Finding Points on a Line at a Specific Distance

We are given a line L that passes through the point A with coordinates (1, -1, 1) and has direction ratios <3, -2, 6>. We need to find the coordinates of the points on this line L whose distance from point A is 2 units.

The equation of a line passing through a point &\((x_0, y_0, z_0)\)& with direction ratios &\((a, b, c)\)& can be written in parametric form. A point &\((x, y, z)\)& on the line can be represented as:

• \(x = x_0 + at\) 
•  \(y = y_0 + bt\) 
•  \(z = z_0 + ct\) 

Here, &\((x_0, y_0, z_0) = (1, -1, 1)\)& and &\((a, b, c) = (3, -2, 6)\)&. So, a point &\((x, y, z)\)& on line L is given by:

•  \(x = 1 + 3t\) 
•  \(y = -1 - 2t\) 
•  \(z = 1 + 6t\) 

for some parameter  \(t\) .

The distance between the point A &\((x_0, y_0, z_0)\)& and any point &\((x, y, z)\)& on the line is given by:

 \(\text{Distance} = \sqrt{(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2}\) 

Substituting the parametric equations, we get:

 \(\text{Distance} = \sqrt{((1 + 3t) - 1)^2 + ((-1 - 2t) - (-1))^2 + ((1 + 6t) - 1)^2}\) 

 \(\text{Distance} = \sqrt{(3t)^2 + (-2t)^2 + (6t)^2}\) 

 \(\text{Distance} = \sqrt{9t^2 + 4t^2 + 36t^2}\) 

 \(\text{Distance} = \sqrt{49t^2}\) 

 \(\text{Distance} = \sqrt{49} \sqrt{t^2} = 7|t|\) 

We are given that the distance from (1, -1, 1) is 2 units. So,

 \(7|t| = 2\) 

 \(|t| = \frac{2}{7}\) 

This gives two possible values for  \(t\) :

•  \(t = \frac{2}{7}\) 
•  \(t = -\frac{2}{7}\) 

Now, we substitute these values of  \(t\)  back into the parametric equations of the line to find the coordinates of the points.

Case 1:  \(t = \frac{2}{7}\) 

•  \(x = 1 + 3 \left(\frac{2}{7}\right) = 1 + \frac{6}{7} = \frac{7+6}{7} = \frac{13}{7}\) 
•  \(y = -1 - 2 \left(\frac{2}{7}\right) = -1 - \frac{4}{7} = \frac{-7-4}{7} = -\frac{11}{7}\) 
•  \(z = 1 + 6 \left(\frac{2}{7}\right) = 1 + \frac{12}{7} = \frac{7+12}{7} = \frac{19}{7}\) 

The first point is  \(\left(\frac{13}{7}, -\frac{11}{7}, \frac{19}{7}\right)\) .

Case 2:  \(t = -\frac{2}{7}\) 

•  \(x = 1 + 3 \left(-\frac{2}{7}\right) = 1 - \frac{6}{7} = \frac{7-6}{7} = \frac{1}{7}\) 
•  \(y = -1 - 2 \left(-\frac{2}{7}\right) = -1 + \frac{4}{7} = \frac{-7+4}{7} = -\frac{3}{7}\) 
•  \(z = 1 + 6 \left(-\frac{2}{7}\right) = 1 - \frac{12}{7} = \frac{7-12}{7} = -\frac{5}{7}\) 

The second point is  \(\left(\frac{1}{7}, -\frac{3}{7}, -\frac{5}{7}\right)\) .

Thus, the coordinates of the points on line L whose distance from (1, -1, 1) is 2 units are  \(\left(\frac{13}{7}, -\frac{11}{7}, \frac{19}{7}\right)\)  and  \(\left(\frac{1}{7}, -\frac{3}{7}, -\frac{5}{7}\right)\) .

Revision Table: Key Concepts

Additional Information: Lines in 3D Geometry

Understanding lines in three-dimensional space is fundamental in vector algebra and coordinate geometry. A line can be uniquely determined if we know either:

• A point on the line and its direction.
• Two points on the line.

The direction of a line is typically represented by direction ratios or direction cosines. Direction cosines <l, m, n> are the cosines of the angles the line makes with the positive x, y, and z axes, respectively. They satisfy the property  \(l^2 + m^2 + n^2 = 1\) . Direction ratios <a, b, c> are proportional to direction cosines, i.e.,  \(l = ka, m = kb, n = kc\)  for some constant  \(k\) . The value of  \(k\)  can be  \(\pm \frac{1}{\sqrt{a^2+b^2+c^2}}\) .

The symmetric form of the equation of a line passing through  \((x_0, y_0, z_0)\)  with direction ratios  \((a, b, c)\)  is:

 \(\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = t\) 

where  \(t\)  is the parameter. This is equivalent to the parametric form used in the solution.

The distance  \(r\)  of a point  \((x, y, z)\)  on the line from  \((x_0, y_0, z_0)\)  is related to the parameter  \(t\)  and the direction cosines  \((l, m, n)\)  by the formula  \((x - x_0) = lt, (y - y_0) = mt, (z - z_0) = nt\) . The distance formula then gives  \(r = \sqrt{(lt)^2 + (mt)^2 + (nt)^2} = \sqrt{t^2(l^2 + m^2 + n^2)} = \sqrt{t^2 \cdot 1} = |t|\)  (assuming  \(t\)  is the parameter based on unit direction vector). If using direction ratios  \((a, b, c)\)  and the parameter  \(t\)  as in  \((x_0+at, y_0+bt, z_0+ct)\) , the distance is  \(|t| \sqrt{a^2 + b^2 + c^2}\) , as calculated in the solution.