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Let's do the following example to analytically find the distance of a point to a line.
Let's find the distance of point C to line AB.
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Let \(A\) be the point \((a_1,a_2,a_3)\) and \(B\) be \((b_1,b_2,b_3)\) of the line \(\overline{AB}\). Let's find the distance of point \(C\) \((c_1,c_2,c_3)\) to this line.
Vector representation of points; \(A\) point is \(\overrightarrow{a}=a_1\overrightarrow{i}+a_2\overrightarrow{j}+a_3\overrightarrow{k}\),
\(B\) point is \(\overrightarrow{b}=b_1\overrightarrow{i}+b_2\overrightarrow{j}+b_3\overrightarrow{k}\) and \(C\) point is \(\overrightarrow{c}=c_1\overrightarrow{i}+c_2\overrightarrow{j}+c_3\overrightarrow{k}\)
Method 1: Finding Distance by Vector Multiplication
\(\overrightarrow{AC}=\displaystyle \overrightarrow{c}-\overrightarrow{a}=(c_1-a_1)\overrightarrow{i}+(c_2-a_2)\overrightarrow{j}+(c_3-a_3)\overrightarrow{k}\)
\(\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=(b_1-a_1)\overrightarrow{i}+(b_2-a_2)\overrightarrow{j}+(b_3-a_3)\overrightarrow{k}\)
Let's multiply the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) vectorwise.
Determinant of matrix \(\overrightarrow{AC}x\overrightarrow{AB}=\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k} \\
(c_1-a_1) & (c_2-a_2) &(c_3-a_3) \\
(b_1-a_1) &(b_2-a_2) &(b_3-a_3)
\end{vmatrix}\)
is found. This determinant is a vector.
The norm of the vector found as a result of vector multiplication is found. \(\left \|\overrightarrow{AC}x\overrightarrow{AB} \right \|\) .
The norm of the line \(\overline{AB}\) is \(\left \|AB \right \|\).
The
\(\displaystyle \left \|CD \right \|=\displaystyle \frac{\displaystyle \left \|\overrightarrow{AC}x\overrightarrow{AB} \right \|}{\displaystyle \left \| AB \right \|}\) value gives the distance of point \(C\) to line \(\overline{AB}\).
Example:
Let \(A\) be the point \((1,3,-1)\) and \(B\) be \((3,6,0)\) of the line \(\overline{AB}\). Let's find the distance of point \(C\) \((-2,4,-3)\) to this line.
Vector representation of points; \(A\) point is \(\overrightarrow{a}=\overrightarrow{i}+3\overrightarrow{j}-\overrightarrow{k}\),
\(B\) point is \(\overrightarrow{b}=3\overrightarrow{i}+6\overrightarrow{j}\) and
\(C\) point is \(\overrightarrow{c}=-2\overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k}\)
\(\overrightarrow{AC}=\displaystyle \overrightarrow{c}-\overrightarrow{a}=-3\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}\)
\(\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=2\overrightarrow{i}+3\overrightarrow{j}+\overrightarrow{k}\)
Let's multiply the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) vectorwise
\(\overrightarrow{AC}x\overrightarrow{AB}=\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k} \\
-3 & 1 &-2 \\
2 & 3 & 1
\end{vmatrix}=7\overrightarrow{i}-\overrightarrow{j}-11\overrightarrow{k}\) is found.
\(\left \|\overrightarrow{AC}x\overrightarrow{AB} \right \|=\sqrt{171}\)
\(\left \|AB \right \|=\sqrt{14}\)
\(\overline{CD}\) distance;
\(\left \| CD \right \|=\displaystyle\frac{\sqrt{171}}{\sqrt{14}}= \displaystyle \sqrt{\frac{171}{14}}\)
Method 2: Using any point
We can write any point \(D\) on the line \(\overline{AB}\) with the parameter \(t\) as follows.
\(\overrightarrow{OD}=\overrightarrow{a}+t\cdot\overrightarrow{AB}\)
\(\overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}\)
he t value that gives the zero value of the scaler product \(\overrightarrow{CD}\cdot \overrightarrow{AB}\) is found. Point D is found from here.
The OD length is calculated.
Example:
\(\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=2\overrightarrow{i}+3\overrightarrow{j}+\overrightarrow{k}\)
\(\overrightarrow{OD}=\overrightarrow{a}+t\cdot\overrightarrow{AB}=\overrightarrow{i}+3\overrightarrow{j}-\overrightarrow{k}+t(2\overrightarrow{i}+3\overrightarrow{j}+\overrightarrow{k})\)
\(\overrightarrow{OD}=(2t+1)\overrightarrow{i}+(3t+3)\overrightarrow{j}+(t-1)\overrightarrow{k}\)
\(\overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}\)
\(\overrightarrow{CD}=(2t+3)\overrightarrow{i}+(3t-1)\overrightarrow{j}+(t+2)\overrightarrow{k}\)
\(\overrightarrow{CD}\cdot \overrightarrow{AB}=0 \)
\( \therefore 2(2t+3)+3(3t-1)+1(t+2)=0\)
\( \therefore t=-\displaystyle\frac{5}{14}\)
Let's substitute the \(t\) value in equation \(\left | CD \right |=\displaystyle\sqrt{\left(2t+3\right)^2+\left(3t-1\right)^2+\left(t+2\right)^2}\).
\(=\displaystyle\sqrt{\left[2\left (-\displaystyle\frac{5}{14}\right)+3\right]^2+\left[3\left (-\displaystyle\frac{5}{14}\right)-1\right]^2+\left[\left (-\displaystyle\frac{5}{14}\right)+2\right]^2}\)
\(=\displaystyle\sqrt{\displaystyle\frac{171}{14}}\)
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