Let point \(A\) of the line \(\overline{AB}\) be \((a_1,a_2,a_3)\), and point \(B\) be \((b_1,b_2,b_3)\).
Let us find the distance from point \(C\), \((c_1,c_2,c_3)\), to this line.
In vector form, point \(A\) is
\(\overrightarrow{a}=a_1\overrightarrow{i}+a_2\overrightarrow{j}+a_3\overrightarrow{k}\),
point \(B\) is
\(\overrightarrow{b}=b_1\overrightarrow{i}+b_2\overrightarrow{j}+b_3\overrightarrow{k}\),
and point \(C\) is
\(\overrightarrow{c}=c_1\overrightarrow{i}+c_2\overrightarrow{j}+c_3\overrightarrow{k}\).
Method 1: Finding the Distance Using the Cross Product
Let us take the cross product of the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
The determinant of the matrix is calculated. This determinant is a vector.
The norm of the vector obtained from the cross product is then found:
\(\left \|\overrightarrow{AC}x\overrightarrow{AB} \right \|\).
The norm of the line segment \(\overline{AB}\) is \(\left \|AB \right \|\).
This value gives the distance from point \(C\) to the line \(\overline{AB}\).
Example:
Let point \(A\) of the line \(\overline{AB}\) be \((1,3,-1)\), and point \(B\) be \((3,6,0)\).
Let us find the distance from point \(C\), \((-2,4,-3)\), to this line.
In vector form, point \(A\) is
\(\overrightarrow{a}=\overrightarrow{i}+3\overrightarrow{j}-\overrightarrow{k}\),
point \(B\) is \(\overrightarrow{b}=3\overrightarrow{i}+6\overrightarrow{j}\),
and point \(C\) is
\(\overrightarrow{c}=-2\overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k}\).
Let us take the cross product of the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
This gives:
Method 2: Using an Arbitrary Point
Any point \(D\) on the line \(\overline{AB}\) can be written using the parameter \(t\) as follows.
The value of \(t\) is found by setting the dot product
\(\overrightarrow{CD}\cdot \overrightarrow{AB}\) equal to zero.
From this, point \(D\) is obtained. Then the length \(CD\) is calculated.
Example:
Substitute the calculated value of \(t\) into the equation.