Yes, you are correct that AxB = -31j.
To find the magnitude |AxB|, you can use the formula you mentioned:
|AxB| = sqrt((-31)^2) = 31
Now, to find the angle between the two vectors, use the following formula:
sin(theta) = |AxB| / (|A|*|B|)
First, find the magnitude of A and B:
|A| = sqrt((2.00)^2 + (3.00)^2) = sqrt(13)
|B| = sqrt((-9.00)^2 + (2.00)^2) = sqrt(81+4) = sqrt(85)
Now substitute the magnitudes into the formula:
sin(theta) = 31 / (sqrt(13) * sqrt(85))
theta = arcsin(31 / (sqrt(13) * sqrt(85)))
Now you can use a calculator to find the angle in radians or degrees, depending on the units you need.
Two vectors are lying in the xz plane: A=2.00i + 3.00k and B= -9.00i + 2.00k.
a) What is the value of AxB?
Check my solution: I found through calculations that the value is 23j.
b) What is the magnitude |AxB|?
I could do ABsin theta, but I have no angle between. So am confused here.
C) What is the ganle between the two vectors.
Here I could do the:
23
---- = sin theta.
AB
Please help. I really would like to know this concept before class today in about 1 hr. Check all my answers, please.
Thank you!
(2.00i + 3.00k)(-9.00i + 2.00k)
ixi=0
ixk= -j
kxi=j
from that, 4(-j) + 27(-j)
magnitude: (sqrt (4^2 + 27^2))
angle between vectors?
arc sin=magnitudeabove/magnitudeA*magnitudeB
I'm still somewhat confused. Does 4(-j) + 27(-j) equal -31j?
also, if you can, please elaborate a bit more on proper steps to conceptualize prob(the rest).
yes, I got -31j
if the cross product is equal to magnitude A * magnitude B * sinTheta, then
crossproduct/(magnitudeA*MagnitudeB)= sinTheta.
Solve for Theta.
1 answer
1 answer