If L is the line with direction ratios < 3, -2, 6 > and passing through (1, -1, 1), then what are the coordinates of the points on L whose distance from (1, -1, 1) is 2 units?
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
| Concept | Description |
|---|---|
| Direction Ratios | Numbers proportional to the direction cosines of a line. Indicate the direction of the line. |
| Parametric Equation of a Line | Represents the coordinates of any point on a line in terms of a single parameter (e.g., \(t\) ). |
| Distance Formula in 3D | \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\) used to calculate the distance between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) . |
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.
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