[Vectors] Cross Product

2021-10-23 21:08 作者:AoiSTZ23 0人讀過 | 我要投稿

By: Tao Steven Zheng (鄭濤)

【Problem】

Consider two vectors $%5Cboldsymbol%7BA%7D%20%3D%20%5Clangle%201%2C2%2C3%5Crangle%20$ and $%5Cboldsymbol%7BB%7D%20%3D%20%5Clangle%202%2C0%2C-1%5Crangle%20$ .

Part 1: Calculate the cross product $%5Cboldsymbol%7BA%20%5Ctimes%20B%7D$ .

Part 2: Determine the norm of $%5Cboldsymbol%7BA%20%5Ctimes%20B%7D$ . Then calculate angle between $%5Cboldsymbol%7BA%7D%20$ and $%5Cboldsymbol%7BB%7D%20$ .

Part 3: Show that the cross-product $%20%5Cboldsymbol%7BA%20%5Ctimes%20B%7D$ is orthogonal to $%5Cboldsymbol%7BA%7D%20$ and $%5Cboldsymbol%7BB%7D%20$ .

【Solution】

Part 1

$%5Cboldsymbol%7BA%20%5Ctimes%20B%7D%20%3D%0A%5Cbegin%7Bvmatrix%7D%0A%5Cboldsymbol%7Bi%7D%20%26%20%5Cboldsymbol%7Bj%7D%20%26%20%5Cboldsymbol%7Bk%7D%20%5C%5C%0A1%20%26%202%20%26%203%20%5C%5C%0A2%20%26%200%20%26%20-1%0A%5Cend%7Bvmatrix%7D$

$%5Cboldsymbol%7BA%20%5Ctimes%20B%7D%20%3D%0A%5Cbegin%7Bvmatrix%7D%0A2%20%26%203%20%5C%5C%0A0%20%26%20-1%0A%5Cend%7Bvmatrix%7D%20%5Cboldsymbol%7Bi%7D%20-%0A%5Cbegin%7Bvmatrix%7D%0A1%20%26%203%20%5C%5C%0A2%20%26%20-1%0A%5Cend%7Bvmatrix%7D%20%5Cboldsymbol%7Bj%7D%20%2B%0A%5Cbegin%7Bvmatrix%7D%0A1%20%26%202%20%5C%5C%0A2%20%26%200%0A%5Cend%7Bvmatrix%7D%20%5Cboldsymbol%7Bk%7D$

$%5Cboldsymbol%7BA%20%5Ctimes%20B%7D%20%3D%20%5B(2%20%5Ccdot%20-1)%20-%20(3%5Ccdot%200)%5D%5Cboldsymbol%7Bi%7D%20-%20%5B(1%5Ccdot%20-1)%20-%20(3%5Ccdot%202)%5D%5Cboldsymbol%7Bj%7D%20%2B%20%5B(1%20%5Ccdot%200)%20-%20(2%20%5Ccdot%202)%5D%20%5Cboldsymbol%7Bk%7D$

$%5Cboldsymbol%7BA%20%5Ctimes%20B%7D%20%3D%20-2%5Cboldsymbol%7Bi%7D%20%2B%207%5Cboldsymbol%7Bj%7D%20-%204%5Cboldsymbol%7Bk%7D$

Part 2

The norm of the cross product of two vectors $%5Cboldsymbol%7BA%7D%2C%20%5Cboldsymbol%7BB%7D$ is the area of the parallelogram with side lengths of the norms of $%5Cboldsymbol%7BA%7D%20$ and $%5Cboldsymbol%7BB%7D%20$ .

$%7C%5Cboldsymbol%7BA%20%5Ctimes%20B%7D%7C%20%3D%20%5Csqrt%7B(-2)%5E2%2B(7)%5E2%2B(-4)%5E2%7D$

$%7C%5Cboldsymbol%7BA%20%5Ctimes%20B%7D%7C%20%3D%20%5Csqrt%7B69%7D$

To determine the angle $%5Ctheta$ between the vectors $%5Cboldsymbol%7BA%7D%20$ and $%5Cboldsymbol%7BB%7D%20$ , use the formula

$%7C%5Cboldsymbol%7BA%20%5Ctimes%20B%7D%7C%20%3D%20%7C%5Cboldsymbol%7BA%7D%7C%20%7C%5Cboldsymbol%7BB%7D%7C%20%5Csin%7B%5Ctheta%7D$

$%7C%5Cboldsymbol%7BA%7D%7C%20%3D%20%5Csqrt%7B(1)%5E2%2B(2)%5E2%2B(3)%5E2%7D%20%3D%20%5Csqrt%7B14%7D$

$%7C%5Cboldsymbol%7BB%7D%7C%20%3D%20%5Csqrt%7B(2)%5E2%2B(0)%5E2%2B(-1)%5E2%7D%20%3D%20%5Csqrt%7B5%7D$

Thus,

$%20%5Csin%7B%5Ctheta%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B69%7D%7D%7B%5Csqrt%7B14%7D%5Csqrt%7B5%7D%7D$

$%5Ctheta%20%3D%5Csin%5E%7B-1%7D%7B%5Cleft(%5Cfrac%7B%5Csqrt%7B69%7D%7D%7B%5Csqrt%7B70%7D%7D%5Cright)%7D$

$%5Ctheta%20%3D%2083.1354%5E%5Ccirc$

Part 3
The fastest way to show that two vectors are orthogonal (perpendicular) is to take the dot-product. If the dot-product is zero, then the two vectors are orthogonal. This is the basis of the right-hand rule.

Let $%20%5Cboldsymbol%7BC%7D%20%3D%20%5Clangle%20-2%2C7%2C-4%5Crangle$ , $%5Cboldsymbol%7BA%7D%20%3D%20%5Clangle%201%2C2%2C3%5Crangle$ , and $%5Cboldsymbol%7BB%7D%20%3D%20%5Clangle%202%2C0%2C-1%5Crangle$ .

$%5Cboldsymbol%7BC%20%5Ccdot%20A%7D%20%3D%20(-2)(1)%20%2B%20(7)(2)%20%2B%20(-4)(3)%20%3D%200$