Three-dimensional geometry plays a major role as a lot of questions are included in the JEE exam. Here is a list of all the three-dimensional geometry formulas which will help students to go through and revise them quickly before the exam.
Three-dimensional Geometry Formulas
1. Vector representation of a point: Position vector of a point P(x, y, z) is
\(\begin{array}{l}x\hat{i}+y\hat{j}+z\hat{k}\end{array} \)
2. Distance formula:
Distance between two points P(x1, y1, z1) and Q(x2, y2, z2) is
\(\begin{array}{l}PQ = \sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}\end{array} \)
\(\begin{array}{l}AB = \left | \vec{OB} -\vec{OA}\right |\end{array} \)
3. Distance of P from coordinate axes:
\(\begin{array}{l}PA = \sqrt{(y^{2}+z^{2})}\end{array} \)
\(\begin{array}{l}PB = \sqrt{(z^{2}+x^{2})}\end{array} \)
\(\begin{array}{l}PC = \sqrt{(x^{2}+y^{2})}\end{array} \)
4. Section Formula:
\(\begin{array}{l}x = \frac{mx_{2}+nx_{1}}{m+n}\end{array} \)
\(\begin{array}{l}y = \frac{my_{2}+ny_{1}}{m+n}\end{array} \)
\(\begin{array}{l}z = \frac{mz_{2}+nz_{1}}{m+n}\end{array} \)
Midpoint:
\(\begin{array}{l}x = \frac{x_{1}+x_{2}}{2}\end{array} \)
\(\begin{array}{l}y = \frac{y_{1}+y_{2}}{2}\end{array} \)
\(\begin{array}{l}z = \frac{z_{1}+z_{2}}{2}\end{array} \)
5. Centroid of a triangle:
\(\begin{array}{l}G =\left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3},\frac{z_{1}+z_{2}+z_{3}}{3} \right )\end{array} \)
6. Incentre of triangle ABC:
\(\begin{array}{l}\left ( \frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\frac{ay_{1}+by_{2}+cy_{3}}{a+b+c},\frac{az_{1}+bz_{2}+cz_{3}}{a+b+c} \right )\end{array} \)
7. Centroid of a tetrahedron:
\(\begin{array}{l}\left ( \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4},\frac{z_{1}+z_{2}+z_{3}+z_{4}}{4} \right )\end{array} \)
8. Direction cosines and direction ratios:
(i) Direction cosines: let α, β, γ be the angles which a directed line makes with the positive directions of the axes of x, y and z respectively, then cos α, cos β and cos γ are called the direction cosines of the line. The direction cosines are usually denoted by (l, m, n).
Therefore, l = cos α, m = cos β, n = cos γ.
(ii) l2+m2+n2 = 1
(iii) If a, b, c are the direction ratios of any line L then
\(\begin{array}{l}a\hat{i}+b\hat{j}+c\hat{k}\end{array} \)
will be a vector parallel to the line L.
(iv) If l, m, and n are the direction cosines of any line L, then
\(\begin{array}{l}l\hat{i}+m\hat{j}+n\hat{k}\end{array} \)
is a unit vector parallel to the line L.
(v) If l, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then
\(\begin{array}{l}l = \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
\(\begin{array}{l}m = \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
\(\begin{array}{l}n = \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
or
\(\begin{array}{l}l =\frac{-a}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
,
\(\begin{array}{l}m = \frac{-b}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
\(\begin{array}{l}n = \frac{-c}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
(vi) If OP = r, the direction cosines of OP are l, m, n then the coordinates of P are (lr, mr, nr).
If the direction cosines of the line AB are l, m, n, |AB| = r and the coordinates of A is (x1, y1, z1) then the coordinates of B are given as (x1 + rl, y1+ rm, z1 + rn).
(vii) If the coordinates P and Q are (x1, y1, z1) and (x2, y2, z2) then the direction ratios of line PQ are a = x2 – x1 , b = y2 – y1 and c = z2 – z1 and the direction cosines of line PQ are:
\(\begin{array}{l}l = \frac{x_{2}-x_{1}}{\left | PQ \right |},\ m = \frac{y_{2}-y_{1}}{\left | PQ \right |}\ n = \frac{z_{2}-z_{1}}{\left | PQ \right |}\end{array} \)
(vii) Direction cosines of the x-axis is (1, 0, 0).
Direction cosines of the y-axis is (0, 1, 0).
Direction cosines of the z-axis is (0, 0, 1).
9. Angle between two line segments:
If a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines and θ is the acute angle between them, then
\(\begin{array}{l}cos\ \theta = \left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right |\end{array} \)
The line will be perpendicular if a1a2 + b1b2 + c1c2 = 0, and parallel if a1/a2 = b1/b2 = c1/c2.
10. Projection of a line segment on a line:
If P(x1, y1, z1) and Q(x2, y2, z2) then the projection of PQ on a line having direction cosines l, m, n is |l(x2 – x1) + m(y2 – y1) + n(z2 – z1)|
11. Equation of a plane: General form: ax + by + cz + d = 0, where a, b, c are not all zero, a, b, c, d ∈ R.
(i) Normal form: lx + my + nz = p
(ii) Plane through the point (x1, y1, z1): a(x – x1) + b(y – y1) + c(z – z1) = 0
(iii) Intercept form:
\(\begin{array}{l}\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1\end{array} \)
(iv) vector form:
\(\begin{array}{l}(\vec{r}-\vec{a} ).\vec{n}= 0\end{array} \)
or \(\begin{array}{l}\vec{r}.\: \vec{n}= \vec{a}.\: \vec{n}\end{array} \)
(v) Planes parallel to the axes :
(a) plane parallel to X-axis is by + cz + d = 0
(b) plane parallel to Y-axis is ax + cz + d = 0
(c) plane parallel to Z-axis is ax + by + d = 0
(vi) Plane through origin: Equation of the plane passing through the origin is ax + by + cz = 0.
(vii) Transformation of the equation of a plane to the normal form: ax+by+cz-d = 0 in normal form is
\(\begin{array}{l}\frac{ax}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}+\frac{by}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}+\frac{cz}{\pm \sqrt{a^{2}+b^{2}+c^{2}}} = \frac{d}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
(viii) Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz + λ = 0.
Distance between ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is
\(\begin{array}{l}\frac{\left | d_{1}-d_{2} \right |}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
(ix) A plane ax + by + cz + d = 0 divides the line segment joining (x1, y1, z1) and (x2, y2, z2) in the ratio
\(\begin{array}{l}\left ( -\frac{ax_{1}+by_{1}+cz_{1}+d}{ax_{2}+by_{2}+cz_{2}+d}\right )\end{array} \)
(x) Coplanarity of four points: The points A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) are coplanar if
\(\begin{array}{l}\begin{vmatrix} x_{2}-x_{1}& y_{2}-y_{1} & z_{2}-z_{1}\\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \\ x_{4}-x_{1}& y_{4}-y_{1} & z_{4}-z_{1} \end{vmatrix}= 0\end{array} \)
12. A point and a plane:
(i) distance of the point (x’, y’, z’) from the plane ax+by+cz+d = 0 is given by
\(\begin{array}{l}\frac{ax^{‘}+by^{‘}+cz^{‘}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
(ii)
\(\begin{array}{l}\text{Length of the perpendicular from a point}\ \vec{a} \text{to the plane}\ \vec{r}.\vec{n} = d\ \text{is given by}\ p = \frac{\left |\vec{a} .\: \vec{n} -d\right |}{\left | \vec{n} \right |}\end{array} \)
.
(iii) Foot (x’, y’, z’) of perpendicular drawn from the point (x1, y1, z1) to the plane ax+by+cz+d = 0 is given by
\(\begin{array}{l}\frac{x^{‘}-x_{1}}{a} = \frac{y^{‘}-y_{1}}{b} =\frac{z^{‘}-z_{1}}{c}= -\frac{ax_{1}+by_{1}+cz_{1}+d}{a^{2}+b^{2}+c^{2}}\end{array} \)
(iv) To find image of a point with respect to a plane:
Let P (x1, y1, z1) be a given point and ax + by + cz + d = 0 is given plane. Let (x’, y’, z’) is the image point. Then,
\(\begin{array}{l}\frac{x^{‘}-x_{1}}{a} = \frac{y^{‘}-y_{1}}{b} =\frac{z^{‘}-z_{1}}{c}= -2\frac{(ax_{1}+by_{1}+cz_{1}+d)}{a^{2}+b^{2}+c^{2}}\end{array} \)
(v) The distance between two parallel planes ax + by + cz + d = 0 and ax + by + cz + d’ = 0 is given by
\(\begin{array}{l}\frac{\left | d-d{}’ \right |}{\sqrt{a^{2}+b^{2}+c^{2}}}\end{array} \)
13. Angle between two planes:
(i)
\(\begin{array}{l}\cos \theta =\left | \frac{aa{}’+bb{}’+cc{}’}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{a{}’^{2}+b{}’^{2}+c{}’^{2}}}\right |\end{array} \)
(ii) Planes are perpendicular if aa’ + bb’ + cc’ = 0 and parallel if a/a’ = b/b’ = c/c’.
(iii) The angle θ between the planes
\(\begin{array}{l}\vec{r}.\vec{n_{1}} = d_{1}\end{array} \)
and \(\begin{array}{l}\vec{r}.\vec{n_{2}} = d_{2}\end{array} \)
is given by \(\begin{array}{l}cos\ \theta = \frac{\vec{n_{1}}.\vec{n_{2}} }{\left | \vec{n_{1}} \right |.\left | \vec{n_{2}} \right |}\end{array} \)
(iv) Planes are perpendicular if
\(\begin{array}{l}\vec{n_{1}}.\vec{n_{2}} = 0\end{array} \)
and planes are parallel if \(\begin{array}{l}\vec{n_{1}}=\lambda \vec{n_{2}}\end{array} \)
, where λ is a scalar.
14. Angle bisectors:
(i) The equations of a planes bisecting the angle between two given planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are
\(\begin{array}{l}\frac{a_{1}x+b_{1}y+c_{1}z+d_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}} = \pm \frac{a_{2}x+b_{2}y+c_{2}z+d_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\end{array} \)
(ii) Bisector of acute or obtuse angle: First, make both the constant terms positive. Then,
a1a2 + b1b2 + c1c2 > 0 ⇒ origin lies on obtuse angle.
a1a2 + b1b2 + c1c2 < 0 ⇒ origin lies on acute angle.
15. Area of a triangle:
Let A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) be the vertices of a triangle, then
\(\begin{array}{l}\Delta = \sqrt{\Delta_{x}^{2}+\Delta _{y}^{2}+\Delta _{z}^{2}}\end{array} \)
where \(\begin{array}{l}\Delta _{x} = \frac{1}{2}\begin{vmatrix} y_{1} & z_{1} & 1\\ y_{2} &z_{2} & 1\\ y_{3}& z_{3} & 1 \end{vmatrix}\end{array} \)
, \(\begin{array}{l}\Delta _{y} = \frac{1}{2}\begin{vmatrix} z_{1} & x_{1} & 1\\ z_{2} &x_{2} & 1\\ z_{3}& x_{3} & 1 \end{vmatrix}\end{array} \)
and \(\begin{array}{l}\Delta _{z} = \frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2} &y_{2} & 1\\ x_{3}& y_{3} & 1 \end{vmatrix}\end{array} \)
Vector method:
\(\begin{array}{l}\text{From two vector}\ \vec{AB}\ \text{and}\ \vec{AC}\ \text{the area is given by}\ \frac{1}{2}\left |\vec{AB} \times \vec{AC} \right |\end{array} \)
16. Volume of a tetrahedron:
Volume of a tetrahedron with vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) is given by
\(\begin{array}{l}V =\frac{1}{6}\begin{vmatrix} x_{1} & y_{1} & z_{1} & 1\\ x_{2} & y_{2} & z_{2} & 1\\ x_{3}& y_{3} & z_{3} & 1\\ x_{4}& y_{4} & z_{4} & 1 \end{vmatrix}\end{array} \)
17. Equation of a line:
(i) A straight line is the intersection of two planes. It is represented by two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 .
(ii) Symmetric form:
\(\begin{array}{l}\frac{(x-x_{1})}{a} = \frac{(y-y_{1})}{b}= \frac{(z-z_{1})}{c} = r\end{array} \)
(iii) vector equation:
\(\begin{array}{l}\vec{r} = \vec{a}+\lambda \vec{b}\end{array} \)
(iv) Reduction of cartesian form of equation of a line to vector form and vice versa
\(\begin{array}{l}\frac{x-x_{1}}{a} = \frac{y-y_{1}}{b}= \frac{z-z_{1}}{c}\end{array} \)
⇔\(\begin{array}{l}\vec{r} = (x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k})+\lambda (a\hat{i}+b\hat{j}+c\hat{k})\end{array} \)
18. To find image of a point with respect to a line:
\(\begin{array}{l}\text{Let}\ L = \frac{x-x_{2}}{a} = \frac{y-y_{2}}{b}= \frac{z-z_{2}}{c}\ \text{be the given line.}\end{array} \)
Let (x’, y’, z’) be the image of the point (x1, y1, z1) with respect to the line L. Then
(i) a(x1 – x’) + b(y1 – y’) + c(z1 – z’) = 0
(ii)
\(\begin{array}{l}\frac{\frac{x_{1}+x{}’}{2}-x_{2}}{a}= \frac{\frac{y_{1}+y{}’}{2}-y_{2}}{b}= \frac{\frac{z_{1}+z{}’}{2}-z_{2}}{c} = \lambda\end{array} \)
From (ii) get the value of x’, y’, z’ in terms of λ as x’= 2aλ + 2x2 – x1, y’= 2bλ + 2y2 – y1,
z’= 2cλ+2z2-z1 . Then put the values of x’, y’, z’ in (i) to get λ and substitute the value of λ to get (x’, y’,z’).
19. Angle between a line and a plane:
(i) If θ is the angle between a line
\(\begin{array}{l}\frac{x-x_{1}}{l}= \frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}\end{array} \)
and the plane ax+by+cz+d = 0, then \(\begin{array}{l}\sin \theta = \left | \frac{al+bm+cn}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{l^{2}+m^{2}+n^{2}}} \right |\end{array} \)
(ii) Vector form : If θ is the angle between a line
\(\begin{array}{l}\vec{r}= \vec{a}+\lambda \vec{b}\end{array} \)
and \(\begin{array}{l}\vec{r}.\vec{n }= d\end{array} \)
then \(\begin{array}{l}sin\ \theta = \frac{\vec{b}.\vec{n}}{\left | \vec{b} \right |\left | \vec{n} \right |}\end{array} \)
(iii) condition for perpendicularity l/a = m/b = n/c ,
\(\begin{array}{l}\vec{b}\times \vec{n}= 0\end{array} \)
(iv) condition for parallel : al + bm + cn = o,
\(\begin{array}{l}\vec{b}.\vec{n}= 0\end{array} \)
20. Condition for a line to lie in a plane:
(i) cartesian form: Line
\(\begin{array}{l}\frac{x-x_{1}}{l}= \frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}\end{array} \)
would lie in a plane ax+by+cz+d = 0, if ax1 + by1 + cz1 + d = 0 and al + bm + cn = 0
(ii) Vector form:
\(\begin{array}{l}\text{Line}\ \vec{r}= \vec{a}+\lambda b\ \text{would lie in the plane}\ \vec{r}.\vec{n}= d\ \text{if}\ \vec{b}.\vec{n}= 0\ \text{and}\ \vec{a}.\vec{n}= d\end{array} \)
21. Skew lines:
(i) The straight lines which are not parallel and non-coplanar are called skew lines.
If
\(\begin{array}{l}\text{If}\ \Delta = \begin{vmatrix} \alpha ^{‘} -\alpha & \beta {}’-\beta &\gamma {}’-\gamma \\ l & m &n \\ l’& m’&n’ \end{vmatrix} \neq 0,\ \text{then the lines are skew.}\end{array} \)
(ii) Shortest distance:
\(\begin{array}{l}SD =\frac{\begin{vmatrix} \alpha ^{‘} -\alpha & \beta {}’-\beta &\gamma {}’-\gamma \\ l & m &n \\ l’& m’&n’ \end{vmatrix} }{\sqrt{\sum (mn’-m’n)^{2}}}\end{array} \)
(iii) Vector form:
\(\begin{array}{l}\text{For lines}\ \vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}}\ \text{and}\ \vec{r} = \vec{a_{2}}+\lambda \vec{b_{2}}\ \text{to be skew}\ (\vec{b_{1}}\times \vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})\neq 0\end{array} \)
(iv) Shortest distance between lines
\(\begin{array}{l}\vec{r} = \vec{a}+\lambda \vec{b}\end{array} \)
and
\(\begin{array}{l}\vec{r} = \vec{a_{2}}+\mu \vec{b}\end{array} \)
is \(\begin{array}{l}d= \left | \frac{\vec{b}\times (\vec{a_{2}}-\vec{a}_{1})}{\left | \vec{b} \right |} \right |\end{array} \)
22. Family of planes:
(i) Any plane through the intersection of a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is a1x + b1y + c1z + d1 + λ(a2x + b2y + c2z + d2) = 0
(ii) The equation of plane passing through the intersection of the planes
\(\begin{array}{l}\vec{r}.\vec{n_{1}}= d_{1}\end{array} \)
and \(\begin{array}{l}\vec{r}.\vec{n_{2}}= d_{2}\end{array} \)
is \(\begin{array}{l}\vec{r}.(\vec{n_{1}}+\lambda \vec{n_{2}})=d_{1}+ \lambda d_{2}\end{array} \)
where λ is an arbitrary scalar.
Up Next: Important Straight Line Formulas For JEE Maths
Vector Algebra and 3D Geometry – Important Topics
Vector Algebra and 3D Geometry – Important Questions
Vector Algebra and 3D Geometry – Important Topics Part 2
Vector Algebra and 3D Geometry – Important Questions Part 2
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