Cross Products
Calculate the cross product of the following two vectors:
U
= 2
i
+ 4
j
and
V
=
i
+
j
–
k
Expand Hint
$$$\vec{a} \times \vec{b}=\begin{bmatrix}a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1b_2-a_2b_1\end{bmatrix}$$$
Hint 2
$$$\vec{a}=(a_{1}, a_{2}, a_{3})$$$
$$$\vec{b}=(b_{1}, b_{2}, b_{3})$$$
Finding the cross product determines the perpendicular vector of U and V. Using the shown formula to get started:
$$$\vec{a} \times \vec{b}=\begin{bmatrix}a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1b_2-a_2b_1\end{bmatrix}$$$
where
$$\vec{a}=(a_{1}, a_{2}, a_{3})$$
and
$$\vec{b}=(b_{1}, b_{2}, b_{3})$$
.
Thus,
$$$\vec{U} \times \vec{V}=\begin{bmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ U_x & U_y & U_z\\ V_x & V_y & V_z\end{bmatrix}=\begin{bmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 4 & 0\\ 1 & 1 & -1\end{bmatrix}$$$
$$$=\mathbf{i}[(4)(-1)-(0)(1)] - \mathbf{j}[(2)(-1)-(0)(1)] + \mathbf{k}[(2)(1)-(4)(1)]$$$
$$$=\mathbf{i}(-4-0) - \mathbf{j}(-2-0) + \mathbf{k}(2-4)=\{-4;\:2;\:-2 \}$$$
{-4; 2; -2}
Time Analysis
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