3-Dimensional Geometry: Important Questions and Solutions
Here are some important questions and solutions from 3-dimensional geometry that frequently appear in exams.
1. Equation of a Line Passing Through Two Points
Question: Find the equation of a line which passes through the points \( (3,6,-7) \) and \( (5,-1,4) \).
Solution:
The equations of line in Cartesian form are:
\[ \frac{x – x_1}{x_2 – x_1} = \frac{y – y_1}{y_2 – y_1} = \frac{z – z_1}{z_2 – z_1} \]
Substituting the given points \( (3,6,-7) \) and \( (5,-1,4) \):
\[ \frac{x – 3}{5 – 3} = \frac{y – 6}{-1 – 6} = \frac{z + 7}{4 + 7} \]
\[ \frac{x – 3}{2} = \frac{y – 6}{-7} = \frac{z + 7}{11} \quad \text{(Cartesian form)} \]
In vector form:
\[ \vec{r} = 3\hat{i} + 6\hat{j} – 7\hat{k} + \mu(2\hat{i} – 7\hat{j} + 11\hat{k}) \]
2. Direction Cosines of a Line
Question: Find the direction cosines of a line passing through points \( (7,-1,8) \) and \( (3,4,-7) \).
Solution:
The direction ratios of line AB are:
\[ \langle x_2 – x_1, y_2 – y_1, z_2 – z_1 \rangle = \langle (3 – 7), (4 + 1), (-7 – 8) \rangle = \langle -4, 5, -9 \rangle \]
Direction cosines of line AB are:
\[ \left\langle \frac{-4}{\sqrt{16 + 25 + 81}}, \frac{5}{\sqrt{16 + 25 + 81}}, \frac{-9}{\sqrt{16 + 25 + 81}} \right\rangle = \left\langle \frac{-4}{\sqrt{122}}, \frac{5}{\sqrt{122}}, \frac{-9}{\sqrt{122}} \right\rangle \]
3. Direction Cosines of a Line Making Equal Angles with Axes
Question: Find the direction ratios and direction cosines of a line which makes equal angles with the coordinate axes.
Solution:
Let α be the angle that the line makes with each coordinate axis. Then:
\[ \cos^2 α + \cos^2 α + \cos^2 α = 1 \\ 3 \cos^2 α = 1 \\ \cos α = \frac{1}{\sqrt{3}} \]
Therefore, the direction cosines are:
\[ \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle \]
And the direction ratios are:
\[ \langle 1, 1, 1 \rangle \]
4. Angle Between Two Lines
Question: Find the angle between the lines:
\[ \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{1} \quad \text{and} \quad \frac{x-2}{2} = \frac{y+1}{1} = \frac{z+1}{2} \]
Solution:
The direction ratios of the lines are \( \langle 1, -1, 1 \rangle \) and \( \langle 2, 1, 2 \rangle \) respectively.
Let θ be the angle between the two lines:
\[ \cos θ = \frac{(1)(2) + (-1)(1) + (1)(2)}{\sqrt{(1)^2 + (-1)^2 + (1)^2} \sqrt{(2)^2 + (1)^2 + (2)^2}} = \frac{2 – 1 + 2}{\sqrt{3} \sqrt{9}} = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} \]
Therefore:
\[ θ = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \]
5. Finding the Value of m for Perpendicular Lines
Question: Find the value of m, if the lines
\[ \frac{x+2}{3} = \frac{y-1}{2m} = \frac{z-2}{7} \quad \text{and} \quad \frac{x-3}{4} = \frac{y-2}{7} = \frac{z+5}{8m} \]
are perpendicular to each other.
Solution:
The direction ratios of the lines are \( \langle 3, 2m, 7 \rangle \) and \( \langle 4, 7, 8m \rangle \) respectively.
Since the lines are perpendicular:
\[ (3)(4) + (2m)(7) + (7)(8m) = 0 \\ 12 + 14m + 56m = 0 \\ 70m = -12 \\ m = \frac{-12}{70} = \frac{-6}{35} \]