What is the cross product of #(- 5 i + 4 j - 5 k)# and #(4 i + 4 j + 2 k)#?

1 Answer
Jan 14, 2016

Answer:

If we call the first vector #vec a# and the second #vec b#, the cross product, #vec a xx vec b# is #(28veci-10vecj-36veck)#.

Explanation:

Sal Khan of Khan academy does a nice job of calculating a cross product in this video: https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/linear-algebra-cross-product-introduction

It's something that's easier to do visually, but I'll try to do it justice here:

#vec a = (-5veci+4vecj-5veck)#
#vec b = (4veci+4vecj+2veck)#

We can refer to the coefficient of #i# in #vec a# as #a_i#, the coefficient of #j# in #vec b# as #b_j# and so on.

#vec a xx vec b = (-5veci+4vecj-5veck) xx (4veci+4vecj+2veck)#

Sal's video above and the Wikipedia article on the cross product will do a better job of explaining why the next step is as follows than I can here:

#vec a xx vec b = (a_jb_k-a_kb_j)vec i + (a_kb_i-a_ib_k)vec j+(a_ib_j-a_jb_i)vec k#

#= (4*2-(-5)*4)vec i + ((-5)*4-(-5)*2)vec j+((-5)*4-4*4)vec k = 28vec i -10 vec j -36vec k#