## Unit Vectors

Find the unit vector perpendicular to the plane formed by the two vectors: U = 5 i + 7 j and V = 1 i + 2 j – 3 k

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 we first must determine the perpendicular vector. Then solve for the unit vector, which is a vector that has a magnitude of 1. By finding the cross product, we are calculating the perpendicular vector of U and V, and can use the below 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:
$$\vec{U} \times \vec{V}=\begin{bmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ U_x & U_y & U_z\\ V_x & V_y & V_z\end{bmatrix}=\begin{bmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 5 & 7 & 0\\ 1 & 2 & -3\end{bmatrix}$$$$$=\mathbf{i}[(7)(-3)-(0)(2)] - \mathbf{j}[(5)(-3)-(0)(1)] + \mathbf{k}[(5)(2)-(7)(1)]$$$
$$=\mathbf{i}(-21-0) - \mathbf{j}(-15-0) + \mathbf{k}(10-7)=\{-21;\:15;\:3 \}$$$To find the unit vector, divide by the magnitude. Using the below formula, we can solve for the magnitude: $$|\vec{a}|=\sqrt{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}=\sqrt{(-21)^2+15^2+3^2}=\sqrt{441+225+9}$$$
$$=\sqrt{675}=15\sqrt{3}\approx 25.98$$$Finally, the unit vector perpendicular to the plane formed by vectors U and V : $$\frac{-21i+15j+3k}{15\sqrt{3}}$$$
$$\frac{-21i+15j+3k}{15\sqrt{3}}$$\$