Important Three Dimensional Geometry Formulas For JEE Maths
This entire article gives the important three-dimensional geometry formulas for jee maths. The jee formula of three dimensional geometry mainly depends on the 3D shapes that occupy the XYZ planes. Here, the important formulas for jee maths three dimensional geometry include vector points, coordinate axes, the centroid of 3D shapes, the equation of plans, planes parallel to the axes....etc.
JEE Formula of Three Dimensional Geometry
Vector Representation of a Point
The position vector of point A(x,y,z) can be represented by \[x\widehat{i} + y \widehat {j} + z \widehat {k}\]
Distance Formula
The distance between the two points A(x1, y1, z1) and B(x2, y2, z2) is
\[AB = \sqrt{(x_{1}-x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1}-z_{2})^{2}} \]
If the distance between two points is in the form of a vector equation.
Distance of P From a Coordinate Axes (A, B, C)
\[PA = \sqrt{(y^{2} + z^{2}}) \]
\[PB = \sqrt{(z^{2} + x^{2}}) \]
\[PC = \sqrt{(x^{2} + y^{2}}) \]
Section Formula
The section formulas are mainly used to find the ratio of the line segment, which is divided by a point. The section formula is also used to find the incenter, centroid and excenters of a triangle.
\[ x = \frac{mx_{2}+nx_{1}}{m+n} \]
\[ y = \frac{my_{2}+ny_{1}}{m+n} \]
\[ z = \frac{ mz_{2}+nz_{1}} { m+n} \]
To find the midpoint of the two points, the below formula is used.
\[ x= \frac{x_{1}+x_{2}}{2} \]
\[ x= \frac{y_{1}+y_{2}}{2} \]
\[ z= \frac{z_{1}+z_{2}}{2} \]
Centroid of a Triangle
Calculating the centroid of a triangle is an important 3d geometry formula for jee. The centroid of a triangle is a point, where the intersection of the three medians of the triangle occurs.
\[ G = (\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}) \]
Incentre of Triangle ABC
The incentre of a triangle is the point, where the intersection of the internal bisector of the angles meets. The formula for calculating the centre of Triangle ABC is given below.
\[ (\frac{ax_{1}+ax_{2} +ax_{3}}{a+b+c}, \frac{ay_{1}+ay_{2}+ay_{3}}{a+b+c}, \frac{az_{1}+az_{2}+az_{3}}{a+b+c})\]
Centroid of a Tetrahedron
The centroid of a tetrahedron can be calculated with the midpoint of the Monge point and circumcenter. Lets the coordinates of the centroid of the tetrahedron whose vertices are (x1, y1, z1 ), (x2, y2, z2), (x3, y3, z3) and (x4, y4, z4 ). Then the formula for calculating the centroid of a tetrahedron is given below
\[ (\frac{x_{1}+x_{2}+x_{3} +x_{4}}{4}, \frac{y_{1}+y_{2}+y_{3} + y_{4}}{4}, \frac{z_{1}+z_{2}+z_{3}+z_{4}}{4}) \]
Direction Cosines and Direction Ratios
If l, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then the formula for calculating direction cosines and direction ratios are given below.
\[ l = \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}} } \]
\[ m = \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}} } \]
\[ n = \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}} } \]
or \[ l = \frac{-a}{\sqrt{a^{2}+b^{2}+c^{2}} } \]
\[ m = \frac{-b}{\sqrt{a^{2}+b^{2}+c^{2}} } \]
\[ n = \frac{-c}{\sqrt{a^{2}+b^{2}+c^{2}} } \]
Angle Between Two Line Segment
If a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines and joints together and make an acute angle θ between them, then the formula for calculating Cos θ is given below.
\[cos\theta = \begin{vmatrix} \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} }{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \end{vmatrix} \]
The line is perpendicular to each other if a1a2 + b1b2 + c1c2 = 0, and the lines are parallel if a1/a2 = b1/b2 = c1/c2.
Projection of a Line Segment on a Line
If P(x1, y1, z1) and Q(x2, y2, z2) are the two points present in the XYZ plane. Then the projection of point PQ on a line having direction cosines l,m, n. Then the projection of a line segment on a line is |l(x2-x1)+m(y2-y1)+n(z2-z1)|
Equation of a Plane
The general equation of a plane is ax+by+cz+d = 0, where a, b, c are not all zero, a, b, c, d ∈ Real numbers.
In normal form, the equation of a plane P = lx+my+nz
If the plane passes through a point (x1, y1, z1) then the equation of the plane through a point is
a (x-x1) + b (y-y1) + c (z-z1) = 0
The equation of the plane in intercept form is given below
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
The equation of the plane in vector form is given below.
\[ (\overrightarrow{r }- \overrightarrow{a}). \overrightarrow{n} = 0\] or \[\overrightarrow{r} . \overrightarrow{n} = \overrightarrow {a} . \overrightarrow{n} \]
If a plane parallel to X-axis is then the equation is by + cz + d = 0
If a plane parallel to Y-axis is then the equation is ax + cz + d = 0
If a plane parallel to z-axis is then the equation is by + ax + d = 0
The equation of a plane passing through the origin (0, 0, 0) is ax + by + cz = 0.
The points A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) are the points on the plain. Then the coplanarity of four points is given below.
\[\begin{vmatrix}x_{2}-x_{1} & y_{2}-y_{1}& z_{2}-z_{1} \\x_{3}-x_{1} & y_{3}-y_{1}&z_{3}-z_{1} \\x_{4}-x_{1}&y_{4}-y_{1}&z_4-z_{1}\end{vmatrix} = 0\]
A plane ax+by+cz+d = 0 divides the line segment joining the points (x1, y1, z1) and (x2, y2, z2) in the ratio. Then the equation of the plane is
\[ \begin{pmatrix} - \frac{ax_{1}+by_{1}+cz_{1}+d}{ax_{2}+by_{2}+cz_{2}+d} \end{pmatrix} \]
A Point and A Plane
The distance between the point (x’, y’, z’) from the plane ax+by+cz+d = 0 can be calculated by the formula
\[ \frac{ax’+by’+cz’+d}{\sqrt{a^{2}+b^{2}+c^{2}}} \]
If the foot (x’, y’, z’) is perpendicularly drawn from the point (x1, y1, z1) to the plane ax+by+cz+d = 0 then the distance between them can calculate through the below formula
\[ \frac{x’-x_{1}}{a} = \frac{y’-y_{1}}{b} = \frac{z’-z_{1}}{c} = - \frac{ax_{1}+by_{1}+cz_{1}+d}{a^{2}+b^{2}+c^{2}} \]
If the planes are perpendicular to each other \[\overrightarrow{n_{1}} . \overrightarrow{n_{2}} = 0 \]
If the planes are parallel then \[ \overrightarrow{n_{1}} = \lambda \overrightarrow{n_{2}} \]
Angle Between Two Planes
If a plane bisecting the angle between two given planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0. Then the equation of the plane is
\[ \frac{a_{1}x+b_{1}y+c_{1}z+d_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}} = \pm \frac{a_{2}x+b_{2}y+c_{2}z+d_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} \]
The plane bisecting the angle may be either acute or obtuse angle: First, make both the constant terms positive. If the a1a2+b1b2+c1c2 > 0 then the origin lies on an obtuse angle. If the a1a2+b1b2+c1c2 < 0 then the origin lies on an acute angle.
Area of Triangle
The vertices of a triangle \[\triangle {ABC}\] are A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3)
Then the area of a triangle \[\triangle = \sqrt{\triangle ^{2}_{x} + \triangle ^{2}_{y} + \triangle ^{2}_{z}} \]
Further,
\[\triangle _{x} = \frac{1}{2} \begin{vmatrix}y_1 &z_1 &1 \\ y_2&z_2 &1 \\ y_3&z_3&1\end{vmatrix},\] \[\triangle _{y} = \frac{1}{2} \begin{vmatrix}z_1 &x_1 &1 \\ z_2&x_2 &1 \\ z_3&x_3&1\end{vmatrix} \] and \[\triangle _{z} = \frac{1}{2} \begin{vmatrix}x_1 &y_1 &1 \\ x_2&y_2 &1 \\ x_3&y_3&1\end{vmatrix} \]
As per the vector method, the area of a triangle can be calculated by \[ \frac{1}{2} \begin{vmatrix} \vec{AB} \times \vec{AC} \end{vmatrix} \]
Volume of a Tetrahedron
If A (x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) and D (x4, y4, z4) are the vertex of the tetrahedron.
Then the volume of a tetrahedron is V = \[\frac{1}{6} \begin{bmatrix}x_{1} & y_{1}&z_{1}&1 \\x_{2} & y_{2}&z_{2}&1 \\x_{3}&y_{3}&z_{3}&1 \\x_{4}&y_{4}&z_{4}&1 \end{bmatrix} \]
Equation of A Line
A straight line is the intersection of two points. If the equations of the planes are represented by a1x+b1y+c1z+d1 = 0 and a2x+b2y+c2z+d2 = 0 .
the equation of a line in symmetric form is \[\frac{(x-x_{1})}{a} = \frac{(y-y_{1})}{b} = \frac{(z-z_{1})}{c} = r \]
The equation of the line in vector form is \[\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b} \]
FAQs on Important Three Dimensional Geometry Formulas
1. What is 3D coordinate geometry?
A 3-dimensional geometry brings the 3D shapes under mathematics and involves them in three coordinates in the XYZ plane. The shapes, which occupy the x-coordinate, y-coordinate and z-coordinate are called 3D shapes. These three dimensions are chosen from the terms width, length, depth, height, and breadth.
2. Do humans see in 3D or 4d?
Human eyes can see 3D creatures. Humans are living in the 3D world, but they can view only two dimensions at a time. The depth that humans can think can see is merely a trick that the human brains have learned; a byproduct of evolution putting our eyes on the front of our faces.
