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3D Geometry JEE Notes: Important Concepts
Coordinate system: We work in a right-handed Cartesian system with origin $$O(0,0,0)$$ and mutually perpendicular $$X, Y, Z$$ axes.
- Point $$P(x,y,z)$$: ordered triple of real numbers.
- Vector $$\vec r = x\hat i + y\hat j + z\hat k$$ corresponds one-to-one with point $$P$$.
- Distance between points $$P(x_1,y_1,z_1)$$ and $$Q(x_2,y_2,z_2):$$
$$PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$ - Direction cosines (d.c.’s) of a line are $$\ell = \cos\alpha, m = \cos\beta, n = \cos\gamma$$ where $$\alpha,\beta,\gamma$$ are the angles the line makes with $$X,Y,Z$$ axes. They satisfy $$\ell^2 + m^2 + n^2 = 1$$.
- Direction ratios (d.r.’s) $$a,b,c$$ are any three numbers proportional to $$\ell,m,n$$. When they are integers JEE often calls them “the vector” along the line.
Straight Lines in 3-D: Forms, Angles and Shortest Distance
Standard equations of a line
| Form | Equation | When to prefer |
|---|---|---|
| Vector form | $$\vec r = \vec a + \lambda \vec b$$ | Dot or cross product problems |
| Symmetric form | $$\dfrac{x-x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c}$$ | Quick comparison of two lines |
| Two-point form | $$\dfrac{x-x_1}{x_2-x_1} = \dfrac{y-y_1}{y_2-y_1} = \dfrac{z-z_1}{z_2-z_1}$$ | Line through $$P_1,P_2$$ directly |
Angle between two lines
If $$\vec b_1 = (a_1,b_1,c_1)$$ and $$\vec b_2 = (a_2,b_2,c_2)$$ are direction ratios,
$$\cos\theta = \dfrac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}$$
Parallel and perpendicular criteria
- Parallel: cross product $$\vec b_1 \times \vec b_2 = \vec 0$$ or proportional ratios.
- Perpendicular: dot product $$\vec b_1 \cdot \vec b_2 = 0$$.
Skew lines and shortest distance (S.D.)
Two lines are skew if they are neither parallel nor intersecting. The shortest distance between skew lines $$L_1$$ and $$L_2$$ is
$$\text{S.D.} = \dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1 \times \vec b_2)|}{|\vec b_1 \times \vec b_2|}$$
where $$\vec a_1,\vec a_2$$ are any position vectors on the respective lines.
Solved Example 1
Find the symmetric equation of the line through $$A(1,2,3)$$ with direction ratios $$2,-1,4$$. Also obtain its direction cosines.
Symmetric form:
$$\dfrac{x-1}{2} = \dfrac{y-2}{-1} = \dfrac{z-3}{4}$$
Direction cosines:
Magnitude $$=\sqrt{2^2+(-1)^2+4^2}= \sqrt{4+1+16}= \sqrt{21}$$.
$$\ell = \dfrac{2}{\sqrt{21}},\; m = \dfrac{-1}{\sqrt{21}},\; n = \dfrac{4}{\sqrt{21}}$$.
Check $$\ell^2+m^2+n^2 = 1$$ verified.
Practise similar construction questions in the extensive JEE Questions repository to build speed.
Planes in 3-D: Forms, Angles, Distance and Intersection
Standard equations of a plane
| Form | Equation | Quick use |
|---|---|---|
| General form | $$Ax+By+Cz+D=0$$ | Most NCERT exercise problems |
| Normal (vector) form | $$(\vec r-\vec a)\cdot \hat n = 0$$ | When unit normal is given |
| Intercept form | $$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$$ | Intercept questions, image of a point |
| Three-point form | Determinant $$\begin{vmatrix}x&y;&z;&1\\x_1&y;_1&z;_1&1\\x_2&y;_2&z;_2&1\\x_3&y;_3&z;_3&1\end{vmatrix}=0$$ | Coordinates of all three vertices known |
Angle between two planes
$$\cos\theta = \dfrac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\,\sqrt{A_2^2+B_2^2+C_2^2}}$$ where $$\hat n_1=(A_1,B_1,C_1)$$ and $$\hat n_2=(A_2,B_2,C_2)$$ are normals.
Line of intersection of two planes
Solve the two plane equations simultaneously: treat $$z$$ as parameter or use cross-product method. JEE favours the determinant approach with parameter $$\lambda$$.
Distance of a point from a plane
$$d = \dfrac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2+B^2+C^2}}$$
Solved Example 2
Find the distance of point $$P(2,-1,4)$$ from the plane $$2x-3y+6z-12=0$$.
$$d = \dfrac{| 2(2) -3(-1) +6(4) -12|}{\sqrt{2^2+(-3)^2+6^2}} = \dfrac{|4+3+24-12|}{\sqrt{4+9+36}} = \dfrac{19}{7}$$
Distance = $$\dfrac{19}{7}$$ units.
Shortest Distance, Skew Lines and Line–Plane Interaction
Condition of perpendicularity between line and plane
A line is perpendicular to a plane if its direction ratios are proportional to the normal of the plane.
Angle between a line and a plane
If $$\theta$$ is the acute angle between line $$\vec b(a,b,c)$$ and plane $$Ax+By+Cz+D=0$$, then
$$\sin\theta = \dfrac{|Aa+Bb+Cc|}{\sqrt{A^2+B^2+C^2}\,\sqrt{a^2+b^2+c^2}}$$.
Solved Example 3
The lines $$\dfrac{x+1}{2}=\dfrac{y-3}{1}=\dfrac{z}{-2} \;(L_1)$$ and $$\dfrac{x-4}{-3}=\dfrac{y+2}{2}=\dfrac{z+1}{1}\;(L_2)$$ are skew. Find their shortest distance.
Take $$\vec a_1 = (-1,3,0), \vec a_2 = (4,-2,-1)$$ and direction vectors $$\vec b_1 = (2,1,-2), \vec b_2 = (-3,2,1)$$.
$$\vec b_1 \times \vec b_2 = \begin{vmatrix}\hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ -3 & 2 & 1\end{vmatrix} = (1(1)-(-2)(2))\hat i - (2(1)-(-2)(-3))\hat j + (2(2)-1(-3))\hat k = (1+4, -(2-6), (4+3)) = (5,4,7)$$.
Magnitude $$|\vec b_1 \times \vec b_2| = \sqrt{5^2+4^2+7^2}= \sqrt{25+16+49}= \sqrt{90}=3\sqrt{10}$$.
Vector between points: $$\vec a_2-\vec a_1 = (5,-5,-1)$$.
Numerator $$|(\vec a_2-\vec a_1)\cdot(\vec b_1 \times \vec b_2)| = |5(5)+(-5)(4)+(-1)(7)| = |25-20-7| = |-2| = 2$$.
S.D. $$=\dfrac{2}{3\sqrt{10}}$$ units.
Sphere and Other Surfaces: Basics Needed for JEE
Sphere
- Standard form: $$x^2+y^2+z^2+2ux+2vy+2wz+d = 0$$.
- Centre: $$(-u,-v,-w)$$, Radius: $$\sqrt{u^2+v^2+w^2-d}$$.
- Diameter form: Endpoints $$A(x_1,y_1,z_1), B(x_2,y_2,z_2)$$ – substitute the midpoint and use $$\vec{PA}\cdot\vec{PB}=0$$.
- Condition for orthogonality of two spheres: $$2(u_1u_2+v_1v_2+w_1w_2) = d_1+d_2$$.
Right circular cylinder (quick recap)
Axis along $$Z$$-axis: $$(x-h)^2 + (y-k)^2 = R^2$$, no $$z$$ term.
Right circular cone (vertex at origin)
$$x^2 + y^2 + z^2 = (x\cos\alpha + y\cos\beta + z\cos\gamma)^2$$ where $$\alpha,\beta,\gamma$$ fix the axis direction.
Solved Example 4
The sphere $$x^2+y^2+z^2-4x+6y-8z+5=0$$ is cut by the plane $$x-2y+2z=3$$. Find the radius of the circle of intersection.
Centre $$C(2,-3,4)$$, radius $$R=\sqrt{2^2+(-3)^2+4^2-5}= \sqrt{4+9+16-5}= \sqrt{24}=2\sqrt6$$.
Distance of centre from plane:
$$d = \dfrac{|2-2(-3)+2(4)-3|}{\sqrt{1^2+(-2)^2+2^2}} = \dfrac{|2+6+8-3|}{3} = \dfrac{13}{3}$$.
Radius of section $$r = \sqrt{R^2-d^2}= \sqrt{24-\dfrac{169}{9}}= \sqrt{\dfrac{216-169}{9}}= \sqrt{\dfrac{47}{9}} = \dfrac{\sqrt{47}}{3}$$.
The required radius is $$\dfrac{\sqrt{47}}{3}$$.
Important Formulas and Results at a Glance
| Concept | Formula | Remember tip |
|---|---|---|
| Distance between two points | $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$ | Pythagoras in 3D |
| Direction cosine relation | $$\ell^2+m^2+n^2=1$$ | Same as unit vector |
| Angle between lines | $$\cos\theta = \dfrac{\sum a_ib_i}{\sqrt{\sum a_i^2}\sqrt{\sum b_i^2}}$$ | Dot product |
| Point-plane distance | $$\dfrac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$$ | Absolute value counts |
| Shortest distance | $$\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1 \times \vec b_2)|}{|\vec b_1 \times \vec b_2|}$$ | Volume over area |
| Sphere radius | $$\sqrt{u^2+v^2+w^2-d}$$ | Compare with $$x^2+y^2+z^2+2ux+2vy+2wz+d=0$$ |
| Angle line–plane | $$\sin\theta = \dfrac{|Aa+Bb+Cc|}{\sqrt{A^2+B^2+C^2}\sqrt{a^2+b^2+c^2}}$$ | Dot of direction and normal |
After revising the table, download the full maths JEE Formula Sheets so you do not miss any result on exam day.
JEE Important Points, Common Mistakes and Quick Revision
- Most 3D questions in JEE Main are single-formula applications. Keep your calculations organised to avoid sign errors.
- Confuse d.c.’s with d.r.’s? Normalise the ratios right at the start.
- Skew lines must fail both intersection and parallel tests. Checking only one loses marks.
- Write the plane normal clearly. Students often copy wrong coefficients leading to wrong angle or distance.
- For sphere problems, confirm that $$u^2+v^2+w^2-d$$ is positive before square rooting. A negative value means the data are inconsistent.
- Once you finish the chapter, solve ten past exam problems back-to-back from the JEE Mains Previous Papers PDF to gauge your readiness.
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