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# How do I do this?

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Find the distance between $$Q = (3, -7, -1)$$and the line through $$A = (1, 1, 2)$$ and $$B = (2, 3, 4).$$ This distance is equal to $$\dfrac{\sqrt{d}}{3}$$for some integer d. What is d?

May 6, 2020

### 1+0 Answers

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Find the distance between $$Q = (3, -7, -1)$$ and the line through $$A = (1, 1, 2)$$ and $$B = (2, 3, 4)$$.
This distance is equal to $$\dfrac{\sqrt{d}}{3}$$ for some integer $$d$$.
What is $$d$$?

Line through $$A = (1, 1, 2)$$ and $$B = (2, 3, 4)$$

$$\small{ \begin{array}{|rclrcl|} \hline \vec{x} &=& \vec{A}+t(\vec{B}-\vec{A}) & \vec{r} &=& \vec{B}-\vec{A} \quad | \quad \vec{A} = (1, 1, 2)\quad \vec{B} = (2, 3, 4)\\ &&& \vec{r} &=& \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}-\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \\ &&&\vec{r} &=& \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \\ \mathbf{\vec{x}} &=& \mathbf{\vec{A}+t\vec{r}} \\ \hline (\vec{x}-\vec{Q})\cdot \vec{r} &=& 0 \quad | \quad \vec{PQ} \text{ is } \perp \text{ to } \vec{r} \\ (\mathbf{\vec{A}+t\vec{r}}-\vec{Q})\cdot \vec{r} &=& 0 \\ (\vec{A}-\vec{Q} + t\vec{r})\cdot \vec{r} &=& 0 & \vec{A}-\vec{Q}&=& \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}-\begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix} \quad | \quad \vec{Q} = (3, -7, -1)\\ & & & \vec{A}-\vec{Q}&=& \begin{pmatrix} -2 \\ 8 \\ 3 \end{pmatrix} \\\\ \left[\begin{pmatrix} -2 \\ 8 \\ 3 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \right]\cdot\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} &=& 0 \\\\ \left[\begin{pmatrix} -2+ t \\ 8+2t \\ 3+2t \end{pmatrix} \right]\cdot\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} &=& 0 \\\\ (-2+ t)\cdot 1 + (8+2t)\cdot 2+ (3+2t)\cdot 2 &=& 0 \\ -2+t+16+4t+6+4t &=& 0 \\ 9t+20 &=& 0 \\ \mathbf{t} &=& \mathbf{ -\dfrac{20}{9} } \\ \hline \end{array} }$$

$$\begin{array}{|rcll|} \hline \mathbf{\vec{P}} &=& \vec{A}+t\vec{r} \\ &=& \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}-\mathbf{ \dfrac{20}{9} }\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \\\\ &=& \begin{pmatrix} 1-\dfrac{20}{9} \\ 1-\dfrac{40}{9} \\ 2-\dfrac{20}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} \mathbf{ -\dfrac{11}{9}} \\ \mathbf{-\dfrac{31}{9}} \\ \mathbf{ -\dfrac{22}{9}} \end{pmatrix} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\vec{PQ}} &=& \vec{Q}- \vec{P} \\ &=& \begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix}- \begin{pmatrix} \mathbf{ -\dfrac{11}{9}} \\ \mathbf{-\dfrac{31}{9}} \\ \mathbf{ -\dfrac{22}{9}} \end{pmatrix} \\\\ &=& \begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix}+ \begin{pmatrix} \dfrac{11}{9} \\ \dfrac{31}{9} \\ \dfrac{22}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} 3 +\dfrac{11}{9} \\ -7+\dfrac{31}{9} \\ -1+\dfrac{22}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} \mathbf{ \dfrac{38}{9}} \\ \mathbf{-\dfrac{32}{9}} \\ \mathbf{ \dfrac{13}{9}} \end{pmatrix} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline |\vec{PQ}| &=& \sqrt{ \mathbf{ \left( \dfrac{38}{9} \right)^2} +\mathbf{\left(-\dfrac{32}{9}\right)^2 } +\mathbf{\left( \dfrac{13}{9}\right)^2 } } \\ &=& \sqrt{ \dfrac{38^2}{9^2} + \dfrac{32^2}{9^2} + \dfrac{13^2}{9^2} } \\\\ &=& \sqrt{ \dfrac{1}{9}\left( \dfrac{38^2}{9} + \dfrac{32^2}{9} + \dfrac{13^2}{9} \right) } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{38^2}{9} + \dfrac{32^2}{9} + \dfrac{13^2}{9} } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{1}{9}\left( 38^2+32^2+13^2 \right) } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{2637}{9} } \\\\ &=& \dfrac{1}{3}\sqrt{293} \\\\ &=& \mathbf{ \dfrac{\sqrt{293}}{3} } \quad | \quad \dfrac{\sqrt{d}}{3} \\ \\ \mathbf{d} &=& \mathbf{293}\\ \hline \end{array}$$

May 7, 2020