Which of the following is a unit vector perpendicular to the plane determined by the vectors A=2i+3j and B=2i+5j-k?

First, we know that the cross product of two vectors gives us a 3rd vector that is perpendicular to both of the originals. Therefore, we first want to find the cross product, AxB. As a reminder, these are the formulas we will use to find the cross product:

$$\begin{gathered} Equation 1: \mathit{a} x \mathit{b} = \left.open\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right| \\ Equation 2: \mathit{a} x \mathit{b} = \mathit{i}·\left.open\begin{array}{cc}{a}_2& a_3\\ b_2& b_3\end{array}\right|- \mathit{j}·\left.open\begin{array}{cc}{a}_1& a_3\\ b_1& b_3\end{array}\right|+\mathit{k}·\left.open\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right| \\ Equation 3:\left.open\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc \end{gathered}$$

Using these formulas and the vectors that we were given in the problem:

$$\begin{gathered} \mathit{A} x \mathit{B} =\left.open\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 3& 0\\ 2& 5& -1\end{array}\right| \\ \mathit{A} x \mathit{B}=\mathit{i}·\left.open\begin{array}{cc}3& 0\\ 5& -1\end{array}\right|-\mathbf{ }\mathit{j}·\left.open\begin{array}{cc}2& 0\\ 2& -1\end{array}\right|+\mathit{k}·\left.open\begin{array}{cc}2& 3\\ 2& 5\end{array}\right| \\ \mathit{A} x \mathit{B}=\mathit{i}·\left(\right(3*-1)-(0*5\left)\right)- \mathit{j}·\left(\right(2*-1)-(0*2\left)\right)+\mathit{k}·\left(\right(2*5)-(3*2\left)\right) \\ \mathit{A}x \mathit{B}=\mathit{i}·(-3-0)- \mathit{j}·(-2-0)+\mathit{k}·(10-6) \\ \mathit{A} x \mathit{B}=-3\mathit{i}+ 2\mathit{j}+4\mathit{k} \end{gathered}$$

We now have a vector that is perpendicular to the plane determined by vectors A and B. However, we are not at our final solution yet. The problem asks us the find a unit vector. We know that a unit vector must have a magnitude of 1. Let's find the magnitude of the vector we got by calculating the cross product:

$$\begin{gathered} opena||=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2+{\left(a_3\right)}^2} \\ open\mathit{A}x\mathit{B}||=\sqrt{{(-3)}^2+{\left(2\right)}^2+{\left(4\right)}^2} \\ open\mathit{A}x\mathit{B}||=\sqrt{9+4+16} \\ open\mathit{A}x\mathit{B}||=\sqrt{29} \end{gathered}$$

Therefore, the magnitude of our current vector is $\sqrt{29}$. To get a vector with the same direction but with a magnitude of 1, we simply need to multiply by $\frac{1}{\sqrt{29}}(\sin ce (\sqrt{29}·\frac{1}{\sqrt{29}}=1).$

$$\begin{gathered} (-3\mathit{i}+ 2\mathit{j}+4\mathit{k}\mathbf{)}\mathbf{·}\left(\frac{1}{\sqrt{29}}\right) \\ \frac{-3\mathit{i}}{\sqrt{29}}+\frac{2\mathit{j}}{\sqrt{29}}+\frac{4\mathit{k}}{\sqrt{29}} \end{gathered}$$

This final vector above is our answer.