What is the cross product of << -1, -1, 2 >>⟨−1,−1,2⟩ and << 4,3,6 >> ⟨4,3,6⟩?
1 Answer
Well, you have at least two ways to do it.
The first way:
Let
color(blue)(vecu xx vecv) = << u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 >>→u×→v=⟨u2v3−u3v2,u3v1−u1v3,u1v2−u2v1⟩
= << -1*6 - 2*3, 2*4 - (-1*6), -1*3 - (-1*4) >>=⟨−1⋅6−2⋅3,2⋅4−(−1⋅6),−1⋅3−(−1⋅4)⟩
= color(blue)(<< -12, 14, 1 >>)=⟨−12,14,1⟩
Assuming you didn't know that formula, the second way (which is a little more foolproof) is recognizing that:
hati xx hatj = hatkˆi׈j=ˆk
hatj xx hatk = hatiˆj׈k=ˆi
hatk xx hati = hatjˆk׈i=ˆj
hatA xx hatA = vec0ˆA׈A=→0
hatA xx hatB = -hatB xx hatAˆA׈B=−ˆB׈A where
hati = << 1,0,0 >>ˆi=⟨1,0,0⟩ ,hatj = << 0,1,0 >>ˆj=⟨0,1,0⟩ , andhatk = << 0,0,1 >>ˆk=⟨0,0,1⟩ .
Thus, rewriting the vectors in unit vector form:
(-hati - hatj + 2hatk)xx(4hati + 3hatj + 6hatk)(−ˆi−ˆj+2ˆk)×(4ˆi+3ˆj+6ˆk)
= cancel(-4(hati xx hati))^(0) - 3(hati xx hatj) - 6(hati xx hatk) - 4(hatj xx hati) - cancel(3(hatj xx hatj))^(0) - 6(hatj xx hatk) + 8(hatk xx hati) + 6(hatk xx hatj) + cancel(12(hatk xx hatk))^(0)
= -3hatk + 6hatj + 4hatk - 6hati + 8hatj - 6hati
= - 12hati + 14hatj + hatk
= color(blue)(<< -12, 14, 1 >>)
as expected.
