Introduction
3D Geometry is a crucial topic in the JEE Main Mathematics syllabus. It involves understanding and visualizing objects in three dimensions, which is essential for solving problems related to lines, planes, and their intersections. This study note will break down the complexities of 3D Geometry into manageable sections, ensuring a clear understanding of each concept.
Coordinate System in 3D
The 3D coordinate system extends the 2D Cartesian system by adding a third axis, the z-axis, perpendicular to both the x-axis and y-axis.
Points in 3D Space
A point in 3D space is represented as $(x, y, z)$, where $x$, $y$, and $z$ are the coordinates along the x-axis, y-axis, and z-axis, respectively.
ExampleConsider the point $A(2, -3, 5)$. This means:
- 2 units along the x-axis
- -3 units along the y-axis
- 5 units along the z-axis
Distance Formula
The distance between two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is given by:
$$ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} $$
ExampleFind the distance between the points $A(1, 2, 3)$ and $B(4, 6, 8)$.
$$ AB = \sqrt{(4-1)^2 + (6-2)^2 + (8-3)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50} = 5\sqrt{2} $$
Section Formula
The coordinates of a point dividing the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio $m:n$ are:
$$ \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right) $$
NoteThis formula is useful for finding midpoints when $m = n$.
Direction Ratios and Direction Cosines
Direction Ratios
Direction ratios (DRs) of a line are proportional to the differences in the coordinates of any two points on the line.
Direction Cosines
Direction cosines (DCs) are the cosines of the angles that a line makes with the coordinate axes. If $\alpha$, $\beta$, and $\gamma$ are the angles with the x-axis, y-axis, and z-axis, respectively, then the direction cosines are $\cos \alpha$, $\cos \beta$, and $\cos \gamma$.
The relationship between direction cosines is:
$$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $$
ExampleIf a line makes angles $45^\circ$, $60^\circ$, and $\theta$ with the x, y, and z axes respectively, find $\theta$.
$$ \cos^2 45^\circ + \cos^2 60^\circ + \cos^2 \theta = 1 \ \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1 \ \frac{1}{2} + \frac{1}{4} + \cos^2 \theta = 1 \ \cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4} \ \cos \theta = \frac{1}{2} \implies \theta = 60^\circ $$
Equations of Lines in 3D
There are various forms to represent the equation of a line in 3D space.
Vector Form
If $\mathbf{a}$ is a position vector of a point on the line and $\mathbf{b}$ is a direction vector, then the vector equation of the line is:
$$ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} $$
Cartesian Form
If a line passes through a point $(x_1, y_1, z_1)$ and has direction ratios $a, b, c$, its Cartesian equation is:
$$ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} $$
Common MistakeA common mistake is to confuse direction ratios and direction cosines. Remember, direction ratios are not necessarily unit vectors, while direction cosines are.
Equations of Planes in 3D
General Form
The general equation of a plane is:
$$ ax + by + cz + d = 0 $$
Normal Form
If the normal to the plane makes angles $\alpha$, $\beta$, and $\gamma$ with the x, y, and z axes, and the perpendicular distance from the origin to the plane is $p$, then the equation is:
$$ x \cos \alpha + y \cos \beta + z \cos \gamma = p $$
Intercept Form
If a plane intercepts the x, y, and z axes at $a$, $b$, and $c$ respectively, its equation is:
$$ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 $$
TipFor quick identification of intercepts, set two variables to zero and solve for the third.
Angle Between Two Lines
If two lines have direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$, the cosine of the angle $\theta$ between them is given by:
$$ \cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 $$
Angle Between a Line and a Plane
If a line with direction ratios $a, b, c$ makes an angle $\theta$ with a plane with normal vector $\mathbf{n} = (A, B, C)$, then:
$$ \sin \theta = \frac{|aA + bB + cC|}{\sqrt{A^2 + B^2 + C^2} \sqrt{a^2 + b^2 + c^2}} $$
Distance of a Point from a Plane
The perpendicular distance of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is:
$$ \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} $$
ExampleFind the distance of the point $(1, 2, 3)$ from the plane $2x + 3y + 6z + 4 = 0$.
$$ \text{Distance} = \frac{|2(1) + 3(2) + 6(3) + 4|}{\sqrt{2^2 + 3^2 + 6^2}} = \frac{|2 + 6 + 18 + 4|}{\sqrt{4 + 9 + 36}} = \frac{30}{7} = \frac{30}{7} $$
Conclusion
Understanding 3D Geometry is essential for solving complex problems in JEE Main Mathematics. By mastering the concepts of points, lines, and planes in three dimensions, students can approach these problems with confidence and precision. Practice regularly and make use of these study notes to reinforce your understanding.