Consider the points A = (0,1,1), B = (1,1,0) and C = (1,0,3). What are vectors \(\overrightarrow{AB} \) and \(\overrightarrow{AC} \).
Let Q = (-2, 3, 4). Then the distance between Q and the plane through the points A = (0,1,1), B = (1,1,0) and C = (1,0,3) is \(\dfrac{d}{\sqrt{11}}\) for some value of d. What's d?
+6248 \(\vec{AB}= B-A = (1,1,0)-(0,1,1) = (1,0,-1)\\ \vec{AC}=C-A = (1,0,3) - (0,1,1) = (1,-1, 2)\)
\(\text{Using $\vec{AB}$ and $\vec{AC}$ we can find the normal to the plane as $\vec{AB}\times \vec{AC}$}\\~\\ n = \left|\begin{pmatrix}i&j&k\\1&0&-1\\1&-1&2\end{pmatrix}\right| = (-1,-3,-1)\\ \text{we want the unit normal so we divide by it's length}\\ \hat{n} = \dfrac{1}{\sqrt{11}}(-1,-3,-1)\)
\(\text{now we find the vector from $A$ to $Q$}\\ \vec{AQ}=(-2,3,4)-(0,1,1)=(-2,2,3)\\ \text{and we dot this with the unit normal vector}\\ \vec{AQ}\cdot \hat{n} = (-2,2,3) \cdot \dfrac{1}{\sqrt{11}}(-1,-3,-1) = \\ \dfrac{1}{\sqrt{11}}(2-6-3)=\dfrac{-7}{\sqrt{11}}\\ d = \left |\vec{AQ}\cdot \hat{n}\right| = \dfrac{7}{\sqrt{11}}\)
\(d=7\)
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