 # Important Three Dimensional Geometry Formulas for JEE Maths

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 $x\hat{i}+y\hat{j}+z\hat{k}$

2.Distance formula:

Distance between two points P(x1, y1, z1) and Q(x2, y2, z2) is PQ =

$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$

AB = $\left | \vec{OB} -\vec{OA}\right |$

3. Distance of P from coordinate axes: PA = $\sqrt{(y^{2}+z^{2})}$

PB = $\sqrt{(z^{2}+x^{2})}$

PC = $\sqrt{(x^{2}+y^{2})}$

4.Section Formula:

x = $\frac{mx_{2}+nx_{1}}{m+n}$

y = $\frac{my_{2}+ny_{1}}{m+n}$

z = $\frac{mz_{2}+nz_{1}}{m+n}$

Midpoint:

x = $\frac{x_{1}+x_{2}}{2}$

y = $\frac{y_{1}+y_{2}}{2}$

z = $\frac{z_{1}+z_{2}}{2}$

5. Centroid of a triangle:

G = $\left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3},\frac{z_{1}+z_{2}+z_{3}}{3} \right )$

6. Incentre of triangle ABC: $\left ( \frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\frac{ay_{1}+by_{2}+cy_{3}}{a+b+c},\frac{az_{1}+bz_{2}+cz_{3}}{a+b+c} \right )$

7. Centroid of a tetrahedron: $\left ( \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4},\frac{z_{1}+z_{2}+z_{3}+z_{4}}{4} \right )$

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 $a\hat{i}+b\hat{j}+c\hat{k}$ will be a vector parallel to the line L.

(iv) If l, m, n are the direction cosines of any line L then $l\hat{i}+m\hat{j}+n\hat{k}$ 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

l = $\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}$

m = $\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}$

n = $\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$

or l = $\frac{-a}{\sqrt{a^{2}+b^{2}+c^{2}}}$,

m = $\frac{-b}{\sqrt{a^{2}+b^{2}+c^{2}}}$

n = $\frac{-c}{\sqrt{a^{2}+b^{2}+c^{2}}}$

(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 l = $\frac{x_{2}-x_{1}}{\left | PQ \right |}$, m = $\frac{y_{2}-y_{1}}{\left | PQ \right |}$ and n = $\frac{z_{2}-z_{1}}{\left | PQ \right |}$

(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

cos θ = $\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right |$

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: $\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1$

(iv) vector form: $(\vec{r}-\vec{a} ).\vec{n}= 0$ or $\vec{r}.\: \vec{n}= \vec{a}.\: \vec{n}$

(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 $\frac{ax}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}+\frac{by}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}+\frac{cz}{\pm \sqrt{a^{2}+b^{2}+c^{2}}} = \frac{d}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}$

(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 $\frac{\left | d_{1}-d_{2} \right |}{\sqrt{a^{2}+b^{2}+c^{2}}}$

(ix) A plane ax+by+cz+d = 0 divides the line segment joining (x1, y1, z1) and (x2, y2, z2) in the ratio $\left ( -\frac{ax_{1}+by_{1}+cz_{1}+d}{ax_{2}+by_{2}+cz_{2}+d}\right )$

(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 $\begin{vmatrix} x_{2}-x_{1}& y_{2}-y_{1} & z_{2}-z_{1}\\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \\ x_{4}-x_{1}& y_{4}-y_{1} & z_{4}-z_{1} \end{vmatrix}= 0$

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 $\frac{ax^{‘}+by^{‘}+cz^{‘}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}$

(ii) Length of the perpendicular from a point $\vec{a}$ to the plane $\vec{r}.\vec{n} = d$ is given by p = $\frac{\left |\vec{a} .\: \vec{n} -d\right |}{\left | \vec{n} \right |}$.

(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 $\frac{x^{‘}-x_{1}}{a} = \frac{y^{‘}-y_{1}}{b} =\frac{z^{‘}-z_{1}}{c}= -\frac{ax_{1}+by_{1}+cz_{1}+d}{a^{2}+b^{2}+c^{2}}$

(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 $\frac{x^{‘}-x_{1}}{a} = \frac{y^{‘}-y_{1}}{b} =\frac{z^{‘}-z_{1}}{c}= -2\frac{(ax_{1}+by_{1}+cz_{1}+d)}{a^{2}+b^{2}+c^{2}}$.

(v) The distance between two parallel planes ax+by+cz+d = 0 and ax+by+cz+d’ = 0 is given by $\frac{\left | d-d{}’ \right |}{\sqrt{a^{2}+b^{2}+c^{2}}}$

13. Angle between two planes:

(i) $\cos \theta =\left | \frac{aa{}’+bb{}’+cc{}’}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{a{}’^{2}+b{}’^{2}+c{}’^{2}}}\right |$

(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 $\vec{r}.\vec{n_{1}} = d_{1}$ and $\vec{r}.\vec{n_{2}} = d_{2}$ is given by cos θ = $\frac{\vec{n_{1}}.\vec{n_{2}} }{\left | \vec{n_{1}} \right |.\left | \vec{n_{2}} \right |}$

(iv) Planes are perpendicular if $\vec{n_{1}}.\vec{n_{2}}$ = 0 and planes are parallel if $\vec{n_{1}}=\lambda \vec{n_{2}}$, 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 $\frac{a_{1}x+b_{1}y+c_{1}z+d_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}$ = $\pm \frac{a_{2}x+b_{2}y+c_{2}z+d_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$

(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 Δ = $\sqrt{\Delta_{x}^{2}+\Delta _{y}^{2}+\Delta _{z}^{2}}$ where $\Delta _{x} = \frac{1}{2}\begin{vmatrix} y_{1} & z_{1} & 1\\ y_{2} &z_{2} & 1\\ y_{3}& z_{3} & 1 \end{vmatrix}$, $\Delta _{y} = \frac{1}{2}\begin{vmatrix} z_{1} & x_{1} & 1\\ z_{2} &x_{2} & 1\\ z_{3}& x_{3} & 1 \end{vmatrix}$ and $\Delta _{z} = \frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2} &y_{2} & 1\\ x_{3}& y_{3} & 1 \end{vmatrix}$

Vector method: from two vector $\vec{AB}$ and $\vec{AC}$ the area is given by $\frac{1}{2}\left |\vec{AB} \times \vec{AC} \right |$

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 V = $\frac{1}{6}\begin{vmatrix} x_{1} & y_{1} & z_{1} & 1\\ x_{2} & y_{2} & z_{2} & 1\\ x_{3}& y_{3} & z_{3} & 1\\ x_{4}& y_{4} & z_{4} & 1 \end{vmatrix}$

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: $\frac{(x-x_{1})}{a} = \frac{(y-y_{1})}{b}= \frac{(z-z_{1})}{c} = r$

(iii) vector equation:$\vec{r} = \vec{a}+\lambda \vec{b}$

(iv) Reduction of cartesian form of equation of a line to vector form and vice versa $\frac{x-x_{1}}{a} = \frac{y-y_{1}}{b}= \frac{z-z_{1}}{c}$$\vec{r} = (x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k})+\lambda (a\hat{i}+b\hat{j}+c\hat{k})$

18. To find image of a point with respect to a line:

Let L = $\frac{x-x_{2}}{a} = \frac{y-y_{2}}{b}= \frac{z-z_{2}}{c}$ is a given line.

Let (x’, y’, z’) is 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) $\frac{\frac{x_{1}+x{}’}{2}-x_{2}}{a}= \frac{\frac{y_{1}+y{}’}{2}-y_{2}}{b}= \frac{\frac{z_{1}+z{}’}{2}-z_{2}}{c} = \lambda$

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) 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 $\frac{x-x_{1}}{l}= \frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$ and the plane ax+by+cz+d = 0, then $\sin \theta = \left | \frac{al+bm+cn}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{l^{2}+m^{2}+n^{2}}} \right |$

(ii) Vector form : If θ is the angle between a line $\vec{r}= \vec{a}+\lambda \vec{b}$ and $\vec{r}.\vec{n }= d$ then sin θ = $\frac{\vec{b}.\vec{n}}{\left | \vec{b} \right |\left | \vec{n} \right |}$

(iii) condition for perpendicularity l/a = m/b = n/c , $\vec{b}\times \vec{n}= 0$

(iv) condition for parallel : al+bm+cn = o, $\vec{b}.\vec{n}= 0$

20. Condition for a line to lie in a plane:

(i) cartesian form: Line $\frac{x-x_{1}}{l}= \frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$ would lie in a plane ax+by+cz+d = 0, if ax1+by1+cz1+d = 0 and al+bm+cn = 0

(ii) Vector form: Line $\vec{r}= \vec{a}+\lambda b$ would lie in the plane $\vec{r}.\vec{n}= d$ if $\vec{b}.\vec{n}= 0$ and $\vec{a}.\vec{n}= d$

21. Skew lines:

(i) The straight lines which are not parallel and non-coplanar are called skew lines.

If $\Delta = \begin{vmatrix} \alpha ^{‘} -\alpha & \beta {}’-\beta &\gamma {}’-\gamma \\ l & m &n \\ l’& m’&n’ \end{vmatrix} \neq 0$ , then the lines are skew.

(ii) Shortest distance:

SD = $\frac{\begin{vmatrix} \alpha ^{‘} -\alpha & \beta {}’-\beta &\gamma {}’-\gamma \\ l & m &n \\ l’& m’&n’ \end{vmatrix} }{\sqrt{\sum (mn’-m’n)^{2}}}$

(iii) Vector form: For lines $\vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}}$ and $\vec{r} = \vec{a_{2}}+\lambda \vec{b_{2}}$ to be skew $(\vec{b_{1}}\times \vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})\neq 0$

(iv) Shortest distance between lines $\vec{r} = \vec{a}+\lambda \vec{b}$

and $\vec{r} = \vec{a_{2}}+\mu \vec{b}$ is $d= \left | \frac{\vec{b}\times (\vec{a_{2}}-\vec{a}_{1})}{\left | \vec{b} \right |} \right |$

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 $\vec{r}.\vec{n_{1}}= d_{1}$ and $\vec{r}.\vec{n_{2}}= d_{2}$ is $\vec{r}.(\vec{n_{1}}+\lambda \vec{n_{2}})=d_{1}+ \lambda d_{2}$ where λ is an arbitrary scalar.

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