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.

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.

Yes, you are correct that the value of AxB is -31j. Your calculations are correct.

To find the magnitude |AxB|, you can use the formula |AxB| = √(x^2 + y^2 + z^2), where x, y, and z are the components of the cross product AxB. In this case, since AxB = -31j, we have x = 0, y = -31, and z = 0. Plugging these values into the formula, we get |AxB| = √(0^2 + (-31)^2 + 0^2) = √961 = 31.

To find the angle between the two vectors, you can use the formula sin(theta) = |AxB| / (|A| * |B|), where |A| and |B| are the magnitudes of vectors A and B, respectively. In this case, |A| = √(2^2 + 0^2 + 3^2) = √13 and |B| = √((-9)^2 + 0^2 + 2^2) = √85. Plugging these values into the formula, we get sin(theta) = 31 / (√13 * √85). To find the angle theta, you can take the inverse sine (sin^-1) of this value.

Let's go through each question step by step:

a) To find the cross product A x B, you can use the determinant method:

A x B = | i j k |
| 2 0 3 |
| -9 0 2 |

Calculate this determinant to get the cross product:

A x B = (2 * 2 - 3 * 0)i - (2 * -9 - 3 * 0)j + (0 * 0 - 3 * -9)k
= 4i + 0j + 27k

So, the cross product of A and B is 4i + 27k.

Your calculation of A x B = 23j is incorrect. Please check your calculation.

b) To find the magnitude |A x B|, you can use the formula:

|A x B| = √(A x B) · (A x B)

Substituting the cross product we found earlier:

|A x B| = √((4i + 27k) · (4i + 27k))
= √(4^2 + 27^2)
= √(16 + 729)
= √745

So, the magnitude of A x B is √745.

c) To find the angle between the two vectors, you can use the formula:

sin(θ) = |A x B| / (|A| * |B|)

Substituting the values we have:

sin(θ) = √745 / (|A| * |B|)

|A| = √(2^2 + 3^2) = √13 (magnitude of A)
|B| = √((-9)^2 + 2^2) = √85 (magnitude of B)

sin(θ) = √745 / (√13 * √85)

To find the angle θ, you can take the inverse sine (also known as arcsine) of sin(θ):

θ = arcsin(√745 / (√13 * √85))

You can use a calculator to find the angle, which is approximately 1.074 radians or 61.55 degrees.

I hope this explanation helps you understand the concept better. Let me know if you have any further questions!