+0  
 
+1
135
1
avatar

Find the distance from the point \((1,2,3)\) to the line described by \(\begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}.\)

 Sep 5, 2019

Best Answer 

 #1
avatar+23893 
+3

Find the distance from the point \((1,2,3)\) to the line described by \(\begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}\).

\(\text{Let point $\vec{p} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} $} \\ \text{Let line $\vec{x} = \begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} $} \\ \text{Let direction vector $\vec{r} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} $} \)

 

\(\begin{array}{|rcll|} \hline \left( \vec{x}-\vec{p} \right) \cdot \vec{r} &=& 0 \quad | \quad (\vec{x}-\vec{p}) \perp \vec{r} \\\\ \Bigg(\begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}-\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \Bigg(\begin{pmatrix} 6-1 \\ 7-2 \\ 7-3 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \Bigg(\begin{pmatrix} 5 \\ 5 \\ 4 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \begin{pmatrix} 5+3t \\ 5+2t \\ 4-2t \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ 3(5+3t)+2(5+2t)-2(4-2t)&=& 0 \\ 15+9t+10+4t-8+4t &=& 0 \\ 17+17t &=& 0 \\ 17t &=& -17 \quad | \quad : 17 \\ \mathbf{ t } &=& \mathbf{ -1 } \\ \hline \end{array}\)

 

\(\text{foot (of a perpendicular):}\)

\(\begin{array}{|rcll|} \hline \vec{x}_{f} &=& \begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + (-1) \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \quad | \quad t=-1 \\\\ &=& \begin{pmatrix} 6-3 \\ 7-2 \\ 7+2 \end{pmatrix} \\\\ \mathbf{ \vec{x}_{f} } &=& \begin{pmatrix} \mathbf{3} \\ \mathbf{5} \\ \mathbf{9} \end{pmatrix} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \vec{PL} &=& \vec{x}_{f} - \vec{p} \\\\ &=& \begin{pmatrix} 3\\ 5 \\ 9\end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \\\\ &=& \begin{pmatrix} 3-1\\ 5-2 \\ 9-3\end{pmatrix} \\\\ \mathbf{\vec{PL}} &=& \begin{pmatrix} \mathbf{2}\\ \mathbf{3} \\ \mathbf{6}\end{pmatrix} \\ \hline \end{array}\)

 

\(\text{distance} = |\vec{PL}|\)

\(\begin{array}{|rcll|} \hline |\vec{PL}| &=& \sqrt{2^2+3^2+6^2} \\ &=& \sqrt{4+9+36} \\ &=& \sqrt{49} \\ \mathbf{|\vec{PL}|} &=& \mathbf{7} \\ \hline \end{array}\)

 

The distance from the point \((1,2,3)\) to the line is 7.

 

laugh

 Sep 6, 2019
 #1
avatar+23893 
+3
Best Answer

Find the distance from the point \((1,2,3)\) to the line described by \(\begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}\).

\(\text{Let point $\vec{p} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} $} \\ \text{Let line $\vec{x} = \begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} $} \\ \text{Let direction vector $\vec{r} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} $} \)

 

\(\begin{array}{|rcll|} \hline \left( \vec{x}-\vec{p} \right) \cdot \vec{r} &=& 0 \quad | \quad (\vec{x}-\vec{p}) \perp \vec{r} \\\\ \Bigg(\begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}-\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \Bigg(\begin{pmatrix} 6-1 \\ 7-2 \\ 7-3 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \Bigg(\begin{pmatrix} 5 \\ 5 \\ 4 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \begin{pmatrix} 5+3t \\ 5+2t \\ 4-2t \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ 3(5+3t)+2(5+2t)-2(4-2t)&=& 0 \\ 15+9t+10+4t-8+4t &=& 0 \\ 17+17t &=& 0 \\ 17t &=& -17 \quad | \quad : 17 \\ \mathbf{ t } &=& \mathbf{ -1 } \\ \hline \end{array}\)

 

\(\text{foot (of a perpendicular):}\)

\(\begin{array}{|rcll|} \hline \vec{x}_{f} &=& \begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + (-1) \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \quad | \quad t=-1 \\\\ &=& \begin{pmatrix} 6-3 \\ 7-2 \\ 7+2 \end{pmatrix} \\\\ \mathbf{ \vec{x}_{f} } &=& \begin{pmatrix} \mathbf{3} \\ \mathbf{5} \\ \mathbf{9} \end{pmatrix} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \vec{PL} &=& \vec{x}_{f} - \vec{p} \\\\ &=& \begin{pmatrix} 3\\ 5 \\ 9\end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \\\\ &=& \begin{pmatrix} 3-1\\ 5-2 \\ 9-3\end{pmatrix} \\\\ \mathbf{\vec{PL}} &=& \begin{pmatrix} \mathbf{2}\\ \mathbf{3} \\ \mathbf{6}\end{pmatrix} \\ \hline \end{array}\)

 

\(\text{distance} = |\vec{PL}|\)

\(\begin{array}{|rcll|} \hline |\vec{PL}| &=& \sqrt{2^2+3^2+6^2} \\ &=& \sqrt{4+9+36} \\ &=& \sqrt{49} \\ \mathbf{|\vec{PL}|} &=& \mathbf{7} \\ \hline \end{array}\)

 

The distance from the point \((1,2,3)\) to the line is 7.

 

laugh

heureka Sep 6, 2019

7 Online Users