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### Section 11.4 : Cross Product

3. Find a vector that is orthogonal to the plane containing the points $$P = \left( {3,0,1} \right)$$, $$Q = \left( {4, - 2,1} \right)$$ and $$R = \left( {5,3, - 1} \right)$$.

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We first need two vectors that are both parallel to the plane. Using the points that we are given (all in the plane) we can quickly get quite a few vectors that are parallel to the plane. We’ll use the following two vectors.

$\overrightarrow {PQ} = \left\langle {1, - 2,0} \right\rangle \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\overrightarrow {PR} = \left\langle {2,3, - 2} \right\rangle$ Show Step 2

Now we know that the cross product of any two vectors will be orthogonal to the two original vectors. Since the two vectors from Step 1 are parallel to the plane (they actually lie in the plane in this case!) we know that the cross product must then also be orthogonal, or normal, to the plane.

So, using the “trick” we used in the notes the cross product is,

\begin{align*}\overrightarrow {PQ} \times \overrightarrow {PR} & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&{ - 2}&0\\2&3&{ - 2}\end{array}} \right|\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\vec i}&{\vec j}\\1&{ - 2}\\2&3\end{array}\\ & = 4\vec i + 0\vec j + 3\vec k - \left( { - 2\vec j} \right) - 0\vec i - \left( { - 4\vec k} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{4\vec i + 2\vec j + 7\vec k}}\end{align*}