Vector Unit
Find the unit vector perpendicular to the plane formed by the two vectors:
U
= 2
i
+ 10
j
and
V
=
i
+ 3
j
– 5
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
$$$\vec{a}=(a_{1}, a_{2}, a_{3})$$$
$$$\vec{b}=(b_{1}, b_{2}, b_{3})$$$
This is a two part problem, where the perpendicular vector must be determined first. Then, the unit vector will be solved next. Recall that a unit vector is a vector that has a magnitude of 1. 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, the cross product is:
$$$\vec{U} \times \vec{V}=\begin{bmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ U_x & U_y & U_z\\ V_x & V_y & V_z\end{bmatrix}=\begin{bmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 10 & 0\\ 1 & 3 & -5\end{bmatrix}$$$
$$$=\mathbf{i}[(10)(-5)-(0)(3)] - \mathbf{j}[(2)(-5)-(0)(1)] + \mathbf{k}[(2)(3)-(10)(1)]$$$
$$$=\mathbf{i}(-50-0) - \mathbf{j}(-10-0) + \mathbf{k}(6-10)=\{-50;\:10;\:-4 \}$$$
To find the unit vector, divide by the magnitude. To find the magnitude:
$$$|\vec{a}|=\sqrt{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}=\sqrt{(-50)^2+10^2+(-4)^2}=\sqrt{2,500+100+16}$$$
$$$=\sqrt{2,616}\approx 51.15$$$
Finally, the unit vector perpendicular to the plane formed by vectors
U
and
V
:
$$$\frac{-50i+10j-4k}{51.15}$$$
$$$\frac{-50i+10j-4k}{51.15}$$$
Time Analysis
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