bzk^ : A? ×B? =(AyBz?AzBy)i^+(AzBx?AxBz)j^+(AxBy?AyBx)k^and A? ×B? =?????i^AxBxj^AyByk^AzBz?????

Part A

Determine the vector product A? ×B? if A? =5.3i^?3.4j^ and B? =?8.6i^+6.0j^+2.5k^. Find the x-component of the vector product.

Part B

Find the y-component of the vector product.

Part C

Find the z-component of the vector product.

Part D

Determine the angle between A? and B? if A? =5.3i^?3.4j^ and B? =?8.6i^+6.0j^+2.5k^.

Answer 1

AxB = (5.3 i - 3.4 j)X(-8.6 i + 6.0 j + 2.5 k)

Using given equation

Ax = 5.3, Ay = -3.4, Az = 0

Bx = -8.6, By = 6.0, Bz = 2.5

AxB = (AyBz?AzBy)i+(AzBx?AxBz)j+(AxBy?AyBx)k

AxB = (-3.4*2.5 - 0) i + (0 - 5.3*2.5) j + (5.3*6.0 - 3.4*8.6) k

AxB = -8.5 i - 13.25 j + 2.56 k

Part A

(AXB)x = -8.5

Part B

(AXB)x = -13.25

Part C

(AXB)x = 2.56

Part D

|AxB| = |A|*|B|*sin theta

|AxB| = sqrt ((-8.5)^2 + (-13.25)^2 + (2.56)^2) = 15.9488

|A| = sqrt ((5.3)^2 + (-3.4)^2 + (0)^2) = 6.2968

|B| = sqrt ((-8.6)^2 + (6.0)^2 + (2.5)^2) = 10.7800

So,

arcsin (|AxB|/(|A|*|B|))

theta = arcsin (15.9488/(6.2968*10.7800))

theta = 13.60 deg

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