Practice: Cross Product

Practice: Cross Product

Recall that iβƒ—=[1, 0, 0]\vec{i}=\left[1,\ 0,\ 0\right], jβƒ—=[0, 1, 0]\vec{j}=\left[0,\ 1,\ 0\right], and kβƒ—=[0, 0, 1]\vec{k}=\left[0,\ 0,\ 1\right].
Match the following cross products with the correct result.
A.
βˆ’jβƒ—-\vec{j}
B.
k⃗\vec{k}
C.
βˆ’iβƒ—-\vec{i}
D.
j⃗\vec{j}
E.
2i⃗2\vec{i}
F.
2iβƒ—βˆ’kβƒ—2\vec{i}-\vec{k}
i⃗×k⃗\vec{i}\times\vec{k}
i⃗×j⃗\vec{i}\times\vec{j}
k⃗×j⃗\vec{k}\times\vec{j}
(βˆ’iβƒ—)Γ—kβƒ—\left(-\vec{i}\right)\times\vec{k}
kβƒ—Γ—(βˆ’2jβƒ—)\vec{k}\times\left(-2\vec{j}\right)

j⃗×(2k⃗+i⃗)\vec{j}\times\left(2\vec{k}+\vec{i}\right)
More Cross Product Questions: